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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{(n+4)!-(n+2)!}{(n+3)!}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+4)!-(n+2)!}{(n+3)!}=\lim _{n \rightarrow \infty}\left(\frac{(n+4)!}{(n+3)!}-\frac{(n+2)!}{(n+3)!}\right)= \\
& =\lim _{n \rightarrow \infty}\left((n+4)-\frac{1}{n+3}\right)=\lim _{n \rightarrow \infty}\left(\frac{(n+4)(n+3)}{n+3}-\frac{1}{n+3}\righ... | \infty | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,104 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow-\frac{1}{3}} \frac{3 x^{2}-2 x-1}{x+\frac{1}{3}}=-4$ | ## 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 \( M \). The number
=\frac{\varepsilon}{3} | Calculus | proof | Yes | Yes | olympiads | false | 46,106 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{\text {): }}$
$$
f(x)=-5 x^{2}-9, x_{0}=3
$$ | ## Solution
By definition, a function $f(x)_{\text {is continuous at the point }} x=x_{0, \text { if }} \forall \varepsilon>0: \exists \delta(\varepsilon)>0$ : $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\right|... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,107 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}-x-2}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-3 x-2}{x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{\left(x^{2}-x-2\right)(x+1)}{x^{2}-x-2}= \\
& =\lim _{x \rightarrow-1}(x+1)=-1+1=0
\end{aligned}
$$
## Problem Kuznetsov Limits 10-9 | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,108 |
## Problem Statement
$\lim _{x \rightarrow 0} \frac{\sqrt[3]{8+3 x+x^{2}}-2}{x+x^{2}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{\sqrt[3]{8+3 x+x^{2}}-2}{x+x^{2}}=\left\{\frac{0}{0}\right\}=$
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt[3]{8+3 x+x^{2}}-2\right)\left(\sqrt[3]{\left(8+3 x+x^{2}\right)^{2}}+2 \sqrt[3]{8+3 x+x^{2}}+4\right)}{x(1+x)\left(\sqrt[3]{\left(8+3 x+x^{2}\right)... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,109 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow u} \frac{2^{x}-1}{\ln (1+2 x)}
$$ | ## Solution
We will use the substitution of equivalent infinitesimals:
\(\ln (1+2 x) \sim 2 x\), as \(x \rightarrow 0\) (i.e., \(2 x \rightarrow 0\))
\(\varepsilon^{x^{2} \ln 2}-1 \sim x \ln 2\), as \(x \rightarrow 0\) (i.e., \(x \ln 2 \rightarrow 0\))
We obtain:
\[
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{... | \frac{\ln2}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,110 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\cos 5 x-\cos 3 x}{\sin ^{2} x}$ | ## Solution
$\lim _{x \rightarrow \pi} \frac{\cos 5 x-\cos 3 x}{\sin ^{2} x}=\lim _{x \rightarrow \pi} \frac{-2 \sin \frac{5 x+3 x}{2} \sin \frac{5 x-3 x}{2}}{\sin ^{2} x}=$
$=\lim _{x \rightarrow \pi} \frac{-2 \sin 4 x \sin x}{\sin ^{2} x}=\lim _{x \rightarrow \pi} \frac{-2 \sin 4 x}{\sin x}=$
Substitution:
$x=y+\... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,111 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{1}{2}} \frac{\ln (4 x-1)}{\sqrt{1-\cos \pi x}-1}$ | ## Solution
Substitution:
$x=y+\frac{1}{2} \Rightarrow y=x-\frac{1}{2}$
$x \rightarrow \frac{1}{2} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{1}{2}} \frac{\ln (4 x-1)}{\sqrt{1-\cos \pi x}-1}=\lim _{y \rightarrow 0} \frac{\ln \left(4\left(y+\frac{1}{2}\right)-1\right)}{\sqr... | \frac{8}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,112 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{12^{x}-5^{-3 x}}{2 \arcsin x-x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{12^{x}-5^{-3 x}}{2 \arcsin x-x}=\lim _{x \rightarrow 0} \frac{\left(12^{x}-1\right)-\left(125^{-x}-1\right)}{2 \arcsin x-x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 12}\right)^{x}-1\right)-\left(\left(e^{\ln 125}\right)^{-x}-1\right)}{2 \arc... | \ln(12\cdot5^3) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,113 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{3}} \frac{1-2 \cos x}{\sin (\pi-3 x)}$ | ## Solution
Substitution:
$x=y+\frac{\pi}{3} \Rightarrow y=x-\frac{\pi}{3}$
$x \rightarrow \frac{\pi}{3} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{3}} \frac{1-2 \cos x}{\sin (\pi-3 x)}=\lim _{y \rightarrow 0} \frac{1-2 \cos \left(y+\frac{\pi}{3}\right)}{\sin \left(\p... | -\frac{\sqrt{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,114 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0}(\cos \pi x)^{\frac{1}{x \cdot \sin \pi x}}$ | Solution
$\lim _{x \rightarrow 0}(\cos \pi x)^{\frac{1}{x \cdot \sin \pi x}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln (\cos \pi x)}\right)^{\frac{1}{x \cdot \sin \pi x}}=$
$=\lim _{x \rightarrow 0} e^{\ln (\cos \pi x) \cdot \frac{1}{x \cdot \sin \pi x}}=$
$=\exp \left\{\lim _{x \rightarrow 0} \ln (\cos \pi x) \cdot ... | e^{-\frac{\pi}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,115 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2 \pi}(\cos x)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}$ | ## Solution
$\lim _{x \rightarrow 2 \pi}(\cos x)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}=\lim _{x \rightarrow 2 \pi}\left(e^{\ln (\cos x)}\right)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}=$
$=\lim _{x \rightarrow 2 \pi} e^{\frac{\operatorname{ctg} 2 x}{\sin 3 x} \cdot \ln (\cos x)}=\exp \left\{\lim _{x \rightarrow ... | e^{-\frac{1}{12}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,117 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(1+e^{x}\right)^{\frac{\sin \pi x}{1-x}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(1+e^{x}\right)^{\frac{\sin \pi x}{1-x}}=\left(\lim _{x \rightarrow 1} 1+e^{x}\right)^{\lim _{x \rightarrow 1} \frac{\sin \pi x}{1-x}}= \\
& =\left(1+e^{1}\right)^{\lim _{x \rightarrow 1} \frac{\sin \pi x}{1-x}}=
\end{aligned}
$$
$x=y+1 \Rightarrow y=x-1$
... | (1+e)^{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,118 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n^{2}-\sqrt{3 n^{5}-7}}{\left(n^{2}-n \cos n+1\right) \sqrt{n}}$ | ## Solution
$\lim _{n \rightarrow x} \frac{n^{2}-\sqrt{3 n n^{5}-\overline{7}}}{\left(n^{2}-n \cos n+1\right) \sqrt{n}}=\lim _{n \rightarrow x} \frac{\frac{1}{n^{2} \sqrt{n}}\left(n^{2}-\sqrt{3 n l^{5}-\bar{i}}\right)}{\frac{1}{n^{2} \sqrt{n}}\left(n^{2}-n \cos n+1\right) \sqrt{n}}=$
$=\lim _{n \rightarrow x} \frac{\... | -\sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,119 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\operatorname{arctg}\left(\frac{3 x}{2}-x^{2} \sin \frac{1}{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... | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,120 |
## Problem Statement
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{1+3 x^{2}}{3+x^{2}}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=\left(\frac{1+3 x^{2}}{3+x^{2}}\right)^{\prime}=\frac{\left(1+3 x^{2}\right)^{\prime}\left(3+x^{2}\right)-\left(1+3 x^{2}\right)\left(3+x^{2}\right)^{\prime}}{\left(3+x^{2}\right)^{2}}=$
$=\frac{6 x\left(3+x^{2}\right)-\left(1+3 x^{2}\right) \cdot 2 x}{\left(3+x^{2}\r... | x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,121 |
## Task Condition
Find the differential $d y$.
