problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
## Task Condition
Find the derivative.
$y=\ln \frac{x^{2}}{\sqrt{1-a x^{4}}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{x^{2}}{\sqrt{1-a x^{4}}}\right)^{\prime}=\frac{\sqrt{1-a x^{4}}}{x^{2}} \cdot\left(\frac{x^{2}}{\sqrt{1-a x^{4}}}\right)^{\prime} \\
& =\frac{\sqrt{1-a x^{4}}}{x^{2}} \cdot \frac{2 x \cdot \sqrt{1-a x^{4}}-x^{2} \cdot \frac{1}{2 \sqrt{1-a x^{4}}} \cdot\left(-... | \frac{2}{x(1-^{4})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,628 |
## Condition of the problem
Find the derivative.
$$
y=\operatorname{ctg} \sqrt[3]{5}-\frac{1}{8} \cdot \frac{\cos ^{2} 4 x}{\sin 8 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{ctg} \sqrt[5]{5}-\frac{1}{8} \cdot \frac{\cos ^{2} 4 x}{\sin 8 x}\right)^{\prime}=-\frac{1}{8} \cdot\left(\frac{\cos ^{2} 4 x}{\sin 8 x}\right)^{\prime}=-\frac{1}{8} \cdot\left(\frac{\cos ^{2} 4 x}{2 \sin 4 x \cdot \cos 4 x}\right)^{\prime}= \\
& =-\frac{... | \frac{1}{4\sin^{2}4x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,629 |
## Task Condition
Find the derivative.
$y=\operatorname{arctg} \frac{\sqrt{1+x^{2}}-1}{x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{\sqrt{1+x^{2}}-1}{x}\right)^{\prime}=\frac{1}{1+\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)^{2}} \cdot\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)^{\prime}= \\
& =\frac{x^{2}}{x^{2}+\left(\sqrt{1+x^{2}}-1\right)^{2}} \cdot \frac{\frac{1}{2 \sqrt{1+x^{... | \frac{-1+\sqrt{1+x^{2}}}{(x^{2}+(\sqrt{1+x^{2}}-1)^{2})\sqrt{1+x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,630 |
## Problem Statement
Find the derivative.
$$
y=\frac{3}{8 \sqrt{2}} \ln \frac{\sqrt{2}+\tanh x}{\sqrt{2}-\tanh x}-\frac{\tanh x}{4\left(2-\tanh^{2} x\right)}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{3}{8 \sqrt{2}} \ln \frac{\sqrt{2}+\operatorname{th} x}{\sqrt{2}-\operatorname{th} x}-\frac{\operatorname{th} x}{4\left(2-\operatorname{th}^{2} x\right)}\right)^{\prime}= \\
& =\frac{3}{8 \sqrt{2}} \cdot \frac{\sqrt{2}-\operatorname{th} x}{\sqrt{2}+\operatorname{t... | \frac{1}{(1+\operatorname{ch}^{2}x)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,631 |
Problem condition
Find the derivative.
$y=\sqrt{9 x^{2}-12 x+5} \cdot \operatorname{arctg}(3 x-2)-\ln \left(3 x-2+\sqrt{9 x^{2}-12 x+5}\right)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{9 x^{2}-12 x+5} \cdot \operatorname{arctg}(3 x-2)-\ln \left(3 x-2+\sqrt{9 x^{2}-12 x+5}\right)\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{9 x^{2}-12 x+5}} \cdot(18 x-12) \cdot \operatorname{arctg}(3 x-2)+\sqrt{9 x^{2}-12 x+5} \cdot \frac{1}{1+(3 x-2)^{2}} \cdot 3-
\... | \frac{(9x-6)\cdot\operatorname{arctg}(3x-2)}{\sqrt{9x^{2}-12x+5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,633 |
## Problem Statement
Find the derivative.
$$
y=x^{3} \arcsin x+\frac{x^{2}+2}{3} \sqrt{1-x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x^{3} \arcsin x+\frac{x^{2}+2}{3} \sqrt{1-x^{2}}\right)^{\prime}= \\
& =3 x^{2} \arcsin x + x^{3} \cdot \frac{1}{\sqrt{1-x^{2}}} + \frac{2 x}{3} \cdot \sqrt{1-x^{2}} + \frac{x^{2}+2}{3} \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot (-2 x)= \\
& =3 x^{2} \arcsin x + \frac{3 x^... | 3x^{2}\arcsinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,634 |
Condition of the problem
Find the derivative.
$$
y=\operatorname{arctg}\left(\frac{\cos x}{\sqrt[4]{\cos 2 x}}\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg}\left(\frac{\cos x}{\sqrt[4]{\cos 2 x}}\right)\right)^{\prime}=\frac{1}{1+\left(\frac{\cos x}{\sqrt[4]{\cos 2 x}}\right)^{2}} \cdot\left(\frac{\cos x}{\sqrt[4]{\cos 2 x}}\right)^{\prime}= \\
& =\frac{\sqrt[2]{\cos 2 x}}{\sqrt[2]{\cos 2 x}+\cos ^{2} ... | -\frac{\sin^{3}x}{(\sqrt[2]{\cos2x}+\cos^{2}x)\cdot\sqrt[4]{(\cos2x)^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,635 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\arcsin (\sin t) \\
y=\arccos (\cos t)
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\arcsin (\sin t))^{\prime}=t^{\prime}=1$
$y_{t}^{\prime}=(\arccos (\cos t))^{\prime}=t^{\prime}=1$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1}{1}=1$
## Kuznetsov Differentiation 16-4 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,636 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=2 t-t^{2} \\
y=3 t-t^{3}
\end{array}
\right.
\]
$t_{0}=1$ | ## Solution
Since $t_{0}=1$, then
$x_{0}=2 \cdot 1-1^{2}=1$
$y_{0}=3 \cdot 1-1^{3}=2$
Let's find the derivatives:
$x_{t}^{\prime}=\left(2 t-t^{2}\right)^{\prime}=2-2 t$
$y_{t}^{\prime}=\left(3 t-t^{3}\right)^{\prime}=3-3 t^{2}$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{3-3 t^{2}}{2-2 t}=\frac{3... | 3x-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,637 |
## Problem Statement
Find the $n$-th order derivative.
