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
Approximately calculate using the differential.
$y=\sqrt{1+x+\sin x}, x=0.01$ | ## Solution
$y\left(x_{0}+\Delta x\right) \approx y\left(x_{0}\right)^{\prime} \cdot \Delta x+y\left(x_{0}\right)$
$\Delta x=x-x_{0}$
$x=0.01$
$x_{0}=0$
$\Delta x=0.01$
$y\left(x_{0}\right)=y(0)=1$
$y^{\prime}=\frac{1}{2 \cdot \sqrt{1+x+\sin x}} \cdot(1+x+\sin x)^{\prime}=\frac{1+\cos x}{2 \cdot \sqrt{1+x+\sin x... | 1.01 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,146 |
## Task Condition
Find the derivative.
$$
y=\frac{x \cdot \sqrt{x+1}}{x^{2}+x+1}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\frac{(x \sqrt{x+1})^{\prime}\left(x^{2}+x+1\right)-x \sqrt{x+1}\left(x^{2}+x+1\right)^{\prime}}{\left(x^{2}+x+1\right)^{2}}=\frac{\left(x^{\prime} \sqrt{x+1}+x(\sqrt{x+1})^{\prime}\right)\left(x^{2}+x+1\right)-x \sqrt{x+1}(2 x+1)}{\left(x^{2}+x+1\right)^{2}}= \\
& =\frac{\l... | \frac{-x^{3}-x^{2}+3x+2}{2\sqrt{x+1}(x^{2}+x+1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,147 |
## Task Condition
Find the derivative.
$$
y=3 e^{\sqrt[3]{x}}\left(\sqrt[3]{x^{5}}-5 \sqrt[3]{x^{4}}+20 x-60 \sqrt[3]{x^{2}}+120 \sqrt[3]{x}-120\right)
$$ | ## Solution
In the process of solving, we will introduce a substitution (to shorten the notation)
$$
\begin{aligned}
& {\left[\sqrt[3]{x^{5}}-5 \sqrt[3]{x^{4}}+20 x-60 \sqrt[3]{x^{2}}+120 \sqrt[3]{x}-120\right]=(\sim)} \\
& y'=\left(3 e^{\sqrt[3]{x}}\right)'(\sim)+3 e^{\sqrt[3]{x}}(\sim)'= \\
& =\frac{3 e^{\sqrt[3]{x... | x\cdote^{\sqrt[3]{x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,148 |
## Problem Statement
Find the derivative.
$$
y=\ln \frac{\sqrt{5}+\tan \frac{x}{2}}{\sqrt{5}-\tan \frac{x}{2}}
$$ | ## Solution
$y=\ln \frac{\sqrt{5}+\tan \frac{x}{2}}{\sqrt{5}-\tan \frac{x}{2}}=\ln \left(\sqrt{5}+\tan \frac{x}{2}\right)-\ln \left(\sqrt{5}-\tan \frac{x}{2}\right)$
$y'=\frac{1}{\sqrt{5}+\tan \frac{x}{2}} \cdot\left(\sqrt{5}+\tan \frac{x}{2}\right)'-\frac{1}{\sqrt{5}-\tan \frac{x}{2}} \cdot\left(\sqrt{5}-\tan \frac{... | \frac{\sqrt{5}}{6\cos^{2}\frac{x}{2}-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,149 |
## Task Condition
Find the derivative.
$$
y=\sqrt[7]{\tan(\cos 2)}+\frac{\sin ^{2} 27 x}{27 \cos 54 x}
$$ | ## Solution
$y^{\prime}=(\sqrt[7]{\operatorname{tg}(\cos 2)})^{\prime}+\left(\frac{\sin ^{2} 27 x}{27 \cos 54 x}\right)^{\prime}=0+\frac{1}{27}\left(\frac{\sin ^{2} 27 x}{\cos 54 x}\right)^{\prime}=\frac{1}{27}\left(\frac{\left(\sin ^{2} 27 x\right)^{\prime} \cos 54 x-\sin ^{2} 27 x(\cos 54 x)^{\prime}}{\cos 54 x}\rig... | \operatorname{tg}54x\cdot\54x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,150 |
## problem statement
Find the derivative.