$$
y=x(\sin (\ln x)-\cos (\ln x))
$$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=(x(\sin (\ln x)-\cos (\ln x)))^{\prime} d x= \\
& =\left((\sin (\ln x)-\cos (\ln x))+x(\sin (\ln x)-\cos (\ln x))^{\prime}\right) d x= \\
& =\left(\sin (\ln x)-\cos (\ln x)+x\left((\sin (\ln x))^{\prime}-(\cos (\ln x))^{\prime}\right)\right) d x= \\
& =\left(\s... | 2\sin(\lnx)\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,122 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt[5]{x^{2}}, x=1.03$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$$
f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x
$$
Choose:
$x_{0} = 1$
Then:
$\Delta x = 0.03$
Calculate:
$y(1) = \sqrt[5]{1^{2}} = 1$
$$
\... | 1.012 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,123 |
## Task Condition
Find the derivative.
$$
y=3 \cdot \sqrt[3]{\frac{x+1}{(x-1)^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(3 \cdot \sqrt[3]{\frac{x+1}{(x-1)^{2}}}\right)^{\prime}=3\left(\left(\frac{x+1}{(x-1)^{2}}\right)^{\frac{1}{3}}\right)^{\prime}=3 \cdot \frac{1}{3} \cdot\left(\frac{x+1}{(x-1)^{2}}\right)^{-\frac{2}{3}} \cdot\left(\frac{x+1}{(x-1)^{2}}\right)^{\prime}= \\
& =\sqrt[3]{\... | -\sqrt[3]{\frac{x-1}{(x+1)^{2}}}\cdot\frac{x+3}{(x-1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,124 |
## Task Condition
Find the derivative.
$$
y=\frac{e^{x}}{2}\left(\left(x^{2}-1\right) \cos x+(x-1)^{2} \sin x\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{e^{x}}{2}\left(\left(x^{2}-1\right) \cos x+(x-1)^{2} \sin x\right)\right)^{\prime}= \\
& =\left(\frac{e^{x}}{2}\right)^{\prime} \cdot\left(\left(x^{2}-1\right) \cos x+(x-1)^{2} \sin x\right)+\left(\frac{e^{x}}{2}\right) \cdot\left(\left(x^{2}-1\right) \cos x+(x-1... | x^{2}\cdote^{x}\cdot\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,125 |
## Problem Statement
Find the derivative.
$$
y=\ln \left(\arccos \frac{1}{\sqrt{x}}\right)
$$ | ## Solution
$$
\begin{aligned}
& y'=\left(\ln \left(\arccos \frac{1}{\sqrt{x}}\right)\right)'=\frac{1}{\arccos \frac{1}{\sqrt{x}}} \cdot\left(\arccos \frac{1}{\sqrt{x}}\right)'= \\
& =\frac{1}{\arccos \frac{1}{\sqrt{x}}} \cdot\left(-\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x}}\right)^{2}}}\right) \cdot\left(\frac{1}{\sqr... | \frac{1}{2x\cdot\sqrt{x-1}\cdot\arccos\frac{1}{\sqrt{x}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,126 |
## Task Condition
Find the derivative.
$y=\sin (\ln 2)+\frac{\sin ^{2} 25 x}{25 \cos 50 x}$ | $$
\begin{aligned}
& y^{\prime}=\left(\sin (\ln 2)+\frac{\sin ^{2} 25 x}{25 \cos 50 x}\right)^{\prime}=0+\left(\frac{\sin ^{2} 25 x}{25 \cos 50 x}\right)^{\prime}= \\
& =\frac{1}{25}\left(\frac{\sin ^{2} 25 x}{\cos 50 x}\right)^{\prime}=\frac{1}{25}\left(\frac{\left(\sin ^{2} 25 x\right)^{\prime} \cdot \cos 50 x-\sin ^... | \frac{\sin50x}{\cos^{2}50x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,127 |
## Task Condition
Find the derivative.
$$
y=\operatorname{arctg} \frac{\sqrt{1-x}}{1-\sqrt{x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{\sqrt{1-x}}{1-\sqrt{x}}\right)^{\prime}=\frac{1}{1+\left(\frac{\sqrt{1-x}}{1-\sqrt{x}}\right)^{2}} \cdot\left(\frac{\sqrt{1-x}}{1-\sqrt{x}}\right)^{\prime}= \\
& =\frac{(1-\sqrt{x})^{2}}{(1-\sqrt{x})^{2}+1-x} \cdot\left(\frac{(\sqrt{1-x})^{\p... | \frac{1}{4\sqrt{x(1-x)}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,128 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{2} \cdot \operatorname{arctan}(\operatorname{sinh} x)-\frac{\operatorname{sinh} x}{2 \operatorname{cosh}^{2} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot \operatorname{arctan}(\operatorname{sinh} x)-\frac{\operatorname{sinh} x}{2 \operatorname{cosh}^{2} x}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \frac{1}{1+\operatorname{sinh}^{2} x} \cdot(\operatorname{sinh} x)^{\prime}-\frac{1}{2} \cdot \frac{(\opera... | \frac{\operatorname{sinh}^{2}x}{\operatorname{cosh}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,129 |
## Task Condition
Find the derivative.
