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## Problem Statement
Calculate the indefinite integral:
$$
\int(8-3 x) \cos 5 x \, dx
$$ | ## Solution
$$
\int(8-3 x) \cos 5 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=8-3 x ; d u=-3 d x \\
& d v=\cos 5 x d x ; v=\frac{1}{5} \sin 5 x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(8-3 x) \cdot \frac{1}{5} \sin 5 x-\int \frac{1... | \frac{1}{5}(8-3x)\sin5x-\frac{3}{25}\cos5x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,733 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(x+5) \sin 3 x \, dx
$$ | ## Solution
$$
\int(x+5) \sin 3 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x+5 ; d u=d x \\
& d v=\sin 3 x d x ; v=-\frac{1}{3} \cos 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}
& =(x+5) \cdot\left(-\frac{1}{3} \cos 3 x\right)-\int\le... | -\frac{1}{3}\cdot(x+5)\cos3x+\frac{1}{9}\cdot\sin3x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,734 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(2-3 x) \sin 2 x \, dx
$$ | ## Solution
$$
\int(2-3 x) \sin 2 x d x=
$$
Let:
$$
\begin{aligned}
& u=2-3 x ; d u=-3 d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(2-3 x) \cdot\left(-\frac{1}{2} \cos 2 x\right)-\int\left(-\f... | \frac{1}{2}\cdot(3x-2)\cos2x-\frac{3}{4}\cdot\sin2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,735 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(4 x+3) \sin 5 x \, d x
$$ | ## Solution
$$
\int(4 x+3) \sin 5 x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=4 x+3 ; d u=4 d x \\
& d v=\sin 5 x d x ; v=-\frac{1}{5} \cos 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}
& =(4 x+3) \cdot\left(-\frac{1}{5} \cos 5 x\right)... | -\frac{1}{5}\cdot(4x+3)\cos5x+\frac{4}{25}\cdot\sin5x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,736 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(7 x-10) \sin 4 x \, d x
$$ | ## Solution
$$
\int(7 x-10) \sin 4 x d x=
$$
Let:
$$
\begin{aligned}
u=7 x-10 ; d u & =7 d x \\
d v=\sin 4 x d x ; v & =-\frac{1}{4} \cos 4 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(7 x-10) \cdot\left(-\frac{1}{4} \cos 4 x\right)-\int\left... | \frac{1}{4}\cdot(10-7x)\cos4x+\frac{7}{16}\cdot\sin4x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,737 |
## Problem Statement
Calculate the indefinite integral:
$$
\int(\sqrt{2}-8 x) \sin 3 x \, d x
$$ | ## Solution
$$
\int(\sqrt{2}-8 x) \sin 3 x d x=
$$
Let:
$$
\begin{aligned}
& u=\sqrt{2}-8 x ; d u=-8 d x \\
& d v=\sin 3 x d x ; v=-\frac{1}{3} \cos 3 x
\end{aligned}
$$
Using the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =(\sqrt{2}-8 x) \cdot\left(-\frac{1}{3} \cos 3 x... | \frac{1}{3}\cdot(8x-\sqrt{2})\cos3x-\frac{8}{9}\cdot\sin3x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,738 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x}{\cos ^{2} x} d x
$$ | ## Solution
$$
\int \frac{x}{\cos ^{2} x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x ; d u=d x \\
& d v=\frac{1}{\cos ^{2} x} d x ; v=\tan x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =x \cdot \tan x-\int \tan x d x=x \cdot \tan x-\in... | x\cdot\tanx+\ln|\cosx|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,739 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x}{\sin ^{2} x} d x
$$ | ## Solution
$$
\int \frac{x}{\sin ^{2} x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x ; d u=d x \\
& d v=\frac{1}{\sin ^{2} x} d x ; v=-\cot 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}
& =x \cdot(-\cot x)-\int(-\cot x) d x=-x \cdot \cot... | -x\cdot\cotx+\ln|\sinx|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,740 |
## Problem Statement
Calculate the indefinite integral:
$$
\int x \cdot \sin ^{2} x d x
$$ | ## Solution
$$
\int x \cdot \sin ^{2} x d x=
$$
Let's denote:
$$
\begin{aligned}
& u=x \cdot \sin x ; d u=(\sin x+x \cdot \cos x) d x \\
& d v=\sin x d x ; v=-\cos x
\end{aligned}
$$
We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:
$$
\begin{aligned}
& =-\cos x \cdot x \cdot \sin x... | -\frac{x}{4}\cdot\sin2x-\frac{1}{8}\cdot\cos2x+\frac{x^{2}}{4}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,741 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x \cdot \cos x}{\sin ^{3} x} d x
$$ | ## Solution
$$
\int \frac{x \cdot \cos x}{\sin ^{3} x} d x=
$$
Let's denote:
$$
\begin{aligned}
& u=\frac{x \cdot \cos x}{\sin x}=x \cdot \operatorname{ctg} x ; d u=\left(\operatorname{ctg} x-\frac{x}{\sin ^{2} x}\right) d x \\
& d v=\frac{1}{\sin ^{2} x} d x ; v=-\operatorname{ctg} x
\end{aligned}
$$
Using the int... | -\frac{x+\cosx\cdot\sinx}{2\sin^{2}x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,742 |
## Task Condition
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{2^{\operatorname{tg} x}-2^{\sin x}}{x^{2}}, 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}\left... | \ln\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,743 |
## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=-2(\sqrt[3]{x}+3 \sqrt{x}), x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=(-2(\sqrt[3]{x}+3 \sqrt{x}))^{\prime}=-2\left(x^{\frac{1}{3}}+3 \sqrt{x}\right)^{\prime}= \\
& =-2\left(\frac{1}{3} \cdot x^{-\frac{2}{3}}+\frac{3}{2 \sqrt{x}}\right)=-\frac{2}{3 \sqrt[3]{x^{2}}}-\frac{3}{\sqrt{x}}
\end{aligned}
$$
Then:
$$
\begin... | 3y+11x+13=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,744 |
## problem statement
Find the differential $d y$.
$y=e^{x}(\cos 2 x+2 \sin 2 x)$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(e^{x}(\cos 2 x+2 \sin 2 x)\right)^{\prime} d x= \\
& =\left(\left(e^{x}\right)^{\prime}(\cos 2 x+2 \sin 2 x)+e^{x}(\cos 2 x+2 \sin 2 x)^{\prime}\right) d x= \\
& =\left(e^{x}(\cos 2 x+2 \sin 2 x)+e^{x}(-2 \sin 2 x+4 \cos 2 x)\right) d x= \\
& =e^{x}(\cos ... | 5e^{x}\cdot\cos2x\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,745 |
## Task Condition
Approximately calculate using the differential.