$y=\frac{4 x+7}{2 x+3}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{4 x+7}{2 x+3}\right)^{\prime}=\frac{4 \cdot(2 x+3)-(4 x+7) \cdot 2}{(2 x+3)^{2}}=\frac{8 x+12-8 x-14}{(2 x+3)^{2}}=-\frac{2}{(2 x+3)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{2}{(2 x+3)^{2}}\right)^{\prime}=-\frac{2 \cdot(-2)}{(2 x... | y^{(n)}=\frac{(-1)^{n}\cdot2^{n}\cdotn!}{(2x+3)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,638 |
## Task Condition
Find the derivative of the specified order.
$$
y=\frac{\ln (x-1)}{\sqrt{x-1}}, y^{\prime \prime \prime}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\ln (x-1)}{\sqrt{x-1}}\right)^{\prime}=\frac{\frac{1}{x-1} \cdot \sqrt{x-1}-\ln (x-1) \cdot \frac{1}{2 \sqrt{x-1}}}{x-1}=\frac{2-\ln (x-1)}{2 \sqrt{(x-1)^{3}}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{2-\ln (x-1)}{2 \sqrt{(x-1)^{3}}}\ri... | \frac{46-15\ln(x-1)}{8\sqrt{(x-1)^{7}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,639 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\operatorname{sh}^{2} t \\
y=\frac{1}{\operatorname{ch}^{2} t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\operatorname{sh}^{2} t\right)^{\prime}=2 \operatorname{sh} t \cdot \operatorname{ch} t$
$y_{t}^{\prime}=\left(\frac{1}{\operatorname{ch}^{2} t}\right)^{\prime}=\frac{-2}{\operatorname{ch}^{3} t} \cdot \operatorname{sh} t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^... | \frac{2}{\operatorname{ch}^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,640 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
\[
\begin{aligned}
& y=2+c \sqrt{1-x^{2}} \\
& \left(1-x^{2}\right) y^{\prime}+x y=2 x
\end{aligned}
\] | ## Solution
$y^{\prime}=\left(2+c \sqrt{1-x^{2}}\right)^{\prime}=\frac{c}{2 \sqrt{1-x^{2}}} \cdot(-2 x)=-\frac{c \cdot x}{\sqrt{1-x^{-2}}}$
Substitute into equation (1)
$\left(1-x^{2}\right) \cdot\left(-\frac{c \cdot x}{\sqrt{1-x^{2}}}\right)+x \cdot\left(2+c \sqrt{1-x^{2}}\right)=2 x$
Simplify:
$-c \cdot x \cdot \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 46,641 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\frac{x^{2}}{4}-\frac{\ln x}{2}, 1 \leq x \leq 2
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\frac{x^{2}}{4}-\frac{\ln x}{2}\right)^{\prime}=\frac{x}... | \frac{3}{4}+\frac{1}{2}\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,643 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\sqrt{1-x^{2}}+\arcsin x, 0 \leq x \leq \frac{7}{9}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\sqrt{1-x^{2}}+\arcsin x\right)^{\prime}=-\frac{2 x}{... | \frac{2\sqrt{2}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,644 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\ln \frac{5}{2 x}, \sqrt{3} \leq x \leq \sqrt{8}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\ln \frac{5}{2 x}\right)^{\prime}=\left(\ln \frac{5}{2}-... | 1+\frac{1}{2}\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,645 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=-\ln \cos x, 0 \leq x \leq \frac{\pi}{6}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(-\ln \cos x)^{\prime}=-\frac{1}{\cos x} \cdot(\cos x)^{\prime}=... | \ln\sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,646 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=e^{x}+6, \ln \sqrt{8} \leq x \leq \ln \sqrt{15}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(e^{x}+6\right)^{\prime}=e^{x}
$$
Then, using the formula a... | 1+\frac{1}{2}\ln\frac{6}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,647 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in the Cartesian coordinate system.
$$
y=2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}, \frac{1}{4} \leq x \leq 1
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(2+\arcsin \sqrt{x}+\sqrt{x-x^{2}}\rig... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,648 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\ln \left(x^{2}-1\right), 2 \leq x \leq 3
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\ln \left(x^{2}-1\right)\right)^{\prime}=\frac{1}{x^{2}-1}\... | 1+\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,649 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\sqrt{1-x^{2}}+\arccos x, 0 \leq x \leq \frac{8}{9}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\sqrt{1-x^{2}}+\arccos x\right)^{\prime}=-\frac{2 x}{... | \frac{4\sqrt{2}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,650 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=2+\operatorname{ch} x, 0 \leq x \leq 1
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(2+\operatorname{ch} x)^{\prime}=\operatorname{sh} x
$$
Then, us... | \operatorname{sh}1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,652 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=1-\ln \cos x, 0 \leq x \leq \frac{\pi}{6}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(1-\ln \cos x)^{\prime}=0-\frac{1}{\cos x} \cdot(\cos x)^{\... | \ln\sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,653 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=e^{x}+13, \ln \sqrt{15} \leq x \leq \ln \sqrt{24}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(e^{x}+13\right)^{\prime}=e^{x}
$$
Then, using the formu... | 1+\frac{1}{2}\ln\frac{10}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,654 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=-\arccos \sqrt{x}+\sqrt{x-x^{2}}, 0 \leq x \leq \frac{1}{4}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(-\arccos \sqrt{x}+\sqrt{x-x^{2}}\righ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,655 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=2-e^{x}, \ln \sqrt{3} \leq x \leq \ln \sqrt{8}
$$ | ## Solution
$$
\begin{aligned}
& f^{\prime}(x)=-\mathrm{e}^{\mathrm{x}} \\
& \left(f^{\prime}(x)\right)^{2}=\mathrm{e}^{2 \mathrm{x}}
\end{aligned}
$$
## Then
$$
1=\int_{\ln \sqrt{3}}^{\ln \sqrt{8}} \sqrt{1+\mathrm{e}^{2 \mathrm{x}}} d x
$$
## Using the substitution
$$
\begin{aligned}
& \mathrm{t}=\sqrt{1+\mathrm{... | 1+\frac{1}{2}\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,656 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\arcsin x-\sqrt{1-x^{2}}, 0 \leq x \leq \frac{15}{16}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\arcsin x-\sqrt{1-x^{2}}\right)^{\prime}=(\arcsin x)^{\prim... | \frac{3}{\sqrt{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,657 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=1-\ln \sin x, \frac{\pi}{3} \leq x \leq \frac{\pi}{2}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(1-\ln \sin x)^{\prime}=0-\frac{1}{\sin x} \cdot(\sin x)^{\pri... | \frac{\ln3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,658 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=1-\ln \left(x^{2}-1\right), 3 \leq x \leq 4