$y=x^{e^{\sin x}}$ | ## Solution
$y=x^{e^{\sin x}}$
$\ln y=e^{\sin x} \cdot \ln x$
$\frac{y^{\prime}}{y}=\left(e^{\sin x} \cdot \ln x\right)^{\prime}=\left(e^{\sin x}\right)^{\prime} \cdot \ln x+e^{\sin x} \cdot(\ln x)^{\prime}=$
$=e^{\sin x} \cdot(\sin x)^{\prime} \cdot \ln x+e^{\sin x} \cdot \frac{1}{x}=e^{\sin x} \cdot\left(\cos x \... | x^{e^{\sinx}}\cdote^{\sinx}\cdot(\cosx\cdot\lnx+\frac{1}{x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,130 |
## Problem Statement
Find the derivative.
$$
y=\frac{2}{3 x-2} \sqrt{-3+12 x-9 x^{2}}+\ln \frac{1+\sqrt{-3+12 x-9 x^{2}}}{3 x-2}
$$ | ## Solution
Let $z(x)=\sqrt{-3+12 x-9 x^{2}}$. We obtain:
$$
\begin{aligned}
& z'=\frac{1}{2 \sqrt{-3+12 x-9 x^{2}}} \cdot\left(-3+12 x-9 x^{2}\right)'= \\
& =\frac{12-18 x}{2 \sqrt{-3+12 x-9 x^{2}}}=\frac{-3(3 x-2)}{\sqrt{-3+12 x-9 x^{2}}}=\frac{-3(3 x-2)}{z}
\end{aligned}
$$
Then:
$$
\begin{aligned}
& y'=\left(\f... | \frac{3-9x}{\sqrt{-3+12x-9x^{2}}(3x-2)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,131 |
## Task Condition
Find the derivative.
$y=\frac{\sqrt{1-x^{2}}}{x}+\arcsin x$ | ## Solution
$y^{\prime}=\left(\frac{\sqrt{1-x^{2}}}{x}+\arcsin x\right)^{\prime}=$
$=\frac{\left(\sqrt{1-x^{2}}\right)^{\prime} \cdot x-\sqrt{1-x^{2}} \cdot x^{\prime}}{x^{2}}+\frac{1}{\sqrt{1-x^{2}}}=$
$=\frac{\frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2 x) \cdot x-\sqrt{1-x^{2}}}{x^{2}}+\frac{1}{\sqrt{1-x^{2}}}=$
$=-\frac... | -\frac{\sqrt{1-x^{2}}}{x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,132 |
## Task Condition
Find the derivative.
$$
y=\frac{2^{x}(\sin x+\cos x \cdot \ln 2)}{1+\ln ^{2} 2}
$$ | ## Solution
$y^{\prime}=\left(\frac{2^{x}(\sin x+\cos x \cdot \ln 2)}{1+\ln ^{2} 2}\right)^{\prime}=\frac{1}{1+\ln ^{2} 2}\left(2^{x}(\sin x+\cos x \cdot \ln 2)\right)^{\prime}=$ $=\frac{1}{1+\ln ^{2} 2}\left(\left(2^{x}\right)^{\prime} \cdot(\sin x+\cos x \cdot \ln 2)+2^{x} \cdot(\sin x+\cos x \cdot \ln 2)^{\prime}\r... | 2^{x}\cdot\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,133 |
## Condition of the problem
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\ln \sqrt{\frac{1-\sin t}{1+\sin t}} \\
y=\frac{1}{2} \operatorname{tg}^{2} t+\ln \cos t
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\ln \sqrt{\frac{1-\sin t}{1+\sin t}}\right)^{\prime}=\sqrt{\frac{1+\sin t}{1-\sin t}} \cdot\left(\sqrt{\frac{1-\sin t}{1+\sin t}}\right)^{\prime}= \\
& =\sqrt{\frac{1+\sin t}{1-\sin t}} \cdot \frac{1}{2 \sqrt{\frac{1-\sin t}{1+\sin t}}} \cdot\left(\frac{1-\sin t}{1... | \frac{\sin\cdot\cos-1}{\cos} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,134 |
Condition of the problem
To find the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
$\left\{\begin{array}{l}x=t^{3}+1 \\ y=t^{2}+t+1\end{array}\right.$
$t_{0}=1$ | ## Solution
Since $t_{0}=1$, then
$x_{0}=1^{3}+1=2$
$y_{0}=1^{2}+1+1=3$
Let's find the derivatives:
$x_{t}^{\prime}=\left(t^{3}+1\right)^{\prime}=3 t^{2}$
$y_{t}^{\prime}=\left(t^{2}+t+1\right)^{\prime}=2 t+1$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 t+1}{3 t^{2}}$
Then:
$y_{0}^{\prime}=\f... | x+1-x+5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,135 |
## Task Condition
Find the $n$-th order derivative.
$y=\lg (2 x+7)$ | ## Solution
$y^{\prime}=(\lg (2 x+7))^{\prime}=\frac{2}{(2 x+7) \ln 10}=\frac{2}{\ln 10} \cdot(2 x+7)^{-1}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{2}{\ln 10} \cdot(2 x+7)^{-1}\right)^{\prime}=-\frac{2^{2}}{\ln 10} \cdot(2 x+7)^{-2}$
$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\... | y^{(n)}=(-1)^{n-1}\cdot\frac{2^{n}\cdot(n-1)!}{\ln10}\cdot(2x+7)^{-n} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,136 |
## Task Condition
Find the derivative of the specified order.