$y=x^{5}, x=2,997$ | ## Solution
If the increment $\Delta x = x - x_{0 \text{ of the argument }} x$ is small in absolute value, then $f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
## Let's choose:
$x_{0} = 3$
Then:
$\Delta x = -0.003$
We calculate:
$y(3) = 3^{5} = 24... | 241.785 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,746 |
## Task Condition
Find the derivative.
$y=3 \frac{\sqrt[3]{x^{2}+x+1}}{x+1}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(3 \frac{\sqrt[3]{x^{2}+x+1}}{x+1}\right)^{\prime}=3\left(\frac{\left(x^{2}+x+1\right)^{\frac{1}{3}}}{x+1}\right)^{\prime}= \\
& =3 \cdot \frac{\left(\left(x^{2}+x+1\right)^{\frac{1}{3}}\right)^{\prime}(x+1)-\left(x^{2}+x+1\right)^{\frac{1}{3}}(x+1)^{\prime}}{(x+1)^{2}}... | -\frac{x^{2}+2}{\sqrt[3]{(x^{2}+x+1)^{2}}(x+1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,747 |
## Problem Statement
Find the derivative.
$$
y=e^{\sin x}\left(x-\frac{1}{\cos x}\right)
$$ | ## Solution
$y^{\prime}=\left(e^{\sin x}\left(x-\frac{1}{\cos x}\right)\right)^{\prime}=\left(e^{\sin x}\right)^{\prime}\left(x-\frac{1}{\cos x}\right)+e^{\sin x}\left(x-\frac{1}{\cos x}\right)^{\prime}=$ $=e^{\sin x} \cdot \cos x \cdot\left(x-\frac{1}{\cos x}\right)+e^{\sin x}\left(1+\frac{1}{\cos ^{2} x} \cdot(-\sin... | e^{\sinx}\cdot(x\cdot\cosx-\frac{\sinx}{\cos^2x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,748 |
## Problem Statement
Find the derivative.
$y=\ln \frac{\sqrt{x^{2}+1}+x \sqrt{2}}{\sqrt{x^{2}+1}-x \sqrt{2}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{\sqrt{x^{2}+1}+x \sqrt{2}}{\sqrt{x^{2}+1}-x \sqrt{2}}\right)^{\prime}= \\
& =\frac{\sqrt{x^{2}+1}-x \sqrt{2}}{\sqrt{x^{2}+1}+x \sqrt{2}} \cdot \frac{\left(\sqrt{x^{2}+1}+x \sqrt{2}\right)^{\prime} \cdot\left(\sqrt{x^{2}+1}-x \sqrt{2}\right)-\left(\sqrt{x^{2}+... | \frac{2\sqrt{2}}{(1-x^{2})\sqrt{x^{2}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,749 |
## Condition of the problem
Find the derivative.
$y=\operatorname{ctg}\left(\sin \frac{1}{13}\right)-\frac{1}{48} \cdot \frac{\cos ^{2} 24 x}{\sin 48 x}$ | ## Solution
$y^{\prime}=\left(\operatorname{ctg}\left(\sin \frac{1}{13}\right)-\frac{1}{48} \cdot \frac{\cos ^{2} 24 x}{\sin 48 x}\right)^{\prime}=0-\frac{1}{48} \cdot\left(\frac{\cos ^{2} 24 x}{\sin 48 x}\right)^{\prime}=$
$=-\frac{1}{48} \cdot \frac{\left(\cos ^{2} 24 x\right)^{\prime} \cdot \sin 48 x-\cos ^{2} 24 ... | \frac{1}{4\sin^{2}24x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,750 |
## problem statement
Find the derivative.
$y=\sqrt{x}+\frac{1}{3} \cdot \operatorname{arctg} \sqrt{x}+\frac{8}{3} \cdot \operatorname{arctg} \frac{\sqrt{x}}{2}$ | ## Solution
$y^{\prime}=\left(\sqrt{x}+\frac{1}{3} \cdot \operatorname{arctg} \sqrt{x}+\frac{8}{3} \cdot \operatorname{arctg} \frac{\sqrt{x}}{2}\right)^{\prime}=$
$=\frac{1}{2 \sqrt{x}}+\frac{1}{3} \cdot \frac{1}{1+(\sqrt{x})^{2}} \cdot \frac{1}{2 \sqrt{x}}+\frac{8}{3} \cdot \frac{1}{1+\left(\frac{\sqrt{x}}{2}\right)... | \frac{3x^{2}+16x+32}{6\sqrt{x}(x+1)(x+4)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,751 |
## Task Condition
Find the derivative.
$$
y=\frac{8}{3} \cdot \operatorname{cth} 2 x-\frac{1}{3 \operatorname{ch} x \cdot \operatorname{sh}^{3} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{8}{3} \cdot \operatorname{cth} 2 x-\frac{1}{3 \operatorname{ch} x \cdot \operatorname{sh}^{3} x}\right)^{\prime}=\left(\frac{8}{3} \cdot \frac{1}{\operatorname{th} 2 x}-\frac{1}{3 \operatorname{ch} x \cdot \operatorname{sh}^{3} x}\right)^{\prime}= \\
& =-\frac{8}... | \frac{1-4\operatorname{sh}^{2}x}{\operatorname{ch}^{2}x\cdot\operatorname{sh}^{4}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,752 |
## Task Condition
Find the derivative.
$y=x^{2^{x}} \cdot 5^{x}$ | ## Solution
$y=x^{2^{x}} \cdot 5^{x}$
$\ln y=\ln \left(x^{2^{x}} \cdot 5^{x}\right)=\ln \left(x^{2^{x}}\right)+x \ln 5=2^{x} \cdot \ln (x)+x \ln 5$
$\frac{y^{\prime}}{y}=\left(2^{x} \cdot \ln (x)+x \ln 5\right)^{\prime}=2^{x} \cdot \ln 2 \cdot \ln (x)+2^{x} \cdot \frac{1}{x}+\ln 5$
$y^{\prime}=y \cdot\left(2^{x} \c... | x^{2^{x}}\cdot5^{x}\cdot(2^{x}\cdot\ln2\cdot\ln(x)+\frac{2^{x}}{x}+\ln5) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,753 |
## Task Condition
Find the derivative.
$$
y=\ln \left(5 x+\sqrt{25 x^{2}+1}\right)-\sqrt{25 x^{2}+1} \cdot \operatorname{arctg} 5 x
$$ | ## Solution
$y^{\prime}=\left(\ln \left(5 x+\sqrt{25 x^{2}+1}\right)-\sqrt{25 x^{2}+1} \cdot \operatorname{arctg} 5 x\right)^{\prime}=$
$=\frac{1}{5 x+\sqrt{25 x^{2}+1}} \cdot\left(5+\frac{50 x}{2 \sqrt{25 x^{2}+1}}\right)-\left(\frac{50 x}{2 \sqrt{25 x^{2}+1}} \cdot \operatorname{arctg} 5 x+\sqrt{25 x^{2}+1} \cdot \... | -\frac{25x\cdot\operatorname{arctg}5x}{\sqrt{25x^{2}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,754 |
## Task Condition
Find the derivative.