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(1-\ln \left(x^{2}-1\right)\right)^{\prime}=0-\frac{1}{x... | 1+\ln\frac{6}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,659 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\sqrt{x-x^{2}}-\arccos \sqrt{x}+5, \frac{1}{9} \leq x \leq 1
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(\sqrt{x-x^{2}}-\arccos \sqrt{x}+5\right)^{\... | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,660 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=-\arccos x+\sqrt{1-x^{2}}+1, \quad 0 \leq x \leq \frac{9}{16}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(-\arccos x+\sqrt{1-x^{2}}+1\right)^{\prime}=-\frac{-1}{\... | \frac{1}{\sqrt{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,661 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\ln \sin x, \frac{\pi}{3} \leq x \leq \frac{\pi}{2}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(\ln \sin x)^{\prime}=\frac{1}{\sin x} \cdot(\sin x)^{\prime}... | \frac{1}{2}\ln3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,662 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\ln 7-\ln x, \sqrt{3} \leq x \leq \sqrt{8}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(\ln 7-\ln x)^{\prime}=0-\frac{1}{x}=-\frac{1}{x}
$$
Then, us... | 1+\frac{1}{2}\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,663 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\operatorname{ch} x+3, \quad 0 \leq x \leq 1
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(\operatorname{ch} x+3)^{\prime}=\operatorname{sh} x
$$
Then, us... | \operatorname{sh}1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,664 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=1+\arcsin x-\sqrt{1-x^{2}}, 0 \leq x \leq \frac{3}{4}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(1+\arcsin x-\sqrt{1-x^{2}}\right)^{\prime}=(\arcsin x)^{\pr... | \sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,665 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\ln \cos x+2, \quad 0 \leq x \leq \frac{\pi}{6}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=(\ln \cos x+2)^{\prime}=\frac{1}{\cos x} \cdot(\cos x)^{\prim... | \ln\sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,666 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=e^{x}+26, \ln \sqrt{8} \leq x \leq \ln \sqrt{24}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(e^{x}+26\right)^{\prime}=e^{x}
$$
Then, using the form... | 2+\frac{1}{2}\ln\frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,667 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\frac{e^{x}+e^{-x}}{2}+3, \quad 0 \leq x \leq 2
$$ | ## Solution
$$
l=\int_{0}^{2} \sqrt{1+\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}} d x=\int_{0}^{2} \sqrt{\frac{4 e^{2 x}+e^{4 x}-2 e^{2 x}+1}{4 e^{2 x}}} d x=\int_{0}^{2} \frac{e^{2 x}+1}{2 e^{x}}=\left.\frac{1}{2}\left(e^{x}-e^{-x}\right)\right|_{0} ^{2}=\frac{1}{2}\left(e^{2}-e^{-2}\right)
$$
Source — "http://pluspi.o... | \frac{1}{2}(e^{2}-e^{-2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,668 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by the equations in the Cartesian coordinate system.
$$
y=\arccos \sqrt{x}-\sqrt{x-x^{2}}+4, \quad 0 \leq x \leq \frac{1}{2}
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
\begin{aligned}
f^{\prime}(x)=\left(\arccos \sqrt{x}-\sqrt{x-x^{2}}+4\right)^{\... | \sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,669 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\frac{e^{2 x}+e^{-2 x}+3}{4}, 0 \leq x \leq 2
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\frac{e^{2 x}+e^{-2 x}+3}{4}\right)^{\prime}=\frac{1}{4}... | \frac{1}{2}(e^{4}-e^{-4}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,670 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=e^{x}+e, \ln \sqrt{3} \leq x \leq \ln \sqrt{15}
$$ | ## Solution
The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(e^{x}+e\right)^{\prime}=e^{x}
$$
Then, using the above... | 2+\frac{1}{2}\ln\frac{9}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,671 |
## Problem Statement
Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.
$$
y=\frac{1-e^{x}-e^{-x}}{2}, 0 \leq x \leq 3
$$ | ## Solution
The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula
$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$
Let's find the derivative of the given function:
$$
f^{\prime}(x)=\left(\frac{1-e^{x}-e^{-x}}{2}\right)^{\prime}=\frac{1}{2} \cdot\... | \frac{1}{2}(e^{3}-e^{-3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,672 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
x+\arcsin \left(x^{2} \sin \frac{6}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,673 |
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\sqrt{x}-3 \sqrt[3]{x}, x_{0}=64$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=(\sqrt{x}-3 \sqrt[3]{x})^{\prime}=\left(\sqrt{x}-3 \cdot x^{\frac{1}{3}}\right)^{\prime}=\frac{1}{2 \sqrt{x}}-3 \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}}=\frac{1}{2 \sqrt{x}}-\frac{1}{\sqrt[3]{x^{2}}}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{1}{2 ... | 64 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,674 |
## Condition of the problem
Find the differential $d y$.
$$
y=\frac{\ln |x|}{1+x^{2}}-\frac{1}{2} \ln \frac{x^{2}}{1+x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\frac{\ln |x|}{1+x^{2}}-\frac{1}{2} \ln \frac{x^{2}}{1+x^{2}}\right)^{\prime} d x= \\
& =\left(\frac{\frac{1}{x} \cdot\left(1+x^{2}\right)-\ln |x| \cdot 2 x}{\left(1+x^{2}\right)^{2}}-\frac{1}{2} \cdot \frac{1+x^{2}}{x^{2}} \cdot\left(\frac{x^{2}}{1+x^{2}... | -\frac{2x\cdot\ln|x|}{(1+x^{2})^{2}}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,675 |
## Task Condition
Approximately calculate using the differential.
$y=x^{21}, x=0.998$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 1$
Then
$\Delta x = -0.002$
Calculate:
$y(1) = 1^{21} = 1$
$y^{\prime} = \l... | 0.958 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,676 |
## Problem Statement
Find the derivative.
$y=\frac{x^{6}+x^{3}-2}{\sqrt{1-x^{3}}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{6}+x^{3}-2}{\sqrt{1-x^{3}}}\right)^{\prime}=\frac{\left(6 x^{5}+3 x^{2}\right) \cdot \sqrt{1-x^{3}}-\left(x^{6}+x^{3}-2\right) \cdot \frac{1}{2 \sqrt{1-x^{3}}} \cdot\left(-3 x^{2}\right)}{1-x^{3}}= \\
& =\frac{2 \cdot\left(6 x^{5}+3 x^{2}\right) \cdot\left(1-x... | \frac{9x^{5}}{2\sqrt{1-x^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,677 |
## Problem Statement
Find the derivative.