$y=\left(x^{2}+3 x+1\right) e^{3 x+2}, y^{V}=?$ | ## Solution
$y^{\prime}=\left(\left(x^{2}+3 x+1\right) e^{3 x+2}\right)^{\prime}=(2 x+3) e^{3 x+2}+3\left(x^{2}+3 x+1\right) e^{3 x+2}=$
$=\left(3 x^{2}+11 x+6\right) e^{3 x+2}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\left(3 x^{2}+11 x+6\right) e^{3 x+2}\right)^{\prime}=$
$=(6 x+11) e^{3 x+2}+3\l... | 3^{3}\cdot(9x^{2}+57x+74)e^{3x+2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,137 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\operatorname{ch} t \\
y=\sqrt[3]{\operatorname{sh}^{2} t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\operatorname{ch} t)^{\prime}=\operatorname{sh} t$
$y_{t}^{\prime}=\left(\sqrt[3]{\operatorname{sh}^{2} t}\right)^{\prime}=\left((\operatorname{sh} t)^{\frac{2}{3}}\right)^{\prime}=\frac{2}{3} \cdot(\operatorname{sh} t)^{-\frac{1}{3}} \cdot \operatorname{ch} t=\frac{2 \operatorname{ch} t}... | -\frac{2(3+\operatorname{ch}^{2})}{9\operatorname{sh}^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,138 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
$y=-x \cdot \cos x+3 x$
$x \cdot y^{\prime}=y+x^{2} \sin x .(1)$ | ## Solution
$y^{\prime}=(-x \cdot \cos x+3 x)^{\prime}=-\cos x+x \cdot \sin x+3$
Substitute into equation (1):
$x \cdot(-\cos x+x \cdot \sin x+3)=-x \cdot \cos x+3 x+x^{2} \sin x$ Simplify:
$-x \cdot \cos x+x^{2} \sin x+3 x=-x \cdot \cos x+3 x+x^{2} \sin x$ $0=0$
The equality holds. The function $y_{\text{satisfie... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,139 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{3 n^{2}+2}{4 n^{2}-1}, a=\frac{3}{4}$ | ## 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{12 n^{2}+8-12 n^{2}+3}{4\left(4 n^{2}-1\right)}\right| \\
& \left.\frac{11}{4\left(4 n^{2}-1\right)} \right\rvert\, \\
& \frac{11}{4\left(4 ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,140 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{3}+(n-1)^{3}}{n^{3}+1}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left((n+1)^{3}+(n-1)^{3}\right)}{\frac{1}{n^{3}}\left(n^{3}+1\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,141 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{n^{2}-\sqrt{n^{3}+1}}{\sqrt[3]{n^{6}+2}-n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n^{2}-\sqrt{n^{3}+1}}{\sqrt[3]{n^{6}+2}-n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n^{2}-\sqrt{n^{3}+1}\right)}{\frac{1}{n^{2}}\left(\sqrt[3]{n^{6}+2}-n\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{1-\sqrt{\frac{1}{n}+\frac{1}{n^{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,142 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)
$$ | ## Solution
$\lim _{n \rightarrow \infty} n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)=$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{n\left(\sqrt{n^{4}+3}-\sqrt{n^{4}-2}\right)\left(\sqrt{n^{4}+3}+\sqrt{n^{4}-2}\right)}{\sqrt{n^{4}+3}+\sqrt{n^{4}-2}}= \\
& =\lim _{n \rightarrow \infty} \frac{n\left(n^{4}+... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,143 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{3+6+9+\ldots+3 n}{n^{2}+4}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{3+6+9+\ldots+3 n}{n^{2}+4}=\lim _{n \rightarrow \infty} \frac{\left(\frac{(3+3 n) n}{2}\right)}{n^{2}+4}= \\
& =\lim _{n \rightarrow \infty} \frac{(3+3 n) n}{2\left(n^{2}+4\right)}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}(3+3 n) n}{\frac{1}{... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,144 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2 n+3}{2 n^{2}+2 n+1}\right)^{3 n^{2}-7}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2 n+3}{2 n^{2}+2 n+1}\right)^{3 n^{2}-7}=\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2 n+1+2}{2 n^{2}+2 n+1}\right)^{3 n^{2}-7}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{2}{2 n^{2}+2 n+1}\right)^{3 n^{2}-7}=
\end{aligned}
$$
$... | e^3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,145 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$$
\lim _{x \rightarrow \frac{1}{3}} \frac{3 x^{2}+17 x-6}{x-\frac{1}{3}}=19
$$ | ## 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 \( a \in M^{\prime} \) is a limit point of the set \( M \). The number
_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{)}$:
$f(x)=3 x^{2}+7, x_{0}=6$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\righ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,147 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{x^{2}-1}{2 x^{2}-x-1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{x^{2}-1}{2 x^{2}-x-1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{(x-1)(2 x+1)}= \\
& =\lim _{x \rightarrow 1} \frac{x+1}{2 x+1}=\frac{1+1}{2 \cdot 1+1}=\frac{2}{3}
\end{aligned}
$$
## Problem Kuznetsov Limits 10-29 | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,148 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{x^{2}-16}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{x^{2}-16}}=\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(x-4)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)}}= \\
& =\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{\sqrt[3]{(\sqrt{x}-2)... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,149 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\tan \left(\pi\left(1+\frac{x}{2}\right)\right)}{\ln (x+1)}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
\(\ln (1+x) \sim x\), as \(x \rightarrow 0\)
\(\operatorname{tg} \frac{\pi x}{2} \sim \frac{\pi x}{2}\), as \(\frac{\pi x}{2} \rightarrow 0\)
We get:
\[
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\operatorname{tg}\left(\pi\left(1+\frac{x}{... | \frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,150 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{3-\sqrt{10-x}}{\sin 3 \pi x}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{3-\sqrt{10-x}}{\sin 3 \pi x}=\lim _{y \rightarrow 0} \frac{3-\sqrt{10-(y+1)}}{\sin 3 \pi(y+1)}= \\