$$
y=x(\arcsin x)^{2}+2 \sqrt{1-x^{2}} \arcsin x-2 x
$$ | ## Solution
$y^{\prime}=\left(x(\arcsin x)^{2}+2 \sqrt{1-x^{2}} \arcsin x-2 x\right)^{\prime}=$
$=(\arcsin x)^{2}+x \cdot 2 \cdot \arcsin x \cdot \frac{1}{\sqrt{1-x^{2}}}+\frac{2}{2 \sqrt{1-x^{2}}} \cdot(-2 x) \cdot \arcsin x+2 \sqrt{1-x^{2}} \cdot \frac{1}{\sqrt{1-x^{2}}}-2=$ $=(\arcsin x)^{2}+\frac{2 x \cdot \arcsi... | (\arcsinx)^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,755 |
## Task Condition
Find the derivative.
$$
y=x-\ln \left(1+e^{x}\right)-2 e^{-\frac{x}{2}} \operatorname{arctan} e^{\frac{x}{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x-\ln \left(1+e^{x}\right)-2 e^{-\frac{x}{2}} \operatorname{arctg} e^{\frac{x}{2}}\right)^{\prime}= \\
& =1-\frac{1}{1+e^{x}} \cdot e^{x}-\left(\left(2 e^{-\frac{x}{2}}\right)^{\prime} \cdot \operatorname{arctg} e^{\frac{x}{2}}+2 e^{-\frac{x}{2}} \cdot\left(\operatorna... | x\cdote^{-\frac{x}{2}}\cdot\operatorname{arctg}e^{\frac{x}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,756 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\operatorname{arctg} \frac{t+1}{t-1} \\
y=\arcsin \sqrt{1-t^{2}}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\operatorname{arctg} \frac{t+1}{t-1}\right)^{\prime}=\frac{1}{1+\left(\frac{t+1}{t-1}\right)^{2}} \cdot\left(\frac{t+1}{t-1}\right)^{\prime}= \\
& =\frac{(t-1)^{2}}{(t-1)^{2}+(t+1)^{2}} \cdot \frac{1 \cdot(t-1)-(t+1) \cdot 1}{(t-1)^{2}}= \\
& =\frac{t-1-t-1}{t^{2}-... | \frac{^{2}+1}{\sqrt{1-^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,757 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=t-t^{4} \\
y=t^{2}-t^{3}
\end{array}
\right.
\]
$t_{0}=1$ | ## Solution
Since $t_{0}=1$, then
$x_{0}=1-1^{4}=0$
$y_{0}=1^{2}-1^{3}=0$
Find the derivatives:
$x_{t}^{\prime}=\left(t-t^{4}\right)^{\prime}=1-4 t^{3}$
$y_{t}^{\prime}=\left(t^{2}-t^{3}\right)^{\prime}=2 t-3 t^{2}$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 t-3 t^{2}}{1-4 t^{3}}$
Then:
$y_{0}... | \frac{x}{3},\,-3x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,758 |
## Problem Statement
Find the $n$-th order derivative.
$$
y=\frac{11+12 x}{6 x+5}
$$ | ## Solution
$y^{\prime}=\left(\frac{11+12 x}{6 x+5}\right)^{\prime}=\frac{12 \cdot(6 x+5)-(11+12 x) \cdot 6}{(6 x+5)^{2}}=$
$=\frac{6 \cdot(12 x+10-11-12 x)}{(6 x+5)^{2}}=-6(6 x+5)^{-2}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-6(6 x+5)^{-2}\right)^{\prime}=$
$=-6 \cdot(-2) \cdot(6 x+5)^{-3} \cdot... | y^{(n)}=(-1)^{n}\cdotn!\cdot6^{n}\cdot(6x+5)^{-n-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,759 |
## Task Condition
Find the derivative of the specified order.
$$
y=x \ln (1-3 x), y^{(I V)}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=(x \ln (1-3 x))^{\prime}=\ln (1-3 x)+\frac{x}{1-3 x} \cdot(-3)=\ln (1-3 x)-\frac{3 x}{1-3 x} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\ln (1-3 x)-\frac{3 x}{1-3 x}\right)^{\prime}=\frac{1}{1-3 x} \cdot(-3)-\frac{3 \cdot(1-3 x)-3 x \cdot(-3)}{(1-3 x)^{2}}... | -\frac{54(4-3x)}{(1-3x)^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,760 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\cos t \\
y=\sin ^{4}\left(\frac{t}{2}\right)
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\cos t)^{\prime}=-\sin t$
$y_{t}^{\prime}=\left(\sin ^{4}\left(\frac{t}{2}\right)\right)^{\prime}=4 \sin ^{3}\left(\frac{t}{2}\right) \cdot \frac{1}{2}=2 \sin ^{3}\left(\frac{t}{2}\right)$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 \sin ^{3}(t / 2)}{-\sin t... | \frac{\cos^{2}(/2)+1}{4\cos^{3}(/2)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,761 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
\[
\begin{aligned}
& y=e^{x+x^{2}}+2 e^{x} \\
& y^{\prime}-y=2 x e^{x+x^{2}}
\end{aligned}
\] | ## Solution
$y^{\prime}=\left(e^{x+x^{2}}+2 e^{x}\right)^{\prime}=e^{x+x^{2}} \cdot(1+2 x)+2 e^{x}$
Substitute into equation (1:
$\left(e^{x+x^{2}} \cdot(1+2 x)+2 e^{x}\right)-\left(e^{x+x^{2}}+2 e^{x}\right)=2 x e^{x+x^{2}}$
## Simplify:
$$
\begin{aligned}
& e^{x+x^{2}}+2 x e^{x+x^{2}}+2 e^{x}-e^{x+x^{2}}-2 e^{x}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,762 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}-3 n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}-3 n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\rig... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,764 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n^{8}+6}-\sqrt{n-6}}{\sqrt[8]{n^{8}+6}+\sqrt{n-6}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n^{8}+6}-\sqrt{n-6}}{\sqrt[8]{n^{8}+6}+\sqrt{n-6}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n^{8}+6}-\sqrt{n-6}\right)}{\frac{1}{n}\left(\sqrt[8]{n^{8}+6}+\sqrt{n-6}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{n^{6}+... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,765 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \sqrt{n+2}(\sqrt{n+3}-\sqrt{n-4})
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \sqrt{n+2}(\sqrt{n+3}-\sqrt{n-4})= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}(\sqrt{n+3}-\sqrt{n-4})(\sqrt{n+3}+\sqrt{n-4})}{\sqrt{n+3}+\sqrt{n-4}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}(n+3-(n-4))}{\sqrt{n+3}+\sqrt{n-4}}=\lim _{n \ri... | \frac{7}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,766 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n!+(n+2)!}{(n-1)!+(n+2)!}=\lim _{n \rightarrow \infty} \frac{(n-1)!(n+n(n+1)(n+2))}{(n-1)!(1+n(n+1)(n+2))}= \\
& =\lim _{n \rightarrow \infty} \frac{n+n(n+1)(n+2)}{1+n(n+1)(n+2)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}(n+n(n+1)(n+2))}{\frac... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,767 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{13 n+3}{13 n-10}\right)^{n-3}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{13 n+3}{13 n-10}\right)^{n-3}=\lim _{n \rightarrow \infty}\left(\frac{13 n-10+13}{13 n-10}\right)^{n-3}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{13}{13 n-10}\right)^{n-3}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{13 n-10}{1... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,768 | |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ ):
$\lim _{x \rightarrow-6} \frac{3 x^{2}+17 x-6}{x+6}=-19$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,769 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{\text {): }}$