$y=\frac{e^{\alpha x}(\alpha \cdot \sin \beta x-\beta \cdot \cos \beta x)}{\alpha^{2}+\beta^{2}}$ | ## Solution
$y^{\prime}=\left(\frac{e^{\alpha x}(\alpha \cdot \sin \beta x-\beta \cdot \cos \beta x)}{\alpha^{2}+\beta^{2}}\right)^{\prime}=\frac{1}{\alpha^{2}+\beta^{2}} \cdot\left(e^{\alpha x}(\alpha \cdot \sin \beta x-\beta \cdot \cos \beta x)\right)^{\prime}=$
$=\frac{1}{\alpha^{2}+\beta^{2}} \cdot\left(\alpha \c... | e^{\alphax}\cdot\sin\betax | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,678 |
## Task Condition
Find the derivative.
$$
y=\ln \sqrt[4]{\frac{1+2 x}{1-2 x}}
$$ | ## Solution
$y^{\prime}=\left(\ln \sqrt[4]{\frac{1+2 x}{1-2 x}}\right)^{\prime}=\sqrt[4]{\frac{1-2 x}{1+2 x}} \cdot\left(\sqrt[4]{\frac{1+2 x}{1-2 x}}\right)^{\prime}=$
$$
\begin{aligned}
& =\sqrt[4]{\frac{1-2 x}{1+2 x}} \cdot \frac{1}{4} \cdot \sqrt[4]{\left(\frac{1-2 x}{1+2 x}\right)^{3}} \cdot\left(\frac{1+2 x}{1-... | \frac{1}{1-4x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,679 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{3} \cdot \cos \left(\tan \frac{1}{2}\right)+\frac{1}{10} \cdot \frac{\sin ^{2} 10 x}{\cos 20 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{3} \cdot \cos \left(\operatorname{tg} \frac{1}{2}\right)+\frac{1}{10} \cdot \frac{\sin ^{2} 10 x}{\cos 20 x}\right)^{\prime}=0+\frac{1}{10} \cdot\left(\frac{\sin ^{2} 10 x}{\cos 20 x}\right)^{\prime}= \\
& =\frac{1}{10} \cdot \frac{2 \sin 10 x \cdot \cos 10 x ... | \frac{\tan20x}{\cos20x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,680 |
## Task Condition
Find the derivative.
$y=\frac{1}{2 \sqrt{x}}+\frac{1+x}{2 x} \cdot \operatorname{arctg} \sqrt{x}$ | ## Solution
$y^{\prime}=\left(\frac{1}{2 \sqrt{x}}+\frac{1+x}{2 x} \cdot \operatorname{arctg} \sqrt{x}\right)^{\prime}=$
$=-\frac{1}{2} \cdot \frac{1}{2 \sqrt{x^{3}}}+\frac{1 \cdot x-(1+x) \cdot 1}{2 x^{2}} \cdot \operatorname{arctg} \sqrt{x}+\frac{1+x}{2 x} \cdot \frac{1}{1+(\sqrt{x})^{2}} \cdot \frac{1}{2 \sqrt{x}}=... | -\frac{1}{2x^{2}}\cdot\operatorname{arctg}\sqrt{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,681 |
## Problem Statement
Find the derivative.
$$
y=\sqrt[4]{\frac{1+\operatorname{th} x}{1-\operatorname{th} x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt[4]{\frac{1+\operatorname{th} x}{1-\operatorname{th} x}}\right)^{\prime}= \\
& =\frac{1}{4} \cdot \sqrt[4]{\left(\frac{1-\operatorname{th} x}{1+\operatorname{th} x}\right)^{3}} \cdot \frac{\frac{1}{\operatorname{ch}^{2} x} \cdot(1-\operatorname{th} x)-(1+\operator... | \frac{1}{2\sqrt{\operatorname{ch}x-\operatorname{sh}x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,682 |
## Task Condition
Find the derivative.
$y=(x \sin x)^{8 \ln (x \sin x)}$ | ## Solution
$y=(x \sin x)^{8 \ln (x \sin x)}$
$\ln y=\ln \left((x \sin x)^{8 \ln (x \sin x)}\right)=8 \ln (x \sin x) \cdot \ln (x \sin x)=8 \ln ^{2}(x \sin x)$
$\frac{y^{\prime}}{y}=\left(8 \ln ^{2}(x \sin x)\right)^{\prime}=16 \ln (x \sin x) \cdot \frac{1}{x \sin x} \cdot(\sin x+x \cos x)=$
$=\frac{16 \ln (x \sin ... | \frac{16(x\sinx)^{8\ln(x\sinx)}\cdot\ln(x\sinx)\cdot(1+x\cdot\operatorname{ctg}x)}{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,683 |
## Task Condition
Find the derivative.
$$
y=(2 x+3)^{4} \cdot \arcsin \frac{1}{2 x+3}+\frac{2}{3} \cdot\left(4 x^{2}+12 x+11\right) \cdot \sqrt{x^{2}+3 x+2}, 2 x+3>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left((2 x+3)^{4} \cdot \arcsin \frac{1}{2 x+3}+\frac{2}{3} \cdot\left(4 x^{2}+12 x+11\right) \cdot \sqrt{x^{2}+3 x+2}\right)^{\prime}= \\
& =4(2 x+3)^{3} \cdot 2 \cdot \arcsin \frac{1}{2 x+3}+(2 x+3)^{4} \cdot \frac{1}{\sqrt{1-\left(\frac{1}{2 x+3}\right)^{2}}} \cdot \frac{... | 8(2x+3)^{3}\cdot\arcsin\frac{1}{2x+3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,684 |
## Problem Statement
Find the derivative.
$$
y=\sqrt{(4+x)(1+x)}+3 \ln (\sqrt{4+x}+\sqrt{1+x})
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=(\sqrt{(4+x)(1+x)}+3 \ln (\sqrt{4+x}+\sqrt{1+x}))^{\prime}= \\
& =\left(\sqrt{4+5 x+x^{2}}+3 \ln (\sqrt{4+x}+\sqrt{1+x})\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{4+5 x+x^{2}}} \cdot(5+2 x)+3 \cdot \frac{1}{\sqrt{4+x}+\sqrt{1+x}} \cdot\left(\frac{1}{2 \sqrt{4+x}}+\frac{1}{2 \s... | \sqrt{\frac{4+x}{1+x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,685 |
## Task Condition
Find the derivative.
$$
y=\left(1+x^{2}\right) e^{\operatorname{arctg} x}
$$ | ## Solution
$y^{\prime}=\left(\left(1+x^{2}\right) e^{\operatorname{arctg} x}\right)^{\prime}=2 x \cdot e^{\operatorname{arctg} x}+\left(1+x^{2}\right) e^{\operatorname{arctg} x} \cdot \frac{1}{1+x^{2}}=$ $=2 x \cdot e^{\operatorname{arctg} x}+e^{\operatorname{arctg} x}=(2 x+1) \cdot e^{\operatorname{arctg} x}$
## Pr... | (2x+1)\cdote^{\operatorname{arctg}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,686 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$
$$
\left\{\begin{array}{l}
x=\ln \frac{1}{\sqrt{1-t^{4}}} \\
y=\arcsin \frac{1-t^{2}}{1+t^{2}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\ln \frac{1}{\sqrt{1-t^{4}}}\right)^{\prime}=\sqrt{1-t^{4}} \cdot\left(\frac{1}{\sqrt{1-t^{4}}}\right)^{\prime}=$
$=\sqrt{1-t^{4}} \cdot\left(-\frac{1}{2}\right) \cdot \frac{1}{\sqrt{\left(1-t^{4}\right)^{3}}} \cdot\left(-4 t^{3}\right)=\frac{2 t^{3}}{1-t^{4}}$
$y_{t}^{\prime}=\left... | \frac{^{2}-1}{^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,687 |
## Problem Statement
Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$
\[
\left\{
\begin{array}{l}
x=a \cdot t \cdot \cos t \\
y=a \cdot t \cdot \sin t
\end{array}
\right.