& =\lim _{y \rightarr... | -\frac{1}{18\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,151 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow a \pi} \frac{\ln \left(\cos \left(\frac{x}{a}\right)+2\right)}{a^{a^{2} \pi^{2} / x^{2}-a \pi / x}-a^{a \pi / x-1}}$ | ## Solution
Substitution:
$x=y+a \pi \Rightarrow y=x-a \pi$
$x \rightarrow a \pi \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow a \pi} \frac{\ln \left(\cos \left(\frac{x}{a}\right)+2\right)}{a^{a^{2} \pi^{2} / x^{2}-a \pi / x}-a^{a \pi / x-1}}= \\
& =\lim _{y \rightarrow 0} \frac{\l... | \frac{\pi^{2}}{2\ln} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,152 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{x+\tan x^{2}}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{x}}{x+\tan x^{2}}=\lim _{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)-\left(e^{x}-1\right)}{x+\tan x^{2}}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{2 x}-1\right)-\left(e^{x}-1\right)\right)}{\frac{1}{x}\left(x+\tan x^{2}\right)}=$
$=\frac{\lim _{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,153 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin ^{3} x}{\cos ^{2} x}$ | ## Solution
Substitution:
$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin ^{3} x}{\cos ^{2} x}=\lim _{y \rightarrow 0} \frac{1-\sin ^{3}\left(y+\frac{\pi}{2}\right)}{\cos ^{2}\l... | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,154 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}(1-\ln (\cos x))^{\frac{1}{\operatorname{tg}^{2} x}}
$$ | ## Solution
$\lim _{x \rightarrow 0}(1-\ln (\cos x))^{\frac{1}{\operatorname{tg}^{2} x}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln (1-\ln (\cos x))}\right)^{\frac{1}{\operatorname{tg}^{2} x}}=$
$=\lim _{x \rightarrow 0} e^{\ln (1-\ln (\cos x)) / \operatorname{tg}^{2} x}=$
$=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln ... | e^{\frac{1}{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,155 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\frac{1+8 x}{2+11 x}\right)^{\frac{1}{x^{2}+1}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{1+8 x}{2+11 x}\right)^{\frac{1}{x^{2}+1}}=\left(\frac{1+8 \cdot 0}{2+11 \cdot 0}\right)^{\frac{1}{0^{2}+1}}=\left(\frac{1}{2}\right)^{1}=\frac{1}{2}$
Problem Kuznetsov Limits 18-29 | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,156 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{1}{x}\right)^{\frac{\ln (x+1)}{\ln (2-x)}}$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{1}{x}\right)^{\frac{\ln (x+1)}{\ln (2-x)}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{1}{x}\right)}\right)^{\frac{\ln (x+1)}{\ln (2-x)}}=$
$=\lim _{x \rightarrow 1} e^{\frac{\ln (x+1)}{\ln (2-x)} \cdot \ln \left(\frac{1}{x}\right)}=\exp \left\{\lim _{x \rightarrow 1... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,157 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\frac{x^{2}+2 x-3}{x^{2}+4 x-5}\right)^{\frac{1}{2-x}}
$$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{x^{2}+2 x-3}{x^{2}+4 x-5}\right)^{\frac{1}{2-x}}=\lim _{x \rightarrow 1}\left(\frac{(x-1)(x+3)}{(x-1)(x+5)}\right)^{\frac{1}{2-x}}=$
$=\lim _{x \rightarrow 1}\left(\frac{x+3}{x+5}\right)^{\frac{1}{2-x}}=\left(\frac{1+3}{1+5}\right)^{\frac{1}{2-1}}=\left(\frac{4}{6}\righ... | \frac{2}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,158 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \sqrt{x\left(2+\sin \left(\frac{1}{x}\right)\right)+4 \cos x}$ | ## Solution
Since $2+\sin \left(\frac{1}{x}\right)_{\text { is bounded, then }}$
$x\left(2+\sin \left(\frac{1}{x}\right)\right) \rightarrow 0 \quad$, as $x \rightarrow 0$
Then:
$\lim _{x \rightarrow 0} \sqrt{x\left(2+\sin \left(\frac{1}{x}\right)\right)+4 \cos x}=\sqrt{0+4 \cos 0}=\sqrt{4 \cdot 1}=2$ | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,159 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
1-\cos \left(x \sin \frac{1}{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... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,160 |
## Problem Statement
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=6 \sqrt[3]{x}-\frac{16 \sqrt[4]{x}}{3}, x_{0}=1$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(6 \sqrt[3]{x}-\frac{16 \sqrt[4]{x}}{3}\right)^{\prime}=\left(6 \cdot x^{\frac{1}{3}}-\frac{16}{3} \cdot x^{\frac{1}{4}}\right)^{\prime}= \\
& =6 \cdot \frac{1}{3} \cdot x^{-\frac{2}{3}}-\frac{16}{3} \cdot \frac{1}{4} x^{-\frac{3}{4}}=2 \cdot x... | \frac{2}{3}\cdotx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,161 |
## Problem Statement
Find the differential $d y$.
$y=x \sqrt{x^{2}-1}+\ln \left|x+\sqrt{x^{2}-1}\right|$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(x \sqrt{x^{2}-1}+\ln \left|x+\sqrt{x^{2}-1}\right|\right)^{\prime} \cdot d x= \\
& =\left(\sqrt{x^{2}-1}+x \cdot \frac{1}{2 \sqrt{x^{2}-1}} \cdot 2 x+\frac{1}{x+\sqrt{x^{2}-1}} \cdot\left(1+\frac{1}{2 \sqrt{x^{2}-1}} \cdot 2 x\right)\right) \cdot d x= \\
... | \frac{2x^{2}\cdot}{\sqrt{x^{2}-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,162 |
## Task Condition
Approximately calculate using the differential.
$y=\frac{1}{\sqrt{2 x+1}}, x=1.58$ | ## Solution
If the increment $\Delta x = x - x_{0 \text{ of the argument }} x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 1.5$
Then:
$\Delta x = 0.08$
Calculate:
$y(1.5) = \frac{1}{\sqrt{2 \c... | 0.49 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,163 |
## Task Condition
Find the derivative.
$y=\frac{3 x^{6}+4 x^{4}-x^{2}-2}{15 \sqrt{1+x^{2}}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{3 x^{6}+4 x^{4}-x^{2}-2}{15 \sqrt{1+x^{2}}}\right)^{\prime}= \\
& =\frac{\left(3 x^{6}+4 x^{4}-x^{2}-2\right)^{\prime} \cdot \sqrt{1+x^{2}}-\left(3 x^{6}+4 x^{4}-x^{2}-2\right) \cdot\left(\sqrt{1+x^{2}}\right)^{\prime}}{15\left(1+x^{2}\right)}= \\
& =\frac{\left(... | \frac{x^{3}(x^{2}+1)^{2}}{\sqrt{1+x^{2}}(1+x^{2})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,164 |
## Task Condition
Find the derivative.