$f(x)=2 x^{2}+8, x_{0}=5$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0$ : $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,770 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{(1+x)^{3}-(1+3 x)}{x^{2}+x^{5}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{1^{3}+3 \cdot 1^{2} \cdot x+3 \cdot 1 \cdot x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=$
$=\lim _{x \rightarrow 0} \frac{1+3 x+3 x^{2}+x^{3}-1-3 x}{x^{2}\left(1+x^{3}\right)}=\lim _{x ... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,771 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-2} \frac{\sqrt[3]{x-6}+2}{\sqrt[3]{x^{3}+8}}=\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+2)\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}{\sqrt[3]{x^{3}+8}\left(\sqrt[3]{(x-6)^{2}}-2 \sqrt[3]{x-6}+4\right)}= \\
& =\lim _{x \rightarrow-2} \frac{(\sqrt[3]{x-6}+... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,772 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$\ln \left(1+x^{2}\right) \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$
We get:
$\lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{1-\sqrt{x^{2}+1}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{x^{2}}{1-\sqr... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,773 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{\cos \left(\frac{\pi x}{2}\right)}{1-\sqrt{x}}$ | ## Solution
Substitution:
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow 1} \frac{\cos \left(\frac{\pi x}{2}\right)}{1-\sqrt{x}}=\lim _{y \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}(y+1)\right)}{1-\sqrt{y+1}}=$
$$
\begin{aligned}
& =\lim _{y \rightarrow 0} ... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,774 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow-3} \frac{\sin \left(e^{\frac{\sqrt[3]{1-x^{2}}}{2}}-e^{\sqrt[3]{x+2}}\right)}{\operatorname{arctg}(x+3)}
$$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y-3 \Rightarrow y=x+3 \\
& x \rightarrow-3 \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow-3} \frac{\sin \left(e^{\frac{\sqrt[3]{1-x^{2}}}{2}}-e^{\sqrt[3]{x+2}}\right)}{\operatorname{arctg}(x+3)}=\lim _{y \rightarrow 0} ... | -\frac{1}{12e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,775 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{\sin 2 x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$
$=\frac{\lim _{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,776 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sin b x - \sin a x}{\ln \left(\tan\left(\frac{\pi}{4} + a x\right)\right)}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin b x-\sin a x}{\ln \left(\tan\left(\frac{\pi}{4}+a x\right)\right)}=\lim _{x \rightarrow 0} \frac{2 \sin \frac{b x-a x}{2} \cos \frac{b x+a x}{2}}{\ln \left(\frac{\tan \frac{\pi}{4}+\tan a x}{1-\tan \frac{\pi}{4} \cdot \tan a x}\right)}= \\
& =\lim _{x... | \frac{b-}{2a} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,777 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(1+\tan^{2} x\right)^{\frac{1}{\ln \left(1+3 x^{2}\right)}}
$$ | ## Solution
$\lim _{x \rightarrow 0}\left(1+\tan ^{2} x\right)^{\frac{1}{\ln \left(1+3 x^{2}\right)}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(1+\tan ^{2} x\right)}\right)^{\frac{1}{\ln \left(1+3 x^{2}\right)}}=$
$=\lim _{x \rightarrow 0} e^{\frac{\ln \left(1+\tan ^{2} x\right)}{\ln \left(1+3 x^{2}\right)}}=$
... | e^{\frac{1}{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,778 |
Condition of the problem
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(6-\frac{5}{\cos x}\right)^{\operatorname{tg}^{2} x}=\left(6-\frac{5}{\cos 0}\right)^{\operatorname{tg}^{2} 0}= \\
& =\left(6-\frac{5}{1}\right)^{0^{2}}=1^{0}=1
\end{aligned}
$$
## Problem Kuznetsov Limits $18-28$ | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,779 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2} \\
& x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0
\end{aligned}
$$
We obtain:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}=\lim _{y \rightarrow 0}\le... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,780 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}
$$ | ## Solution
$\lim _{x \rightarrow 1}(\sqrt[3]{x}+x-1)^{\sin \left(\frac{\pi x}{4}\right)}=(\sqrt[3]{1}+1-1)^{\sin \left(\frac{\pi \cdot 1}{4}\right)}=(1)^{\frac{\sqrt{2}}{2}}=1$
## Problem Kuznetsov Limits 20-28 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,781 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1} \tan\left(\cos x+\sin \frac{x-1}{x+1} \cdot \cos \frac{x+1}{x-1}\right)
$$ | ## Solution
Since $\cos \frac{x+1}{x-1}$ is bounded, and
$\lim _{x \rightarrow 1} \sin \frac{x-1}{x+1}=\sin \frac{1-1}{1+1}=\sin 0=0$, then
$\sin \frac{x-1}{x+1} \cdot \cos \frac{x+1}{x-1} \rightarrow 0 \quad$, as $x \rightarrow 1$
Then:
$\lim _{x \rightarrow 1} \operatorname{tg}\left(\cos x+\sin \frac{x-1}{x+1} \c... | \tan(\cos1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,782 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\frac{e^{x^{2}}-\cos 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}\... | 1.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,783 |
## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=3(\sqrt[3]{x}-2 \sqrt{x}), x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=(3(\sqrt[3]{x}-2 \sqrt{x}))^{\prime}=3\left(x^{\frac{1}{3}}-2 \sqrt{x}\right)^{\prime}= \\
& =3\left(\frac{1}{3} \cdot x^{-\frac{2}{3}}-\frac{2}{2 \sqrt{x}}\right)^{\prime}=\frac{1}{\sqrt[3]{x^{2}}}-\frac{3}{\sqrt{x}}
\end{aligned}
$$
Then:
$y_{0}... | -2x-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,784 |
## Task Condition
Find the differential $d y$.
$$
y=\operatorname{arctg} \frac{x^{2}-1}{x}
$$ | ## Solution
$d y=y^{\prime} \cdot d x=\left(\operatorname{arctg} \frac{x^{2}-1}{x}\right)^{\prime} d x=\frac{1}{1+\left(\frac{x^{2}-1}{x}\right)^{2}} \cdot\left(\frac{x^{2}-1}{x}\right)^{\prime} d x=$
$$
\begin{aligned}
& =\frac{x^{2}}{x^{2}+\left(x^{2}-1\right)^{2}} \cdot\left(\frac{2 x \cdot x-\left(x^{2}-1\right)}... | \frac{x^{2}+1}{x^{4}-x^{2}+1}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,785 |
## Task Condition
Find the derivative.