$$
y=\frac{x}{2 \sqrt{1-4 x^{2}}} \arcsin 2 x+\frac{1}{8} \cdot \ln \left(1-4 x^{2}\right)
$$ | ## Solution
$y^{\prime}=\frac{1}{2}\left(\frac{(x \cdot \arcsin 2 x)^{\prime} \cdot \sqrt{1-4 x^{2}}-x \cdot \arcsin 2 x\left(\sqrt{1-4 x^{2}}\right)^{\prime}}{1-4 x^{2}}\right)+\frac{1}{8} \cdot \frac{1}{1-4 x^{2}} \cdot\left(1-4 x^{2}\right)^{\prime}=$ $=\frac{1}{2}\left(\frac{\left(x^{\prime} \cdot \arcsin 2 x+x \c... | \frac{\arcsin2x}{2(1-4x^{2})\cdot\sqrt{1-4x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,151 |
## Problem Statement
Find the derivative.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | ## Solution
$$
\begin{aligned}
& y^{\prime}=-\frac{1}{2} \cdot \frac{(\operatorname{sh} x)^{\prime} \operatorname{ch}^{2} x-\operatorname{sh} x\left(\operatorname{ch}^{2} x\right)^{\prime}}{\operatorname{ch}^{4} x}+\frac{(\operatorname{sh} x)^{\prime}}{\operatorname{sh}^{2} x}-\frac{3}{2} \cdot \frac{(\operatorname{sh... | \frac{\operatorname{ch}x}{\operatorname{sh}^{2}x}-\frac{2-\operatorname{sh}^{2}x}{\operatorname{ch}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,152 |
## Task Condition
Find the derivative.
$y=x^{e^{\operatorname{arctan} x}}$ | ## Solution
$\ln y=e^{\operatorname{arctg} x} \cdot \ln x$
$(\ln y)^{\prime}=\left(e^{\operatorname{arctg} x}\right)^{\prime} \cdot \ln x+e^{\operatorname{arctg} x} \cdot(\ln x)^{\prime}=e^{\operatorname{arctg} x} \cdot(\operatorname{arctg} x)^{\prime} \cdot \ln x+e^{\operatorname{arctg} x} \cdot \frac{1}{x}=$ $=e^{\... | y^{\}=x^{e^{\operatorname{arctg}x}}\cdote^{\operatorname{arctg}x}(\frac{\lnx}{1+x^{2}}+\frac{1}{x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,153 |
## Problem Statement
Find the derivative.
$y=\frac{1}{\sqrt{2}} \cdot \operatorname{arctan} \frac{2 x+1}{\sqrt{2}}+\frac{2 x+1}{4 x^{2}+4 x+3}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\frac{1}{\sqrt{2}} \cdot \frac{1}{1+\left(\frac{2 x+1}{\sqrt{2}}\right)^{2}} \cdot\left(\frac{2 x+1}{\sqrt{2}}\right)^{\prime}+\frac{(2 x+1)^{\prime}\left(4 x^{2}+4 x+3\right)-(2 x+1)\left(4 x^{2}+4 x+3\right)^{\prime}}{\left(4 x^{2}+4 x+3\right)^{2}}= \\
& =\frac{1}{1+\frac... | \frac{8}{(4x^{2}+4x+3)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,154 |
## Problem Statement
Find the derivative.
$$
y=\frac{\sqrt{x^{2}+2}}{x^{2}}-\frac{1}{\sqrt{2}} \ln \frac{\sqrt{2}+\sqrt{x^{2}+2}}{x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\frac{\left(\sqrt{x^{2}+2}\right)^{\prime} \cdot x^{2}-2 x \cdot \sqrt{x^{2}+2}}{x^{4}}-\frac{1}{\sqrt{2}} \cdot \frac{x}{\sqrt{2}+\sqrt{x^{2}+2}} \cdot\left(\frac{\sqrt{2}+\sqrt{x^{2}+2}}{x}\right)^{\prime}= \\
& =\frac{\frac{1}{2 \sqrt{x^{2}+2}} \cdot 2 x \cdot x^{2}-2 x \... | -\frac{4}{x^{3}\cdot\sqrt{x^{2}+2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,155 |
## Problem Statement
Find the derivative.