$y=\frac{e^{x^{2}}}{1+x^{2}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{e^{x^{2}}}{1+x^{2}}\right)^{\prime}=\frac{\left(e^{x^{2}}\right)^{\prime}\left(1+x^{2}\right)-e^{x^{2}}\left(1+x^{2}\right)^{\prime}}{\left(1+x^{2}\right)^{2}}= \\
& =\frac{e^{x^{2}} \cdot 2 x \cdot\left(1+x^{2}\right)-e^{x^{2}} \cdot 2 x}{\left(1+x^{2}\right)^{2... | \frac{2x^{3}\cdote^{x^{2}}}{(1+x^{2})^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,165 |
Condition of the problem
Find the derivative.
$y=\ln \ln ^{2} \ln ^{3} x$ | ## Solution
$y^{\prime}=\left(\ln \ln ^{2} \ln ^{3} x\right)^{\prime}=\frac{1}{\ln ^{2} \ln ^{3} x} \cdot 2 \ln ^{3} \ln ^{3} x \cdot \frac{1}{\ln ^{3} x} \cdot 3 \ln ^{2} x \cdot \frac{1}{x}=$ $=\frac{2}{\ln ^{3} \ln ^{3} x} \cdot \frac{3}{\ln x} \cdot \frac{1}{x}=\frac{6}{x \cdot \ln x \cdot \ln \ln ^{3} x}$
Proble... | \frac{6}{x\cdot\lnx\cdot\ln\ln^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,166 |
## Task Condition
Find the derivative.
$y=\operatorname{tg} \sqrt{\cos \left(\frac{1}{3}\right)}+\frac{\sin ^{2} 31 x}{31 \cos 62 x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{tg} \sqrt{\cos \left(\frac{1}{3}\right)}+\frac{\sin ^{2} 31 x}{31 \cos 62 x}\right)^{\prime}=\left(\frac{\sin ^{2} 31 x}{31 \cos 62 x}\right)^{\prime}= \\
& =\left(\frac{2 \sin ^{2} 31 x}{62 \cos 62 x}\right)^{\prime}=\left(\frac{1-\cos 62 x}{62 \cos 62 x... | \frac{\operatorname{tg}62x}{\cos62x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,167 |
## Task Condition
Find the derivative.
$y=\operatorname{arctg} \frac{\operatorname{tg} \frac{x}{2}+1}{2}$ | ## Solution
$y^{\prime}=\left(\operatorname{arctg} \frac{\operatorname{tg} \frac{x}{2}+1}{2}\right)^{\prime}=\frac{1}{1+\left(\frac{\operatorname{tg}(x / 2)+1}{2}\right)^{2}} \cdot\left(\frac{\operatorname{tg} \frac{x}{2}+1}{2}\right)^{\prime}=$
$=\frac{4}{\operatorname{tg}^{2} \frac{x}{2}+2 \operatorname{tg} \frac{x}... | \frac{1}{\sinx+2\cosx+3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,168 |
## Task Condition
Find the derivative.
$y=\frac{2}{3} \cdot \operatorname{cth} x-\frac{\operatorname{ch} x}{3 \operatorname{sh}^{3} x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2}{3} \cdot \operatorname{cth} x-\frac{\operatorname{ch} x}{3 \operatorname{sh}^{3} x}\right)^{\prime}=-\frac{2}{3} \cdot \frac{1}{\operatorname{sh}^{2} x}-\frac{(\operatorname{ch} x)^{\prime} \cdot \operatorname{sh}^{3} x-\operatorname{ch} x \cdot\left(\operator... | \frac{1}{\operatorname{sh}^{4}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,169 |
## Task Condition
Find the derivative.
$y=x^{e^{x}} \cdot x^{9}$ | ## Solution
$y=x^{e^{x}} \cdot x^{9}$
$\ln y=\ln \left(x^{e^{x}} \cdot x^{9}\right)=e^{x} \cdot \ln x+9 \cdot \ln x=\ln x \cdot\left(e^{x}+9\right)$
$$
\begin{aligned}
& \frac{y^{\prime}}{y}=\left(\ln x \cdot\left(e^{x}+9\right)\right)^{\prime}=\frac{1}{x} \cdot\left(e^{x}+9\right)+\ln x \cdot e^{x}=e^{x} \cdot\left... | y^{\}=x^{e^{x}}\cdotx^{9}\cdot(e^{x}\cdot(\lnx+\frac{1}{x})+\frac{9}{x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,170 |
## Problem Statement
Find the derivative.
$y=\arcsin \left(e^{-2 x}\right)+\ln \left(e^{2 x}+\sqrt{e^{4 x}-1}\right)$ | ## Solution
$y^{\prime}=\left(\arcsin \left(e^{-2 x}\right)+\ln \left(e^{2 x}+\sqrt{e^{4 x}-1}\right)\right)^{\prime}=$
$=\frac{1}{\sqrt{1-\left(e^{-2 x}\right)^{2}}}+\frac{1}{e^{2 x}+\sqrt{e^{4 x}-1}} \cdot\left(e^{2 x} \cdot 2+\frac{1}{2 \sqrt{e^{4 x}-1}} \cdot e^{4 x} \cdot 4\right)=$
$=\frac{1}{\sqrt{1-e^{-4 x}}... | \frac{e^{2x}}{\sqrt{e^{4x}-1}}+\frac{2e^{2x}}{e^{2x}+\sqrt{e^{4x}-1}}\cdot\frac{\sqrt{e^{4x}-1}+e^{2x}}{\sqrt{e^{4x}-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,171 |
## Problem Statement
Find the derivative.
$y=\frac{\arcsin x}{\sqrt{1-x^{2}}}+\frac{1}{2} \ln \frac{1-x}{1+x}$ | ## Solution
$y^{\prime}=\left(\frac{\arcsin x}{\sqrt{1-x^{2}}}+\frac{1}{2} \ln \frac{1-x}{1+x}\right)^{\prime}=$
$=\frac{(\arcsin x)^{\prime} \cdot \sqrt{1-x^{2}}-\arcsin x \cdot\left(\sqrt{1-x^{2}}\right)^{\prime}}{1-x^{2}}+\frac{1}{2} \cdot \frac{1+x}{1-x} \cdot \frac{-1 \cdot(1+x)-(1-x) \cdot 1}{(1+x)^{2}}=$
$=\f... | -\frac{x\cdot\arcsinx}{(1-x^{2})\sqrt{1-x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,172 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\ln \left(t+\sqrt{1+t^{2}}\right) \\
y=\sqrt{1+t^{2}}-\ln \frac{1+\sqrt{1+t^{2}}}{t}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\ln \left(t+\sqrt{1+t^{2}}\right)\right)^{\prime}=\frac{1}{t+\sqrt{1+t^{2}}} \cdot\left(1+\frac{1}{2 \sqrt{1+t^{2}}} \cdot 2 t\right)= \\
& =\frac{1}{t+\sqrt{1+t^{2}}} \cdot \frac{\sqrt{1+t^{2}}+t}{\sqrt{1+t^{2}}}=\frac{1}{t+\sqrt{1+t^{2}}} \cdot \frac{\sqrt{1+t^{2... | \frac{}{^{2}+1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,174 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=2 e^{t} \\
y=e^{-t}
\end{array}\right.