$$
y=\frac{\left(2 x^{2}+3\right) \sqrt{x^{2}-3}}{9 x^{3}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\left(2 x^{2}+3\right) \sqrt{x^{2}-3}}{9 x^{3}}\right)^{\prime}= \\
& =\frac{\left(4 x \sqrt{x^{2}-3}+\left(2 x^{2}+3\right) \frac{1}{2 \sqrt{x^{2}-3}} \cdot 2 x\right) \cdot x^{3}-\left(2 x^{2}+3\right) \sqrt{x^{2}-3} \cdot 3 x^{2}}{9 x^{6}}= \\
& =\frac{\left(4... | \frac{3}{x^{4}\sqrt{x^{2}-3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,787 |
## Task Condition
Find the derivative.
$$
y=x-\ln \left(1+e^{x}\right)-2 e^{-\frac{x}{2}} \cdot \operatorname{arctg} e^{\frac{x}{2}}-\left(\operatorname{arctg} e^{\frac{x}{2}}\right)^{2}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x-\ln \left(1+e^{x}\right)-2 e^{-\frac{x}{2}} \cdot \operatorname{arctg} e^{\frac{x}{2}}-\left(\operatorname{arctg} e^{\frac{x}{2}}\right)^{2}\right)^{\prime}= \\
& =1-\frac{1}{1+e^{x}} \cdot e^{x}-2 \cdot\left(e^{-\frac{x}{2}} \cdot\left(-\frac{1}{2}\right) \cdot \ope... | \frac{\operatorname{arctg}e^{x/2}}{e^{x/2}\cdot(1+e^{x})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,788 |
## Task Condition
Find the derivative.
$$
y=\log _{a}\left(\frac{1}{\sqrt{1-x^{4}}}\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\log _{a}\left(\frac{1}{\sqrt{1-x^{4}}}\right)\right)^{\prime}=\frac{1}{\left(\frac{1}{\sqrt{1-x^{4}}}\right) \cdot \ln a} \cdot\left(\frac{1}{\sqrt{1-x^{4}}}\right)^{\prime}= \\
& =\frac{\sqrt{1-x^{4}}}{\ln a} \cdot\left(-\frac{1}{2}\right) \cdot \frac{1}{\sqrt{\left(... | \frac{2x^{3}}{\ln\cdot(1-x^{4})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,789 |
## Task Condition
Find the derivative.
$$
y=\frac{\operatorname{tg}(\ln 2) \cdot \sin ^{2} 19 x}{19 \cos 38 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\tan(\ln 2) \cdot \sin ^{2} 19 x}{19 \cos 38 x}\right)^{\prime}=\frac{\tan(\ln 2)}{19} \cdot\left(\frac{\sin ^{2} 19 x}{\cos 38 x}\right)^{\prime}= \\
& =\frac{\tan(\ln 2)}{19} \cdot\left(\frac{\sin ^{2} 19 x}{\cos 38 x}\right)^{\prime}= \\
& =\frac{\tan(\ln 2)}{... | \frac{\tan(\ln2)^{2}\cdot\tan38x}{\cos38x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,790 |
## Problem Statement
Find the derivative.
$$
y=\frac{2 \sqrt{1-x} \cdot \arcsin \sqrt{x}}{x}+\frac{2}{\sqrt{x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 \sqrt{1-x} \cdot \arcsin \sqrt{x}}{x}+\frac{2}{\sqrt{x}}\right)^{\prime}=\left(2 \sqrt{\frac{1-x}{x^{2}}} \cdot \arcsin \sqrt{x}+\frac{2}{\sqrt{x}}\right)^{\prime}= \\
& =2 \cdot \frac{1}{2 \sqrt{\frac{1-x}{x^{2}}}} \cdot\left(\frac{-1 \cdot x^{2}-(1-x) \cdot 2... | \frac{x-2}{x^{2}\sqrt{1-x}}\cdot\arcsin\sqrt{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,791 |
## Problem Statement
Find the derivative.
$$
y=\frac{1}{\sqrt{8}} \ln \frac{4+\sqrt{8} \tanh \frac{x}{2}}{4-\sqrt{8} \tanh \frac{x}{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sqrt{8}} \ln \frac{4+\sqrt{8} \tanh \frac{x}{2}}{4-\sqrt{8} \tanh \frac{x}{2}}\right)^{\prime}=\frac{1}{\sqrt{8}} \cdot \frac{4-\sqrt{8} \tanh \frac{x}{2}}{4+\sqrt{8} \tanh \frac{x}{2}} \cdot\left(\frac{4+\sqrt{8} \tanh \frac{x}{2}}{4-\sqrt{8} \tanh \frac{x}{... | \frac{1}{2(\cosh^{2}\frac{x}{2}+1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,792 |
## Task Condition
Find the derivative.
$y=19^{x^{19}} \cdot x^{19}$ | ## Solution
$y=19^{x^{19}} \cdot x^{19}$
$\ln y=\ln \left(19^{x^{19}} \cdot x^{19}\right)=\ln \left(19^{x^{19}}\right)+\ln x^{19}=x^{19} \cdot \ln 19+19 \ln x$
$\frac{y^{\prime}}{y}=\left(x^{19} \cdot \ln 19+19 \ln x\right)^{\prime}=19 x^{18} \cdot \ln 19+19 \cdot \frac{1}{x}=$
$=19\left(x^{18} \cdot \ln 19+\frac{1}... | 19^{x^{19}}\cdotx^{19}\cdot19(x^{18}\cdot\ln19+\frac{1}{x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,793 |
## Task Condition
Find the derivative.
$$
y=\ln \left(2 x-3+\sqrt{4 x^{2}-12 x+10}\right)-\sqrt{4 x^{2}-12 x+10} \cdot \operatorname{arctg}(2 x-3)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \left(2 x-3+\sqrt{4 x^{2}-12 x+10}\right)-\sqrt{4 x^{2}-12 x+10} \cdot \operatorname{arctg}(2 x-3)\right)^{\prime}= \\
& =\frac{1}{2 x-3+\sqrt{4 x^{2}-12 x+10}} \cdot\left(2+\frac{1}{2 \sqrt{4 x^{2}-12 x+10}} \cdot(8 x-12)\right)- \\
& -\left(\frac{1}{2 \sqrt{4 x^{... | -\frac{\operatorname{arctg}(2x-3)}{\sqrt{4x^{2}-12x+10}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,794 |
## Problem Statement
Find the derivative.
$$
y=\ln \sqrt[3]{\frac{x-1}{x+1}}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{x^{2}-1}\right) \operatorname{arctan} x
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \sqrt[3]{\frac{x-1}{x+1}}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{x^{2}-1}\right) \operatorname{arctg} x\right)^{\prime}= \\
& =\sqrt[3]{\frac{x+1}{x-1}} \cdot \frac{1}{3} \cdot\left(\frac{x-1}{x+1}\right)^{-\frac{2}{3}} \cdot \frac{1 \cdot(x+1)-(x-1) \cdot 1}{(x+1)^... | \frac{5x^{2}+17}{12(x^{4}-1)}+\frac{x\cdot\operatorname{arctg}x}{(x^{2}-1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,795 |
## Task Condition
Find the derivative.