$$
y=2 \frac{\cos x}{\sin ^{4} x}+3 \frac{\cos x}{\sin ^{2} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=2 \frac{(\cos x)^{\prime} \cdot \sin ^{4} x-\cos x\left(\sin ^{4} x\right)^{\prime}}{\sin ^{8} x}+3 \frac{(\cos x)^{\prime} \cdot \sin ^{2} x-\cos x\left(\sin ^{2} x\right)^{\prime}}{\sin ^{4} x}= \\
& =2 \frac{-\sin ^{5} x-4 \sin ^{3} x \cdot \cos ^{2} x}{\sin ^{8} x}+3 \fr... | 3\operatorname{cosec}x-8\operatorname{cosec}^{5}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,156 |
## Task Condition
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\ln (\operatorname{tg} t) \\
y=\frac{1}{\sin ^{2} t}
\end{array}\right.
$$ | ## Solution
$y_{x}^{\prime}=\frac{y^{\prime}(t)}{x^{\prime}(t)}$
$y^{\prime}(t)=\left(1 / \sin ^{2} t\right)^{\prime}=-\frac{2 \sin t \cdot \cos t}{\sin ^{4} t}=-\frac{\sin 2 t}{\sin ^{4} t}$
$x^{\prime}(t)=(\ln \operatorname{tg} t)^{\prime}=\frac{1}{\operatorname{tg} t} \cdot(\operatorname{tg} t)^{\prime}=\frac{\co... | -2\operatorname{ctg}^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,157 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=2 \operatorname{tg} t \\
y=2 \sin ^{2} t+\sin 2 t
\end{array}\right.
\]
$t_{0}=\frac{\pi}{4}$ | ## Solution
$x_{0}=x\left(t_{0}\right)=2 \operatorname{tg} \frac{\pi}{4}=2$
$y_{0}=y\left(t_{0}\right)=2 \sin ^{2} \frac{\pi}{4}+\sin \frac{\pi}{2}=2$
$y_{x}^{\prime}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{4 \sin t \cdot \cos t+2 \cos 2 t}{\frac{2}{\cos ^{2} t}}=\frac{4 \sin t \cdot \cos ^{3} t+2 \cos 2 t \cdot \... | \frac{1}{2}x+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,158 |
## Task Condition
Find the $n$-th order derivative.
$$
y=\frac{x}{x+1}
$$ | ## Solution
To derive the formula of the $\mathrm{N}$-th order, we first find several derivatives and then establish a general rule suitable for calculating the derivative of any order:
$$
\begin{aligned}
y^{\prime} & =\frac{x+1-x}{(x+1)^{2}}=\frac{1}{(x+1)^{2}} \\
y^{\prime \prime} & =-\frac{1 \cdot 2(x+1)}{(x+1)^{4... | y^{(n)}=(-1)^{n+1}\cdot\frac{n!}{(x+1)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,159 |
## Task Condition
Find the derivative of the specified order.
$$
y=\frac{\ln (x-2)}{x-2}, y^{V}=?
$$ | ## Solution № 1
Let's represent the original function as a product of 2 functions:
$y=\ln (x-2) \cdot(x-2)^{-1} \Rightarrow\left\{\begin{array}{l}u=\ln (x-2), \\ v=(x-2)^{-1} .\end{array}\right.$
Find the fifth derivative of both functions:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ u = \operatorname { l n }... | \frac{274-120\ln(x-2)}{(x-2)^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,160 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=2(t-\sin t) \\
y=4(2+\cos t)
\end{array}\right.