\]
$t_{0}=\frac{\pi}{2}$ | ## Solution
Since $t_{0}=\frac{\pi}{2}$, then
$x_{0}=a \cdot \frac{\pi}{2} \cdot \cos \frac{\pi}{2}=0$
$y_{0}=a \cdot \frac{\frac{\pi}{2}}{2} \cdot \sin \frac{\frac{\pi}{2}}{2}=\frac{a \cdot \pi}{2}$
Let's find the derivatives:
$x_{t}^{\prime}=(a \cdot t \cdot \cos t)^{\prime}=a \cdot \cos t-a \cdot t \cdot \sin t... | -\frac{2x}{\pi}+\frac{\cdot\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,688 |
## Task Condition
Find the $n$-th order derivative.
$y=2^{3 x+5}$ | ## Solution
$y=2^{3 x+5}=2^{3 x} \cdot 32=32 \cdot\left(e^{\ln 2}\right)^{3 x}=32 e^{3 x \cdot \ln 2}$
$y^{\prime}=\left(32 e^{3 x \cdot \ln 2}\right)^{\prime}=32 e^{3 x \cdot \ln 2} \cdot 3 \cdot \ln 2$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(32 e^{3 x \cdot \ln 2} \cdot 3 \cdot \ln 2\right)^{\pri... | y^{(n)}=2^{3x+5}\cdot3^{n}\cdot\ln^{n}2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,689 |
## Task Condition
Find the derivative of the specified order.
$y=\frac{\ln x}{x^{3}}, y^{IV}=?$ | ## Solution
$y^{\prime}=\left(\frac{\ln x}{x^{3}}\right)^{\prime}=\frac{\frac{1}{x} \cdot x^{3}-\ln x \cdot 3 x^{2}}{x^{6}}=\frac{1-3 \ln x}{x^{4}}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1-3 \ln x}{x^{4}}\right)^{\prime}=\frac{-\frac{3}{x} \cdot x^{4}-(1-3 \ln x) \cdot 4 x^{3}}{x^{8}}=$
$=\... | \frac{-342+360\lnx}{x^{7}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,690 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sqrt{t} \\
y=\sqrt[3]{t-1}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\sqrt{t})^{\prime}=\frac{1}{2 \sqrt{t}}$
$y_{t}^{\prime}=(\sqrt[3]{t-1})^{\prime}=\left((t-1)^{\frac{1}{3}}\right)^{\prime}=\frac{1}{3} \cdot(t-1)^{-\frac{2}{3}}=\frac{1}{3 \sqrt[3]{(t-1)^{2}}}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\... | -\frac{2(+3)}{9\sqrt[3]{(-1)^{5}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,691 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{\cos x-\cos 3 x}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,693 |
## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{x^{2}-2 x-3}{4}, x_{0}=4$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(\frac{x^{2}-2 x-3}{4}\right)^{\prime}=\frac{2 x-2}{4}=\frac{x-1}{2}
$$
Then:
$$
y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{4-1}{2}=\frac{3}{2}
$$
Since the function $y^{\prime}{ }_{\text {at point }} x_{0}$ has a finite derivative, the equation of the... | \frac{3}{2}\cdotx-\frac{19}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,694 |
## Condition of the problem
Find the differential $d y$.
$y=x \cdot \operatorname{arctg} x-\ln \sqrt{1+x^{2}}$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(x \cdot \operatorname{arctg} x-\ln \sqrt{1+x^{2}}\right)^{\prime} d x= \\
& =\left(\operatorname{arctg} x+x \cdot \frac{1}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}} \cdot \frac{1}{2 \sqrt{1+x^{2}}} \cdot 2 x\right) d x= \\
& =\left(\operatorname{arctg} x+\frac{x}... | \operatorname{arctg}x\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,695 |
## problem statement
Approximately calculate using the differential.
$y=\sqrt{x^{2}+5}, x=1.97$ | ## Solution
If the increment $\Delta x=x-x_{0}$ of the argument $x$ is small in absolute value, then $f(x)=f\left(x_{0}+\Delta x\right) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0}=2$
Then:
$\Delta x=-0.03$
Calculate:
$y(2)=\sqrt{2^{2}+5}=\sqrt{9}=3$
$y^{\prime}=\left(... | 2.98 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,696 |
## Task Condition
Find the derivative.
$y=\frac{3 x+\sqrt{x}}{\sqrt{x^{2}+2}}$ | ## Solution
$y^{\prime}=\left(\frac{3 x+\sqrt{x}}{\sqrt{x^{2}+2}}\right)^{\prime}=\frac{\left(3+\frac{1}{2 \sqrt{x}}\right) \sqrt{x^{2}+2}-(3 x+\sqrt{x}) \cdot \frac{1}{2 \sqrt{x^{2}+2}} \cdot 2 x}{x^{2}+2}=$
$=\frac{\left(3+\frac{1}{2 \sqrt{x}}\right)\left(x^{2}+2\right)-(3 x+\sqrt{x}) \cdot x}{\left(x^{2}+2\right) ... | \frac{12\sqrt{x}+2-x^{2}}{2\sqrt{x}(x^{2}+2)\sqrt{x^{2}+2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,697 |
## Task Condition
Find the derivative.
$$
y=-\frac{1}{2} \cdot e^{-x^{2}}\left(x^{4}+2 x^{2}+2\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(-\frac{1}{2} \cdot e^{-x^{2}}\left(x^{4}+2 x^{2}+2\right)\right)^{\prime}= \\
& =-\frac{1}{2} \cdot\left(e^{-x^{2}} \cdot(-2 x) \cdot\left(x^{4}+2 x^{2}+2\right)+e^{-x^{2}}\left(4 x^{3}+4 x\right)\right)= \\
& =e^{-x^{2}} \cdot x \cdot\left(x^{4}+2 x^{2}+2\right)-e^{-x... | x^{5}\cdote^{-x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,698 |
## Task Condition
Find the derivative.