$y=\ln \frac{\sqrt{2}+\operatorname{th} x}{\sqrt{2}-\operatorname{th} x}$ | ## Solution
$y^{\prime}=\left(\ln \frac{\sqrt{2}+\operatorname{th} x}{\sqrt{2}-\operatorname{th} x}\right)^{\prime}=\frac{\sqrt{2}-\operatorname{th} x}{\sqrt{2}+\operatorname{th} x} \cdot \frac{\frac{1}{\operatorname{ch}^{2} x} \cdot(\sqrt{2}-\operatorname{th} x)-(\sqrt{2}+\operatorname{th} x) \cdot\left(-\frac{1}{\op... | \frac{2\sqrt{2}}{\operatorname{ch}^{2}x+1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,796 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\arcsin \sqrt{t} \\
y=\sqrt{1+\sqrt{t}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\arcsin \sqrt{t})^{\prime}=\frac{1}{\sqrt{1-(\sqrt{t})^{2}}} \cdot \frac{1}{2 \sqrt{t}}=\frac{1}{2 \sqrt{t(1-t)}}$
$y_{t}^{\prime}=(\sqrt{1+\sqrt{t}})^{\prime}=\frac{1}{2 \sqrt{1+\sqrt{t}}} \cdot \frac{1}{2 \sqrt{t}}=\frac{1}{4 \sqrt{t(1+\sqrt{t})}}$
We obtain:
$$
\begin{aligned}
& y_{x... | \frac{\sqrt{1-}}{2\sqrt{1+\sqrt{}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,797 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=1-t^{2} \\
y=t-t^{3}
\end{array}
\right.
\]
$t_{0}=2$ | ## Solution
Since $t_{0}=1$, then
$x_{0}=1-2^{2}=-3$
$y_{0}=2-2^{3}=-6$
Let's find the derivatives:
$x_{t}^{\prime}=\left(1-t^{2}\right)^{\prime}=-2 t$
$y_{t}^{\prime}=\left(t-t^{3}\right)^{\prime}=1-3 t^{2}$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1-3 t^{2}}{-2 t}=\frac{3 t^{2}-1}{2 t}$
Then... | \frac{11x}{4}+\frac{9}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,798 |
## Problem Statement
Find the $n$-th order derivative.
$y=\frac{4}{x}$ | ## Solution
$$
\begin{aligned}
y & =\frac{4}{x} \\
y^{\prime} & =\left(\frac{4}{x}\right)^{\prime}=-\frac{4}{x^{2}}
\end{aligned}
$$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{4}{x^{2}}\right)^{\prime}=-\frac{4 \cdot(-2)}{x^{3}}=\frac{4 \cdot 2}{x^{3}}$
$y^{\prime \prime \prime}=\left(y^{\prime... | y^{(n)}=\frac{4\cdot(-1)^{n}\cdotn!}{x^{n}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,799 |
## Task Condition
Find the derivative of the specified order.
$y=(x+7) \ln (x+4), y^{V}=?$ | ## Solution
$y^{\prime}=((x+7) \ln (x+4))^{\prime}=\ln (x+4)+(x+7) \cdot \frac{1}{x+4}=$
$=\ln (x+4)+\frac{x+7}{x+4}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\ln (x+4)+\frac{x+7}{x+4}\right)^{\prime}=\frac{1}{x+4}+\frac{1 \cdot(x+4)-(x+7) \cdot 1}{(x+4)^{2}}=$
$=\frac{x+4}{(x+4)^{2}}+\frac{x+4-x-7... | \frac{-120x+1680}{(x+4)^{7}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,800 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=t+\sin t \\ y=2+\cos t\end{array}\right.$ | ## Solution
$x_{t}^{\prime}=(t+\sin t)^{\prime}=1+\cos t$
$y_{t}^{\prime}=(2+\cos t)^{\prime}=-\sin t$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-\sin t}{1+\cos t}=-\frac{\sin t}{1+\cos t}$
$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(-\frac{\sin t}{1+\cos t}\right)^{\prime}=-\frac{... | -\frac{1}{(1+\cos)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,801 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{16} \sqrt{256-x^{2}} d x
$$ | ## Solution
$$
\int_{0}^{16} \sqrt{256-x^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=16 \sin t \Rightarrow d x=16 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{16}=0 \\
& x=16 \Rightarrow t=\arcsin \frac{16}{16}=\frac{\pi}{2}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 2} \sqrt{256-25... | 64\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,803 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} x^{2} \sqrt{1-x^{2}} d x
$$ | ## Solution
$$
\int_{0}^{1} x^{2} \sqrt{1-x^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=\sin t \Rightarrow d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=1 \Rightarrow t=\arcsin 1=\frac{\pi}{2}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 2} \sin ^{2} t \cdot \sqrt{1-\sin ^{2} t} ... | \frac{\pi}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,804 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{5} \frac{d x}{\left(25+x^{2}\right) \sqrt{25+x^{2}}}
$$ | ## Solution
$$
\int_{0}^{5} \frac{d x}{\left(25+x^{2}\right) \sqrt{25+x^{2}}}=\int_{0}^{5} \frac{d x}{\left(25+x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=5 \operatorname{tg} t \Rightarrow d x=\frac{5 d t}{\cos ^{2} t} \\
& x=0 \Rightarrow t=\operatorname{arctg} \frac{0}{5}=0 \\
& x=5 \Rightarrow ... | \frac{\sqrt{2}}{50} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,805 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{3} \frac{d x}{\left(9+x^{2}\right)^{3 / 2}}
$$ | ## Solution
$$
\int_{0}^{3} \frac{d x}{\left(9+x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=3 \operatorname{tg} t \Rightarrow d x=\frac{3 d t}{\cos ^{2} t} \\
& x=0 \Rightarrow t=\operatorname{arctg} \frac{0}{3}=0 \\
& x=3 \Rightarrow t=\operatorname{arctg} \frac{3}{3}=\frac{\pi}{4}
\end{aligned}
$... | \frac{\sqrt{2}}{18} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,806 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\sqrt{5} / 2} \frac{d x}{\sqrt{\left(5-x^{2}\right)^{3}}}
$$ | ## Solution
$$
\int_{0}^{\sqrt{5} / 2} \frac{d x}{\sqrt{\left(5-x^{2}\right)^{3}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sqrt{5} \sin t ; d x=\sqrt{5} \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{\sqrt{5}}=0 \\
& x=\frac{\sqrt{5}}{2} \Rightarrow t=\arcsin \frac{\left(\frac{\sqrt{5}}{2}\right)}{\sqrt{5}}=\a... | \frac{\sqrt{3}}{15} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,807 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x^{4}} d x
$$ | ## Solution
$$
\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x^{4}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=\frac{1}{\sin t} \Rightarrow d x=-\frac{1}{\sin ^{2} t} \cdot \cos t d t=-\frac{\cos t d t}{\sin ^{2} t} \\
& x=1 \Rightarrow t=\arcsin \frac{1}{1}=\frac{\pi}{2} \\
& x=2 \Rightarrow t=\arcsin \frac{1}{2}=\frac{\pi}... | \frac{\sqrt{3}}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,808 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\sqrt{2} / 2} \frac{x^{4} \cdot d x}{\sqrt{\left(1-x^{2}\right)^{3}}}
$$ | ## Solution
$$