$$ | ## Solution
$y_{x}^{\prime}=\frac{y^{\prime}(t)}{x^{\prime}(t)}$
$y_{x^{2}}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x^{\prime}(t)}$
$y^{\prime}(t)=4(2+\cos (t))^{\prime}=-4 \sin (t)$
$x^{\prime}(t)=2(t-\sin (t))^{\prime}=2(1-\cos (t))$
$y_{x}^{\prime}=\frac{-4 \sin (t)}{2(1-\cos (t))}=\frac... | \frac{1}{(1-\cos())^2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,161 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
\[
\begin{aligned}
& y=\frac{x}{x-1}+x^{2} \\
& x(x-1) y^{\prime}+y=x^{2}(2 x-1)
\end{aligned}
\] | ## Solution
$y^{\prime}=\left(\frac{x}{x-1}+x^{2}\right)^{\prime}=\frac{x-1-x}{(x-1)^{2}}+2 x=2 x-\frac{1}{(x-1)^{2}}$
Substitute $y_{\text {and }} y_{\text {into equation (1) }}^{\prime}$
$x(x-1)\left(2 x-\frac{1}{(x-1)^{2}}\right)+\frac{x}{x-1}+x^{2}=x^{2}(2 x-1)$
$2 x^{2}(x-1)-\frac{x}{x-1}+\frac{x}{x-1}+x^{2}=x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 47,162 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{d x}{x \sqrt{x^{2}+1}}
$$ | ## Solution
Let's introduce the substitution
$$
\begin{aligned}
& x=\frac{1}{t} \\
& d x=-\frac{1}{t^{2}} d t
\end{aligned}
$$
Then
$$
\begin{aligned}
& \int \frac{d x}{x \sqrt{x^{2}+1}}=\int \frac{-\frac{1}{t^{2}} d t}{\frac{1}{t} \sqrt{\left(\frac{1}{t}\right)^{2}+1}}=-\int \frac{d t}{\sqrt{t^{2}+1}}= \\
& =-\ln ... | -\ln|\frac{1+\sqrt{x^{2}+1}}{x}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,163 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{1+\ln x}{x} d x
$$ | ## Solution
## Method 1
$$
\int \frac{1+\ln x}{x} d x=\int(1+\ln x) d(\ln x)=\ln x+\frac{1}{2} \ln ^{2} x+C
$$
## Method 2
$$
\int \frac{1+\ln x}{x} d x=\int \frac{d x}{x}+\int \frac{\ln x d x}{x} d x=\ln |x|+\int \ln x d(\ln x)=\ln |x|+\frac{1}{2} \ln ^{2} x+C
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\... | \lnx+\frac{1}{2}\ln^2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,164 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{d x}{x \sqrt{x^{2}-1}}
$$ | ## Solution
$$
\int \frac{d x}{x \sqrt{x^{2}-1}}=\int \frac{d x}{x \cdot|x| \cdot \sqrt{1-\frac{1}{x^{2}}}}=
$$
First, consider the case $x>0$:
$$
=\int \frac{d x}{x^{2} \sqrt{1-\frac{1}{x^{2}}}}=
$$
Substitution:
$$
\begin{aligned}
& y=\frac{1}{x} \Rightarrow y^{2}=\frac{1}{x^{2}} \\
& d y=-\frac{1}{x^{2}} d x \R... | -\arcsin\frac{1}{|x|}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,165 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{2}+\ln x^{2}}{x} d x
$$ | ## Solution
$$
\int \frac{x^{2}+\ln x^{2}}{x} d x=\int x \cdot d x+\int \frac{\ln x^{2}}{x} d x=\frac{x^{2}}{2}+\int \frac{\ln x^{2}}{x} d x=
$$
Substitution:
$$
\begin{aligned}
& y=\ln x^{2} \\
& d y=\frac{1}{x^{2}} \cdot 2 x \cdot d x=\frac{2}{x} d x \Rightarrow \frac{1}{x} d x=\frac{d y}{2}
\end{aligned}
$$
We g... | \frac{x^{2}}{2}+\ln^{2}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,166 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x}{\sqrt{x^{4}+x^{2}+1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x}{\sqrt{x^{4}+x^{2}+1}} d x=\int \frac{\frac{1}{2} \cdot d\left(x^{2}+\frac{1}{2}\right)}{\sqrt{\left(x^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}}=\frac{1}{2} \cdot \ln \left|x^{2}+\frac{1}{2}+\sqrt{\left(x^{2}+\frac{1}{2}\right)^{2}+\frac{3}{4}}\right|+C= \\
& =\frac{1}{2} \... | \frac{1}{2}\cdot\ln|x^{2}+\frac{1}{2}+\sqrt{x^{4}+x^{2}+1}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,167 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{\arccos ^{3} x-1}{\sqrt{1-x^{2}}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{\arccos ^{3} x-1}{\sqrt{1-x^{2}}} d x=\int \frac{\arccos ^{3} x}{\sqrt{1-x^{2}}} d x+\int \frac{-d x}{\sqrt{1-x^{2}}}= \\