$y=\ln \ln ^{3} \ln ^{2} x$ | ## Solution
$y^{\prime}=\left(\ln \ln ^{3} \ln ^{2} x\right)^{\prime}=\frac{1}{\ln ^{3} \ln ^{2} x} \cdot 3 \ln ^{2} \ln ^{2} x \cdot \frac{1}{\ln ^{2} x} \cdot 2 \ln x \cdot \frac{1}{x}=$
$$
=\frac{3}{\ln ^{2} \ln ^{2} x} \cdot \frac{2}{\ln x} \cdot \frac{1}{x}=\frac{6}{x \cdot \ln x \cdot \ln \ln ^{2} x}
$$
## Pro... | \frac{6}{x\cdot\lnx\cdot\ln\ln^{2}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,699 |
## Task Condition
Find the derivative.
$$
y=\sin ^{3}(\cos 2)-\frac{\cos ^{2} 30 x}{60 \sin 60 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sin ^{3}(\cos 2)-\frac{\cos ^{2} 30 x}{60 \sin 60 x}\right)^{\prime}=0-\left(\frac{\cos ^{2} 30 x}{60 \sin 60 x}\right)^{\prime}= \\
& =-\left(\frac{\cos ^{2} 30 x}{120 \sin 30 x \cdot \cos 30 x}\right)^{\prime}=-\frac{1}{120} \cdot\left(\frac{\cos 30 x}{\sin 30 x}\ri... | \frac{1}{4\sin^{2}30x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,700 |
## Problem Statement
Find the derivative.
$$
y=\sqrt{1+2 x-x^{2}} \arcsin \frac{x \sqrt{2}}{1+x}-\sqrt{2} \cdot \ln (1+x)
$$ | ## Solution
$y^{\prime}=\left(\sqrt{1+2 x-x^{2}} \arcsin \frac{x \sqrt{2}}{1+x}-\sqrt{2} \cdot \ln (1+x)\right)^{\prime}=$
$=\frac{1}{2 \sqrt{1+2 x-x^{2}}} \cdot(2-2 x) \arcsin \frac{x \sqrt{2}}{1+x}+\sqrt{1+2 x-x^{2}} \cdot \frac{1}{\sqrt{1-\left(\frac{x \sqrt{2}}{1+x}\right)^{2}}} \cdot\left(\frac{x \sqrt{2}}{1+x}\... | \frac{1-x}{\sqrt{1+2x-x^{2}}}\cdot\arcsin\frac{x\sqrt{2}}{1+x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,701 |
## Task Condition
Find the derivative.
$y=-\frac{\operatorname{ch} x}{2 \operatorname{sh}^{2} x}-\frac{1}{2} \ln \left(\operatorname{th} \frac{x}{2}\right)$ | ## Solution
$y^{\prime}=\left(-\frac{\operatorname{ch} x}{2 \operatorname{sh}^{2} x}-\frac{1}{2} \ln \left(\operatorname{th} \frac{x}{2}\right)\right)^{\prime}=-\frac{\operatorname{sh} x \cdot \operatorname{sh}^{2} x-\operatorname{ch} x \cdot 2 \operatorname{sh} x \cdot \operatorname{ch} x}{2 \operatorname{sh}^{4} x}-... | \frac{1}{\operatorname{sh}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,702 |
## Task Condition
Find the derivative.
$y=(\cos 2 x)^{\ln (\cos 2 x) / 4}$ | ## Solution
$y=(\cos 2 x)^{\ln (\cos 2 x) / 4}$
$\ln y=\frac{\ln (\cos 2 x)}{4} \cdot \ln (\cos 2 x)$
$\ln y=\frac{\ln ^{2}(\cos 2 x)}{4}$
$\frac{y^{\prime}}{y}=\left(\frac{\ln ^{2}(\cos 2 x)}{4}\right)^{\prime}=\frac{2 \ln (\cos 2 x)}{4} \cdot \frac{1}{\cos 2 x} \cdot(-\sin 2 x) \cdot 2=-\operatorname{tg} 2 x \cdo... | -(\cos2x)^{\ln(\cos2x)/4}\cdot\operatorname{tg}2x\cdot\ln(\cos2x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,703 |
## Problem Statement
Find the derivative.
$y=\frac{1}{x} \sqrt{1-4 x^{2}}+\ln \frac{1+\sqrt{1+4 x^{2}}}{2 x}$ | ## Solution
$y^{\prime}=\left(\frac{1}{x} \sqrt{1-4 x^{2}}+\ln \frac{1+\sqrt{1+4 x^{2}}}{2 x}\right)^{\prime}=$
$$
\begin{aligned}
& =-\frac{1}{x^{2}} \sqrt{1-4 x^{2}}+\frac{1}{x} \cdot \frac{1}{2 \sqrt{1-4 x^{2}}} \cdot(-8 x)+\frac{2 x}{1+\sqrt{1+4 x^{2}}} \cdot \frac{\frac{1}{2 \sqrt{1+4 x^{2}}} \cdot 8 x \cdot x-1... | -\frac{1}{x^{2}\sqrt{1-4x^{2}}}-\frac{1}{x\sqrt{1+4x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,704 |
## Task Condition
Find the derivative.
$y=x \cdot \arcsin \sqrt{\frac{x}{x+1}}-\sqrt{x}+\operatorname{arctg} \sqrt{x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x \cdot \arcsin \sqrt{\frac{x}{x+1}}-\sqrt{x}+\operatorname{arctg} \sqrt{x}\right)^{\prime}= \\
& =\arcsin \sqrt{\frac{x}{x+1}}+x \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{x}{x+1}}\right)^{2}}} \cdot\left(\sqrt{\frac{x}{x+1}}\right)^{\prime}-\frac{1}{2 \sqrt{x}}+\frac{1... | \arcsin\sqrt{\frac{x}{x+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,705 |
## Problem Statement
Find the derivative.
$y=\frac{1}{2} \ln \frac{1+\cos x}{1-\cos x}-\frac{1}{\cos x}-\frac{1}{3 \cos ^{3} x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \ln \frac{1+\cos x}{1-\cos x}-\frac{1}{\cos x}-\frac{1}{3 \cos ^{3} x}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \frac{1-\cos x}{1+\cos x} \cdot \frac{-\sin x \cdot(1-\cos x)-(1+\cos x) \cdot \sin x}{(1-\cos x)^{2}}+\frac{1}{\cos ^{2} x} \cdot(-\sin x)+\frac... | -\frac{1}{\sinx\cdot\cos^{4}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,706 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$
$$
\left\{\begin{array}{l}
x=\frac{t}{\sqrt{1-t^{2}}} \cdot \arcsin t+\ln \sqrt{1-t^{2}} \\
y=\frac{t}{\sqrt{1-t^{2}}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\frac{t}{\sqrt{1-t^{2}}} \cdot \arcsin t+\ln \sqrt{1-t^{2}}\right)^{\prime}=$
$=\frac{\sqrt{1-t^{2}}-t \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)}{1-t^{2}} \cdot \arcsin t+\frac{t}{\sqrt{1-t^{2}}} \cdot \frac{1}{\sqrt{1-t^{2}}}+\frac{1}{\sqrt{1-t^{2}}} \cdot \frac{1}{2 \sqrt{1-t^{2... | \frac{1}{\arcsin} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,707 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=\sin t \\
y=\cos 2 t
\end{array}
\right.