\int_{0}^{\sqrt{2} / 2} \frac{x^{4} \cdot d x}{\sqrt{\left(1-x^{2}\right)^{3}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sin t \Rightarrow d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=\frac{\sqrt{2}}{2} \Rightarrow t=\arcsin \frac{\sqrt{2}}{2}=\frac{\pi}{4}
\end{aligned}
$$
We get:
$$
\... | \frac{5}{4}-\frac{3\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,809 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\sqrt{3}} \frac{d x}{\sqrt{\left(4-x^{2}\right)^{3}}}
$$ | ## Solution
$$
\int_{0}^{\sqrt{3}} \frac{d x}{\sqrt{\left(4-x^{2}\right)^{3}}}=
$$
Substitution:
$$
\begin{aligned}
& x=2 \sin t ; d x=2 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{2}=0 \\
& x=\sqrt{3} \Rightarrow t=\arcsin \frac{\sqrt{3}}{2}=\frac{\pi}{3}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int... | \frac{\sqrt{3}}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,810 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{x^{4} \cdot d x}{\left(2-x^{2}\right)^{3 / 2}}
$$ | ## Solution
$$
\int_{0}^{1} \frac{x^{4} \cdot d x}{\left(2-x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sqrt{2} \sin t \Rightarrow d x=\sqrt{2} \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{\sqrt{2}}=0 \\
& x=1 \Rightarrow t=\arcsin \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$
We get:
... | \frac{5}{2}-\frac{3\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,811 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{x^{2} \cdot d x}{\sqrt{16-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{2} \frac{x^{2} \cdot d x}{\sqrt{16-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=4 \sin t \Rightarrow d x=4 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{4}=0 \\
& x=2 \Rightarrow t=\arcsin \frac{2}{4}=\frac{\pi}{6}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 6} ... | \frac{4\pi}{3}-2\sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,812 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \sqrt{4-x^{2}} d x
$$ | ## Solution
$$
\int_{0}^{2} \sqrt{4-x^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=2 \sin t \Rightarrow d x=2 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{2}=0 \\
& x=2 \Rightarrow t=\arcsin \frac{2}{2}=\frac{\pi}{2}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 2} \sqrt{4-4 \sin ^{2} t... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,813 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{4} \frac{d x}{\left(16+x^{2}\right)^{3 / 2}}
$$ | ## Solution
$$
\int_{0}^{4} \frac{d x}{\left(16+x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=4 \operatorname{tg} t \Rightarrow d x=\frac{4 d t}{\cos ^{2} t} \\
& x=0 \Rightarrow t=\operatorname{arctg} \frac{0}{4}=0 \\
& x=4 \Rightarrow t=\operatorname{arctg} \frac{4}{4}=\frac{\pi}{4}
\end{aligned}
... | \frac{\sqrt{2}}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,814 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{5 / 2} \frac{x^{2} \cdot d x}{\sqrt{25-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{5 / 2} \frac{x^{2} \cdot d x}{\sqrt{25-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=5 \sin t \Rightarrow d x=5 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{5}=0 \\
& x=\frac{5}{2} \Rightarrow t=\arcsin \frac{\left(\frac{5}{2}\right)}{5}=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligne... | \frac{25\pi}{12}-\frac{25\sqrt{3}}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,816 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{5} x^{2} \cdot \sqrt{25-x^{2}} d x
$$ | ## Solution
$$
\int_{0}^{5} x^{2} \cdot \sqrt{25-x^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=5 \sin t \Rightarrow d x=5 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{5}=0 \\
& x=5 \Rightarrow t=\arcsin \frac{5}{5}=\frac{\pi}{2}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 2} 25 \sin ... | \frac{625\pi}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,817 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{4} \sqrt{16-x^{2}} d x
$$ | ## Solution
$$
\int_{0}^{4} \sqrt{16-x^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=4 \sin t \Rightarrow d x=4 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{4}=0 \\
& x=4 \Rightarrow t=\arcsin \frac{4}{4}=\frac{\pi}{2}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 2} \sqrt{16-16 \sin ^{2... | 4\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,818 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{4 \sqrt{3}} \frac{d x}{\sqrt{\left(64-x^{2}\right)^{3}}}
$$ | ## Solution
$$
\int_{0}^{4 \sqrt{3}} \frac{d x}{\sqrt{\left(64-x^{2}\right)^{3}}}=
$$
Substitution:
$$
\begin{aligned}
& x=8 \sin t ; d x=8 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{8}=0 \\
& x=4 \sqrt{3} \Rightarrow t=\arcsin \frac{4 \sqrt{3}}{8}=\frac{\pi}{3}
\end{aligned}
$$
We get:
$$
\begin{aligned}
... | \frac{\sqrt{3}}{64} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,819 |
## Problem Statement
Calculate the definite integral:
$$
\int_{\sqrt{2}}^{2 \sqrt{2}} \frac{\sqrt{x^{2}-2}}{x^{4}} d x
$$ | ## Solution
$$
\int_{\sqrt{2}}^{2 \sqrt{2}} \frac{\sqrt{x^{2}-2}}{x^{4}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=\frac{\sqrt{2}}{\sin t} \Rightarrow d x=-\frac{\sqrt{2}}{\sin ^{2} t} \cdot \cos t d t=-\frac{\sqrt{2} \cos t d t}{\sin ^{2} t} \\
& x=\sqrt{2} \Rightarrow t=\arcsin \frac{\sqrt{2}}{\sqrt{2}}=\frac{\... | \frac{\sqrt{3}}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,820 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2 \sqrt{2}} \frac{x^{4} \cdot d x}{\left(16-x^{2}\right) \sqrt{16-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{2 \sqrt{2}} \frac{x^{4} \cdot d x}{\left(16-x^{2}\right) \sqrt{16-x^{2}}}=\int_{0}^{2 \sqrt{2}} \frac{x^{4} \cdot d x}{\left(16-x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=4 \sin t \Rightarrow d x=4 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{4}=0 \\
& x=2 \sqrt{2} ... | 20-6\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,821 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-3}^{3} x^{2} \cdot \sqrt{9-x^{2}} d x
$$ | ## Solution
$$
\int_{-3}^{3} x^{2} \cdot \sqrt{9-x^{2}} d x=
$$
Integrals 12-20
Substitution:
$$
x=3 \sin t \Rightarrow d x=3 \cos t d t
$$
$$
\begin{aligned}