& =\int-(\arccos x)^{3} d(\arccos x)+\arccos x=-\frac{1}{4} \cdot(\arccos x)^{4}+\arccos x+C
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php... | -\frac{1}{4}\cdot(\arccosx)^{4}+\arccosx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,168 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \tan x \cdot \ln \cos x \, dx
$$ | ## Solution
$$
\int \operatorname{tg} x \cdot \ln \cos x d x=
$$
Substitution:
$$
\begin{aligned}
& y=\ln \cos x \\
& d y=\frac{1}{\cos x} \cdot(-\sin x) \cdot d x=-\operatorname{tg} x \cdot d x
\end{aligned}
$$
We get:
$$
=\int-y \cdot d y=-\frac{y^{2}}{2}+C=
$$
Reverse substitution:
$$
=-\frac{\ln ^{2} \cos x}... | -\frac{\ln^{2}\cosx}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,169 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{\operatorname{tg}(x+1)}{\cos ^{2}(x+1)} d x
$$ | ## Solution
$$
\int \frac{\operatorname{tg}(x+1)}{\cos ^{2}(x+1)} d x=\int \operatorname{tg}(x+1) \cdot d(\operatorname{tg}(x+1))=\frac{\operatorname{tg}^{2}(x+1)}{2}+C
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\... | \frac{\operatorname{tg}^{2}(x+1)}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,170 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x=\frac{1}{2} \cdot \int \frac{x^{2}}{\left(x^{2}+1\right)^{2}} d\left(x^{2}\right)=\frac{1}{2} \cdot \int \frac{x^{2}+1-1}{\left(x^{2}+1\right)^{2}} d\left(x^{2}\right)= \\
& =\frac{1}{2} \cdot \int \frac{1}{x^{2}+1} d\left(x^{2}\right)-\f... | \frac{1}{2}\cdot\ln(x^{2}+1)+\frac{1}{2(x^{2}+1)}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,171 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{1-\cos x}{(x-\sin x)^{2}} d x
$$ | ## Solution
$$
\int \frac{1-\cos x}{(x-\sin x)^{2}} d x=\int \frac{d(x-\sin x)}{(x-\sin x)^{2}}=-\frac{1}{x-\sin x}+C
$$
Source — "http://pluspi.org/wiki/index.php/?%?%?%?%?%?%?%?%?%?%?%? %D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD %D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_3-10" Categories: Kuznets... | -\frac{1}{x-\sinx}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,172 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{\sin x - \cos x}{(\cos x + \sin x)^{5}} d x
$$ | ## Solution
$$
\int \frac{\sin x-\cos x}{(\cos x+\sin x)^{5}} d x=-\int \frac{d(\cos x+\sin x)}{(\cos x+\sin x)^{5}}=\frac{1}{4} \cdot \frac{1}{(\cos x+\sin x)^{4}}+C
$$
Source — "http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_3... | \frac{1}{4}\cdot\frac{1}{(\cosx+\sinx)^{4}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,173 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x \cdot \cos x+\sin x}{(x \cdot \sin x)^{2}} d x
$$ | ## Solution
$$
\int \frac{x \cdot \cos x+\sin x}{(x \cdot \sin x)^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& y=x \cdot \sin x \\
& d y=(\sin x+x \cdot \cos x) d x
\end{aligned}
$$
We get:
$$
=\int \frac{d y}{y^{2}}=-\frac{1}{y}+C=
$$
Reverse substitution:
$$
=-\frac{1}{x \cdot \sin x}+C
$$
Source — «http:/... | -\frac{1}{x\cdot\sinx}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,174 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}+x}{x^{4}+1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x^{3}+x}{x^{4}+1} d x=\int \frac{x^{3}}{x^{4}+1} d x+\int \frac{x}{x^{4}+1} d x= \\
& =\frac{1}{4} \cdot \int \frac{1}{x^{4}+1} d\left(x^{4}+1\right)+\frac{1}{2} \cdot \int \frac{1}{x^{4}+1} d\left(x^{2}\right)= \\