\]
$t_{0}=\frac{\pi}{6}$ | ## Solution
Since $t_{0}=\frac{\pi}{6}$, then
$x_{0}=\sin \frac{\pi}{6}=0.5$
$y_{0}=\cos \left(2 \cdot \frac{\pi}{6}\right)=\cos \frac{\pi}{3}=0.5$
Let's find the derivatives:
$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$
$y_{t}^{\prime}=(\cos 2 t)^{\prime}=-2 \sin 2 t$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\... | -2x+1.5\frac{x}{2}+0.25 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,708 |
## Problem Statement
Find the $n$-th order derivative.
$y=\frac{7 x+1}{17(4 x+3)}$ | ## Solution
$y=\frac{7 x+1}{17(4 x+3)}$
$y^{\prime}=\left(\frac{7 x+1}{17(4 x+3)}\right)^{\prime}=\frac{7 \cdot(4 x+3)-(7 x+1) \cdot 4}{17(4 x+3)^{2}}=$
$=\frac{28 x+21-28 x-4}{17(4 x+3)^{2}}=\frac{17}{17(4 x+3)^{2}}=\frac{1}{(4 x+3)^{2}}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{(4 x+3)^{... | y^{(n)}=\frac{(-1)^{n-1}\cdotn!\cdot4^{n-1}}{(4x+3)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,709 |
## Task Condition
Find the derivative of the specified order.
$y=\frac{\log _{3} x}{x^{2}}, y^{IV}=?$ | ## Solution
$y^{\prime}=\left(\frac{\log _{3} x}{x^{2}}\right)^{\prime}=\left(\frac{\ln x}{x^{2} \cdot \ln 3}\right)^{\prime}=\frac{\frac{1}{x} \cdot x^{2}-\ln x \cdot 2 x}{x^{4} \cdot \ln 3}=\frac{1-2 \ln x}{x^{3} \cdot \ln 3}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1-2 \ln x}{x^{3} \cdot \l... | \frac{-154+120\lnx}{x^{6}\cdot\ln3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,710 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=\cos t+\sin t \\ y=\sin 2 t\end{array}\right.$ | ## Solution
$x_{t}^{\prime}=(\cos t+\sin t)^{\prime}=-\sin t+\cos t$
$y_{t}^{\prime}=(\sin 2 t)^{\prime}=2 \cos 2 t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 \cos 2 t}{-\sin t+\cos t}=2 \cdot \frac{\cos ^{2} t-\sin ^{2} t}{\cos t-\sin t}=2(\sin t+\cos t) \\
& \lef... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,711 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(4-3 x) e^{-3 x} d x
$$ | ## Solution
$$
\int(4-3 x) e^{-3 x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=4-3 x ; d u=-3 d x \\
& d v=e^{-3 x} d x ; v=-\frac{1}{3} e^{-3 x}
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(4-3 x) \cdot\left(-\frac{1}{3} e^{-3 x}\right... | (x-1)e^{-3x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,713 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \operatorname{arctg} \sqrt{4 x-1} \, d x
$$ | ## Solution
$$
\int \operatorname{arctg} \sqrt{4 x-1} d x=
$$
Let:
$$
\begin{aligned}
& u=\operatorname{arctg} \sqrt{4 x-1} ; d u=\frac{1}{1+(\sqrt{4 x-1})^{2}} \cdot \frac{1}{2 \sqrt{4 x-1}} \cdot 4 d x= \\
& =\frac{1}{4 x} \cdot \frac{2}{\sqrt{4 x-1}} \cdot d x=\frac{d x}{2 x \sqrt{4 x-1}} \\
& d v=d x ; v=x
\end{... | x\cdot\operatorname{arctg}\sqrt{4x-1}-\frac{1}{4}\cdot\sqrt{4x-1}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,714 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(3 x+4) e^{3 x} d x
$$ | ## Solution
$$
\int(3 x+4) e^{3 x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=3 x+4 ; d u=3 d x \\
& d v=e^{3 x} d x ; v=\frac{1}{3} e^{3 x}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =(3 x+4) \cdot \frac{1}{3} e^{3 x}-\int \frac{1}{3} e^{3 x... | (x+1)e^{3x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,715 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(4 x-2) \cos 2 x \, d x
$$ | ## Solution
$$
\int(4 x-2) \cos 2 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=4 x-2 ; d u=4 d x \\
& d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =(4 x-2) \cdot \frac{1}{2} \sin 2 x-\int \frac{1}{2} \s... | (2x-1)\sin2x+\cos2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,716 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(4-16 x) \sin 4 x \, dx
$$ | ## Solution
$$
\int(4-16 x) \sin 4 x d x=
$$
Let:
$$
\begin{aligned}
& u=4-16 x ; d u=-16 d x \\
& d v=\sin 4 x d x ; v=-\frac{1}{4} \cos 4 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(4-16 x) \cdot\left(-\frac{1}{4} \cos 4 x\right)-\int\left... | (4x-1)\cos4x-\sin4x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,717 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(5 x-2) e^{3 x} d x
$$ | ## Solution
$$
\int(5 x-2) e^{3 x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=5 x-2 ; d u=5 d x \\
& d v=e^{3 x} d x ; v=\frac{1}{3} e^{3 x}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =(5 x-2) \cdot \frac{1}{3} e^{3 x}-\int \frac{1}{3} e^{3 x... | \frac{1}{9}\cdot(15x-11)e^{3x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,718 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(1-6 x) e^{2 x} d x
$$ | ## Solution
$$
\int(1-6 x) e^{2 x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=1-6 x ; d u=-6 d x \\
& d v=e^{2 x} d x ; v=\frac{1}{2} e^{2 x}
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =(1-6 x) \cdot \frac{1}{2} e^{2 x}-\int \frac{1}{2} e^{2 ... | (2-3x)e^{2x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,719 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \ln \left(x^{2}+4\right) d x
$$ | ## Solution
$$
\int \ln \left(x^{2}+4\right) d x=
$$
Let:
$$
\begin{aligned}
& u=\ln \left(x^{2}+4\right) ; d u=\frac{1}{x^{2}+4} \cdot 2 x \cdot d x=\frac{2 x d x}{x^{2}+4} \\
& d v=d x ; v=x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =\ln \le... | x\cdot\ln(x^{2}+4)-2x+4\operatorname{arctg}\frac{x}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,720 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \ln \left(4 x^{2}+1\right) d x
$$ | ## Solution
$$
\int \ln \left(4 x^{2}+1\right) d x=