\]
$t_{0}=0$ | ## Solution
Since $t_{0}=0$, then
$x_{0}=2 e^{0}=2$
$y_{0}=e^{-0}=1$
Let's find the derivatives:
$x_{t}^{\prime}=\left(2 e^{t}\right)^{\prime}=2 e^{t}$
$y_{t}^{\prime}=\left(e^{-t}\right)^{\prime}=-e^{-t}$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-e^{-t}}{2 e^{t}}=-\frac{1}{2 e^{2 t}}$
Then:
... | -\frac{1}{2}\cdotx+22x-3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,175 |
## Task Condition
Find the $n$-th order derivative.
$y=3^{2 x+5}$ | ## Solution
$y^{\prime}=\left(3^{2 x+5}\right)^{\prime}=3^{2 x+5} \cdot \ln 3 \cdot 2$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(3^{2 x+5} \cdot \ln 3 \cdot 2\right)^{\prime}=3^{2 x+5} \cdot \ln ^{2} 3 \cdot 2^{2}$
...
Obviously,
$y^{(n)}=3^{2 x+5} \cdot \ln ^{n} 3 \cdot 2^{n}=2^{n} \cdot \ln ^{n} ... | 2^{n}\cdot\ln^{n}3\cdot3^{2x+5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,176 |
## Task Condition
Find the derivative of the specified order.
$y=\left(x^{3}+3\right) e^{4 x+3}, y^{IV}=?$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\left(x^{3}+3\right) e^{4 x+3}\right)^{\prime}=3 x^{2} \cdot e^{4 x+3}+\left(x^{3}+3\right) e^{4 x+3} \cdot 4= \\
& =\left(4 x^{3}+3 x^{2}+12\right) \cdot e^{4 x+3} \\
& y^{\prime \prime}=\left(\left(4 x^{3}+3 x^{2}+12\right) \cdot e^{4 x+3}\right)^{\prime}=\left(12 x^... | (256x^{3}+768x^{2}+576x+864)\cdote^{4x+3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,177 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\ln t \\
y=\operatorname{arctg} t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\ln t)^{\prime}=\frac{1}{t}$
$y_{t}^{\prime}=(\operatorname{arctg} t)^{\prime}=\frac{1}{1+t^{2}}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\frac{1}{1+t^{2}}}{\frac{1}{t}}=\frac{t}{1+t^{2}} \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=\l... | \frac{\cdot(1-^2)}{(1+^2)^2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,178 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
$y=-\sqrt{x^{4}-x^{2}}$
$x \cdot y \cdot y^{\prime}-y^{2}=x^{4}$ | ## Solution
$y^{\prime}=\left(-\sqrt{x^{4}-x^{2}}\right)^{\prime}=-\frac{1}{2 \sqrt{x^{4}-x^{2}}} \cdot\left(4 x^{3}-2 x\right)=\frac{x-2 x^{3}}{\sqrt{x^{4}-x^{2}}}$
Substitute into equation (1):
$x \cdot\left(-\sqrt{x^{4}-x^{2}}\right) \cdot \frac{x-2 x^{3}}{\sqrt{x^{4}-x^{2}}}-\left(-\sqrt{x^{4}-x^{2}}\right)^{2}=... | proof | Algebra | proof | Yes | Yes | olympiads | false | 46,179 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{-5 ;-5 ; 5\}$
$p=\{-2 ; 0 ; 1\}$
$q=\{1 ; 3 ;-1\}$
$r=\{0 ; 4 ; 1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | p-3q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,180 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{3 ; 5 ; 4\}$
$b=\{5 ; 9 ; 7\}$
$c_{1}=-2 a+b$
$c_{2}=3 a-2 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=-2 a+b=\{-2 \cdot 3+5 ;-2 \cdot 5+9 ;-2 \cdot 4+7\}=\{-1 ;-1 ;-1\} \\
& c_{2}=3 a-2 b=\{3 \cdot 3-2 \cdot... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,181 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$$
A(-4 ; -2 ; 0), B(-1 ; -2 ; 4), C(3 ; -2 ; 1)
$$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-1-(-4) ;-2-(-2) ; 4-0)=(3 ; 0 ; 4)$
$\overrightarrow{A C}=(3-(-4) ;-2-(-2) ; 1-0)=(7 ; 0 ; 1)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\over... | \frac{1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,182 |
## Problem Statement
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$$
a = p - 2q
$$
$$
\begin{aligned}
& b = 2p + q \\
& |p| = 2 \\
& |q| = 3 \\
& (\widehat{p, q}) = \frac{3 \pi}{4}
\end{aligned}
$$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(p-2 q) \times(2 p+q)=2 \cdot p \times p+p \times q-2 \cdot 2 \cdot q \times p-2 ... | 15\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,183 |
## Condition of the problem
Are the vectors $a, b$ and $c$ coplanar?