& x=-3 \Rightarrow t=\arc... | \frac{81\pi}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,822 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{\sqrt{3}} \frac{d x}{\sqrt{\left(1+x^{2}\right)^{3}}}
$$ | ## Solution
$$
\int_{1}^{\sqrt{3}} \frac{d x}{\sqrt{\left(1+x^{2}\right)^{3}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\operatorname{tg} t \Rightarrow d x=\frac{d t}{\cos ^{2} t} \\
& x=1 \Rightarrow t=\operatorname{arctg} 1=\frac{\pi}{4} \\
& x=\sqrt{3} \Rightarrow t=\operatorname{arctg} \sqrt{3}=\frac{\pi}{3}
\en... | \frac{\sqrt{3}-\sqrt{2}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,823 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{d x}{\sqrt{\left(16-x^{2}\right)^{3}}}
$$ | ## Solution
$$
\int_{0}^{2} \frac{d x}{\sqrt{\left(16-x^{2}\right)^{3}}}=
$$
Substitution:
$$
\begin{aligned}
& x=4 \sin t ; d x=4 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{4}=0 \\
& x=2 \Rightarrow t=\arcsin \frac{2}{4}=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int... | \frac{\sqrt{3}}{48} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,824 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{x^{4} \cdot d x}{\sqrt{\left(8-x^{2}\right)^{3}}}
$$ | ## Solution
$$
\int_{0}^{2} \frac{x^{4} \cdot d x}{\sqrt{\left(8-x^{2}\right)^{3}}}=\int_{0}^{2} \frac{x^{4} \cdot d x}{\left(8-x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=2 \sqrt{2} \sin t \Rightarrow d x=2 \sqrt{2} \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{2 \sqrt{2}}=0 \\
& x=2 \Righta... | 10-3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,825 |
## Problem Statement
Calculate the definite integral:
$$
\int_{3}^{6} \frac{\sqrt{x^{2}-9}}{x^{4}} d x
$$ | ## Solution
$$
\int_{3}^{6} \frac{\sqrt{x^{2}-9}}{x^{4}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=\frac{3}{\sin t} \Rightarrow d x=-\frac{3}{\sin ^{2} t} \cdot \cos t d t=-\frac{3 \cos t d t}{\sin ^{2} t} \\
& x=3 \Rightarrow t=\arcsin \frac{3}{3}=\frac{\pi}{2} \\
& x=6 \Rightarrow t=\arcsin \frac{3}{6}=\frac{\p... | \frac{\sqrt{3}}{72} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,826 |
## Problem Statement
Calculate the definite integral:
$$
\int_{2}^{4} \frac{\sqrt{x^{2}-4}}{x^{4}} d x
$$ | ## Solution
$$
\int_{2}^{4} \frac{\sqrt{x^{2}-4}}{x^{4}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=\frac{2}{\sin t} \Rightarrow d x=-\frac{2}{\sin ^{2} t} \cdot \cos t d t=-\frac{2 \cos t d t}{\sin ^{2} t} \\
& x=2 \Rightarrow t=\arcsin \frac{2}{2}=\frac{\pi}{2} \\
& x=4 \Rightarrow t=\arcsin \frac{2}{4}=\frac{\p... | \frac{\sqrt{3}}{32} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,828 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{d x}{\left(4+x^{2}\right) \sqrt{4+x^{2}}}
$$ | ## Solution
$$
\int_{0}^{2} \frac{d x}{\left(4+x^{2}\right) \sqrt{4+x^{2}}}=\int_{0}^{2} \frac{d x}{\left(4+x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=2 \operatorname{tg} t \Rightarrow d x=\frac{2 d t}{\cos ^{2} t} \\
& x=0 \Rightarrow t=\operatorname{arctg} \frac{0}{2}=0 \\
& x=2 \Rightarrow t=\... | \frac{\sqrt{2}}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,829 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\sqrt{2}} \frac{x^{4} \cdot d x}{\left(4-x^{2}\right)^{3 / 2}}
$$ | ## Solution
$$
\int_{0}^{\sqrt{2}} \frac{x^{4} \cdot d x}{\left(4-x^{2}\right)^{3 / 2}}=
$$
Substitution:
$$
\begin{aligned}
& x=2 \sin t \Rightarrow d x=2 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{2}=0 \\
& x=\sqrt{2} \Rightarrow t=\arcsin \frac{\sqrt{2}}{2}=\frac{\pi}{4}
\end{aligned}
$$
We get:
$$
\beg... | 5-\frac{3\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,830 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{1 / \sqrt{2}} \frac{d x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=\sin t ; d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=\frac{1}{\sqrt{2}} \Rightarrow t=\arcsin \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,831 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} \frac{x^{2} \cdot d x}{\sqrt{4-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{1} \frac{x^{2} \cdot d x}{\sqrt{4-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=2 \sin t \Rightarrow d x=2 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{2}=0 \\
& x=1 \Rightarrow t=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 6} \... | \frac{\pi}{3}-\frac{\sqrt{3}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,832 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{3 / 2} \frac{x^{2} \cdot d x}{\sqrt{9-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{3 / 2} \frac{x^{2} \cdot d x}{\sqrt{9-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=3 \sin t \Rightarrow d x=3 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{3}=0 \\
& x=\frac{3}{2} \Rightarrow t=\arcsin \frac{\left(\frac{3}{2}\right)}{3}=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned... | \frac{3\pi}{4}-\frac{9\sqrt{3}}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,833 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{n}{3 n-1}, a=\frac{1}{3}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left.\frac{1}{3(3 n-1)} \right\rvert\,
\end{aligned}
$$
$\frac{1}{3(3 n-1)}$
$3 n-1>\frac{1}{3 \varepsilon} ;=>$
$n>\frac{1}{3}\left(\frac{1}{3 \var... | N(\varepsilon)=[\frac{1+12\varepsilon}{9\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 46,834 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{1+3+5+\ldots+(2 n-1)}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{1+3+5+\ldots+(2 n-1)}=\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{\frac{(1+(2 n-1)) n}{2}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{\frac{2 n \cdot n}{... | -\sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,838 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow-\frac{7}{2}} \frac{2 x^{2}+13 x+21}{2 x+7}=-\frac{1}{2}
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,840 |
Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=4 x^{2}+4, x_{0}=9$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\righ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,841 |
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