$$
Let:
$$
\begin{aligned}
& u=\ln \left(4 x^{2}+1\right) ; d u=\frac{1}{4 x^{2}+1} \cdot 8 x \cdot d x=\frac{8 x d x}{4 x^{2}+1} \\
& d v=d x ; v=x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& ... | x\cdot\ln(4x^{2}+1)-2x+\operatorname{arctg}2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,721 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(2-4 x) \sin 2 x \, dx
$$ | ## Solution
$$
\int(2-4 x) \sin 2 x d x=
$$
Let:
$$
\begin{aligned}
& u=2-4 x ; d u=-4 d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(2-4 x) \cdot\left(-\frac{1}{2} \cos 2 x\right)-\int\left(-\f... | (2x-1)\cos2x-\sin2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,722 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \operatorname{arctg} \sqrt{6 x-1} \, d x
$$ | ## Solution
$$
\int \operatorname{arctg} \sqrt{6 x-1} d x=
$$
Let:
$$
\begin{aligned}
& u=\operatorname{arctg} \sqrt{6 x-1} ; d u=\frac{1}{1+(\sqrt{6 x-1})^{2}} \cdot \frac{1}{2 \sqrt{6 x-1}} \cdot 6 d x= \\
& =\frac{1}{6 x} \cdot \frac{3}{\sqrt{6 x-1}} \cdot d x=\frac{d x}{2 x \sqrt{6 x-1}} \\
& d v=d x ; v=x
\end{... | x\cdot\operatorname{arctg}\sqrt{6x-1}-\frac{1}{6}\cdot\sqrt{6x-1}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,723 |
## Problem Statement
Calculate the indefinite integral:
$$
\int e^{-2 x}(4 x-3) d x
$$ | ## Solution
$$
\int e^{-2 x}(4 x-3) d x=
$$
Let's denote:
$$
\begin{aligned}
& u=4 x-3 ; d u=4 d x \\
& d v=e^{-2 x} d x ; v=-\frac{1}{2} e^{-2 x}
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(4 x-3) \cdot\left(-\frac{1}{2} e^{-2 x}\right)... | \frac{1}{2}\cdot(1-4x)e^{-2x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,724 |
## Problem Statement
Calculate the indefinite integral:
$$
\int e^{-3 x}(2-9 x) d x
$$ | ## Solution
$$
\int e^{-3 x}(2-9 x) d x=
$$
Let's denote:
$$
\begin{aligned}
& u=2-9 x ; d u=-9 d x \\
& d v=e^{-3 x} d x ; v=-\frac{1}{3} e^{-3 x}
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(2-9 x) \cdot\left(-\frac{1}{3} e^{-3 x}\right... | \frac{1}{3}(1+9x)e^{-3x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,725 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \operatorname{arctg} \sqrt{2 x-1} \, d x
$$ | ## Solution
$$
\int \operatorname{arctg} \sqrt{2 x-1} d x=
$$
Let:
$$
\begin{aligned}
& u=\operatorname{arctg} \sqrt{2 x-1} ; d u=\frac{1}{1+(\sqrt{2 x-1})^{2}} \cdot \frac{1}{2 \sqrt{2 x-1}} \cdot 2 d x= \\
& =\frac{1}{2 x} \cdot \frac{1}{\sqrt{2 x-1}} \cdot d x=\frac{d x}{2 x \sqrt{2 x-1}} \\
& d v=d x ; v=x
\end{... | x\cdot\operatorname{arctg}\sqrt{2x-1}-\frac{1}{2}\cdot\sqrt{2x-1}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,726 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \operatorname{arctg} \sqrt{3 x-1} \, d x
$$ | ## Solution
$$
\int \operatorname{arctg} \sqrt{3 x-1} d x=
$$
Let:
$$
\begin{aligned}
& u=\operatorname{arctg} \sqrt{3 x-1} ; d u=\frac{1}{1+(\sqrt{3 x-1})^{2}} \cdot \frac{1}{2 \sqrt{3 x-1}} \cdot 3 d x= \\
& =\frac{1}{3 x} \cdot \frac{3}{2 \sqrt{3 x-1}} \cdot d x=\frac{d x}{2 x \sqrt{3 x-1}} \\
& d v=d x ; v=x
\en... | x\cdot\operatorname{arctg}\sqrt{3x-1}-\frac{1}{3}\cdot\sqrt{3x-1}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,727 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \operatorname{arctg} \sqrt{5 x-1} \, d x
$$ | ## Solution
$$
\int \operatorname{arctg} \sqrt{5 x-1} d x=
$$
Let:
$$
\begin{aligned}
& u=\operatorname{arctg} \sqrt{5 x-1} ; d u=\frac{1}{1+(\sqrt{5 x-1})^{2}} \cdot \frac{1}{2 \sqrt{5 x-1}} \cdot 5 d x= \\
& =\frac{1}{5 x} \cdot \frac{5}{2 \sqrt{5 x-1}} \cdot d x=\frac{d x}{2 x \sqrt{5 x-1}} \\
& d v=d x ; v=x
\en... | x\cdot\operatorname{arctg}\sqrt{5x-1}-\frac{1}{5}\cdot\sqrt{5x-1}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,728 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(5 x+6) \cos 2 x \, dx
$$ | ## Solution
$$
\int(5 x+6) \cos 2 x d x=
$$
Let:
$$
\begin{aligned}
& u=5 x+6 ; d u=5 d x \\
& d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$, we get:
$$
\begin{aligned}
& =(5 x+6) \cdot \frac{1}{2} \sin 2 x-\int \frac{1}{2} \sin 2 x \c... | \frac{1}{2}\cdot(5x+6)\sin2x+\frac{5}{4}\cdot\cos2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,729 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(3 x-2) \cos 5 x \, d x
$$ | ## Solution
$$
\int(3 x-2) \cos 5 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=3 x-2 ; d u=3 d x \\
& d v=\cos 5 x d x ; v=\frac{1}{5} \sin 5 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(3 x-2) \cdot \frac{1}{5} \sin 5 x-\int \frac{1}... | \frac{1}{5}(3x-2)\sin5x+\frac{3}{25}\cos5x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,730 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(4 x+7) \cos 3 x \, d x
$$ | ## Solution
$$
\int(4 x+7) \cos 3 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=4 x+7 ; d u=4 d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(4 x+7) \cdot \frac{1}{3} \sin 3 x-\int \frac{1}... | \frac{1}{3}\cdot(4x+7)\sin3x+\frac{4}{9}\cdot\cos3x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,731 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(2 x-5) \cos 4 x \, d x
$$ | ## Solution
$$
\int(2 x-5) \cos 4 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=2 x-5 ; d u=2 d x \\
& d v=\cos 4 x d x ; v=\frac{1}{4} \sin 4 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(2 x-5) \cdot \frac{1}{4} \sin 4 x-\int \frac{1}... | \frac{1}{4}(2x-5)\sin4x+\frac{1}{8}\cos4x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,732 |
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