$$
\begin{aligned}
& a=\{3 ; 3 ; 1\} \\
& b=\{1 ;-2 ; 1\} \\
& c=\{1 ; 1 ; 1\}
\end{aligned}
$$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
(a, b, c)=\left|\begin{array}{ccc}
3 & 3 & 1 \\
1 & -2 & 1 \\
1 & 1 & 1
\end{array}\right|=
$$
$=3 \cdot\left|\begin{array}{cc}-2 &... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,184 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(-1 ; -5 ; 2) \)
\( A_{2}(-6 ; 0 ; -3) \)
\( A_{3}(3 ; 6 ; -3) \)
\( A_{4}(-10 ; 6 ; 7) \) | ## Solution
From vertex $A_{1 \text { we draw vectors: }}$
$\overrightarrow{A_{1} A_{2}}=\{-6-(-1) ; 0-(-5) ;-3-2\}=\{-5 ; 5 ;-5\}$
$\overrightarrow{A_{1} A_{3}}=\{3-(-1) ; 6-(-5) ;-3-2\}=\{4 ; 11 ;-5\}$
$\overrightarrow{A_{1} A_{4}}=\{-10-(-1) ; 6-(-5) ; 7-2\}=\{-9 ; 11 ; 5\}$
According to the geometric meaning o... | 2\sqrt{38} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,185 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 2 ; 0)$
$M_{2}(1 ;-1 ; 2)$
$M_{3}(0 ; 1 ;-1)$
$M_{0}(2 ;-1 ; 4)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-1 & y-2 & z-0 \\
1-1 & -1-2 & 2-0 \\
0-1 & 1-2 & -1-0
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z \\
0 & -3 & 2 \\
-1 & -1 & -... | \frac{1}{\sqrt{38}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,186 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(7, -5, 1)$
$B(5, -1, -3)$
$C(3, 0, -4)$ | ## Solution
Let's find the vector $\overrightarrow{B C}:$
$\overrightarrow{B C}=\{3-5 ; 0-(-1) ;-4-(-3)\}=\{-2 ; 1 ;-1\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$$
\begin{aligned}
& -2 \cdot(x... | -2x+y-z+20=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,187 |
## problem statement
Find the angle between the planes:
$6 x+2 y-4 z+17=0$
$9 x+3 y-6 z-4=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{6 ; 2 ;-4\}$
$\overrightarrow{n_{2}}=\{9 ; 3 ;-6\}$
$
$B(-13 ; 4 ; 6)$
$C(10 ;-9 ; 5)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(-13-0)^{2}+(4-0)^{2}+(6-z)^{2}}=\sqrt{169+16+36-12 z+z^{2}}=\sqrt{z^{2}-12 z+221} \\
& A C=\sqrt{(10-0)^{2}+(-9-0)^{2}+(5-z)^{2}}=\sqrt{100+81+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+206}
\end{aligned}
$$
Since by the condition of the prob... | A(0;0;7.5) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,189 |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(1 ; \frac{1}{3} ;-2\right)$
$a: x-3 y+z+6=0$
$k=\frac{1}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: x-3 y+z+2=0$
Substitute the coordinates of point $A$ into the equatio... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,190 |
## Task Condition
Write the canonical equations of the line.
$$
\begin{aligned}
& 2 x+3 y+z+6=0 \\
& x-3 y-2 z+3=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}_{\text {- its direction }}$ vector.
Since the line belongs to both planes simultaneously, its dir... | \frac{x+3}{-3}=\frac{y}{5}=\frac{z}{-9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,191 |
## Problem Statement
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x-5}{1}=\frac{y-3}{-1}=\frac{z-2}{0} \\
& 3 x+y-5 z-12=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$\frac{x-5}{1}=\frac{y-3}{-1}=\frac{z-2}{0}=t \Rightarrow$
$\left\{\begin{array}{l}x=5+t \\ y=3-t \\ z=2\end{array}\right.$
Substitute into the equation of the plane:
$3(5+t)+(3-t)-5 \cdot 2-12=0$
$15+3 t+3-t-10-12=0$
$2 t-4=0$
$t=2$
Find the coordi... | (7;1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,192 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(1 ; 0 ;-1)$
$\frac{x-3.5}{2}=\frac{y-1.5}{2}=\frac{z}{0}$ | ## Solution
We find the equation of the plane that is perpendicular to the given line and passes through the point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:
$\vec{n}=\vec{s}=\{2 ; 2 ; 0\}$
Then the equation of the desired plane is:
$2 ... | M^{\}(2;-1;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,193 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+13 x+9}{(x+1)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+13 x+9}{(x+1)(x+2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+13 x+9}{(x+1)(x+2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x... | \ln|x+1|-\frac{1}{2(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,194 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+13 x+8}{x(x+2)^{3}}=\frac{A}{x}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x+2)^{3}}= ... | \ln|x|-\frac{1}{2(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,195 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+13 x-6}{(x+2)(x-2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+13 x-6}{(x+2)(x-2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+13 x-6}{(x+2)(x-2)^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}... | \ln|x+2|-\frac{1}{2(x-2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,196 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+14 x+10}{(x+1)(x+2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{... | \ln|x+1|-\frac{1}{(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,197 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3... | \ln|x+2|+\frac{1}{2(x-2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,198 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+11 x+7}{(x+1)(x+2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x... | \ln|x+1|+\frac{1}{2(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,199 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+6 x^{2}+7 x+1}{(x-1)(x+1)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+6 x^{2}+7 x+1}{(x-1)(x+1)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+7 x+1}{(x-1)(x+1)^{3}}=\frac{A}{x-1}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3... | 2\cdot\ln|x-1|-\frac{1}{2(x+1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,200 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+10 x+10}{(x-1)(x+2)^{3}}=\frac{A}{x-1}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{... | \ln|x-1|+\frac{1}{(x+2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,201 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+6 x^{2}+7 x+2}{x(x+1)^{3}}=\frac{A}{x}+\frac{B_{1}}{x+1}+\frac{B_{2}}{(x+1)^{2}}+\frac{B_{3}}{(x+1)^{3}}... | 2\cdot\ln|x|-\frac{1}{2(x+1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,202 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}}=\frac{A}{x}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}{(x-2)^{3}}= ... | \ln|x|-\frac{1}{2(x-2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,203 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+13 x-7}{(x+1)(x-2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+13 x-7}{(x+1)(x-2)^{3}} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+13 x-7}{(x+1)(x-2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}{(x... | \ln|x+1|-\frac{1}{2(x-2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,204 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+14 x-6}{(x+1)(x-2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+14 x-6}{(x+1)(x-2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+14 x-6}{(x+1)(x-2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}... | \ln|x+1|-\frac{1}{(x-2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,205 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}-6 x^{2}+10 x-10}{(x+1)(x-2)^{3}} d x
$$ | ## Solution
$$
\int \frac{x^{3}-6 x^{2}+10 x-10}{(x+1)(x-2)^{3}} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+10 x-10}{(x+1)(x-2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3... | \ln|x+1|+\frac{1}{(x-2)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,206 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.