& =\frac{1}{4} \cdot \ln \left|x^{4}+1\right|+\frac{1}{2} \c... | \frac{1}{4}\cdot\ln|x^{4}+1|+\frac{1}{2}\cdot\operatorname{arctg}x^{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,175 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x}{\sqrt{x^{4}-x^{2}-1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x}{\sqrt{x^{4}-x^{2}-1}} d x=\int \frac{\frac{1}{2} \cdot d\left(x^{2}-\frac{1}{2}\right)}{\sqrt{\left(x^{2}-\frac{1}{2}\right)^{2}-\frac{5}{4}}}=\frac{1}{2} \cdot \ln \left|x^{2}-\frac{1}{2}+\sqrt{\left(x^{2}-\frac{1}{2}\right)^{2}-\frac{5}{4}}\right|+C= \\
& =\frac{1}{2} \... | \frac{1}{2}\cdot\ln|x^{2}-\frac{1}{2}+\sqrt{x^{4}-x^{2}-1}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,176 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x}{\sqrt[3]{x-1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x}{\sqrt[3]{x-1}} d x=\int \frac{x-1+1}{\sqrt[3]{x-1}} d x=\int \sqrt[3]{(x-1)^{2}} d x+\int \frac{1}{\sqrt[3]{x-1}} d x= \\
& =\frac{3}{5} \cdot \sqrt[3]{(x-1)^{5}}+\frac{3}{2} \cdot \sqrt[3]{(x-1)^{2}}+C
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�... | \frac{3}{5}\cdot\sqrt[3]{(x-1)^{5}}+\frac{3}{2}\cdot\sqrt[3]{(x-1)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,177 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{1+\ln (x-1)}{x-1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{1+\ln (x-1)}{x-1} d x=\int \frac{d x}{x-1}+\int \frac{\ln (x-1)}{x-1} d x= \\
& =\int \frac{d(x-1)}{x-1}+\int \ln (x-1) d(\ln (x-1))= \\
& =\ln (x-1)+\frac{1}{2} \cdot \ln ^{2}(x-1)+C
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\... | \ln(x-1)+\frac{1}{2}\cdot\ln^{2}(x-1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,178 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{\left(x^{2}+1\right) d x}{\left(x^{3}+3 x+1\right)^{5}}
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{\left(x^{2}+1\right) d x}{\left(x^{3}+3 x+1\right)^{5}}=\int \frac{\frac{1}{3} \cdot\left(3 x^{2}+3\right) d x}{\left(x^{3}+3 x+1\right)^{5}}=\int \frac{\frac{1}{3} \cdot d\left(x^{3}+3 x+1\right)}{\left(x^{3}+3 x+1\right)^{5}}= \\
& =\frac{1}{3} \cdot\left(-\frac{1}{4}\righ... | -\frac{1}{12}\cdot\frac{1}{(x^{3}+3x+1)^{4}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,179 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{4 \operatorname{arctg} x - x}{1 + x^{2}} \, dx
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{4 \operatorname{arctg} x - x}{1 + x^{2}} d x = \int \frac{4 \operatorname{arctg} x}{1 + x^{2}} d x - \int \frac{x}{1 + x^{2}} d x = \\
& = 4 \int \operatorname{arctg} x d(\operatorname{arctg} x) - \frac{1}{2} \int \frac{d\left(1 + x^{2}\right)}{1 + x^{2}} = \frac{1}{2}\left(... | \frac{1}{2}(4\operatorname{arctg}^{2}x-\ln(1+x^{2}))+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,180 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}}{x^{2}+4} d x
$$ | ## Solution
## Solution №1
$$
\begin{aligned}
& \int \frac{x^{3}}{x^{2}+4} d x=\frac{1}{2} \cdot \int \frac{x^{2}}{x^{2}+4} d\left(x^{2}\right)=\frac{1}{2} \cdot \int \frac{x^{2}+4-4}{x^{2}+4} d\left(x^{2}\right)= \\
& =\frac{1}{2} \cdot \int d\left(x^{2}\right)-\frac{1}{2} \cdot \int \frac{4}{x^{2}+4} d\left(x^{2}\r... | \frac{x^2}{2}-2\ln|x^2+4|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,181 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x+\cos x}{x^{2}+2 \sin x} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x+\cos x}{x^{2}+2 \sin x} d x=\int \frac{\frac{1}{2}(2 x+2 \cos x)}{x^{2}+2 \sin x} d x=\frac{1}{2} \cdot \int \frac{d\left(x^{2}+2 \sin x\right)}{x^{2}+2 \sin x}= \\
& =\frac{1}{2} \cdot \ln \left|x^{2}+2 \sin x\right|+C
\end{aligned}
$$
Source — "http://pluspi.org/wiki/in... | \frac{1}{2}\cdot\ln|x^{2}+2\sinx|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,182 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{2 \cos x + 3 \sin x}{(2 \sin x - 3 \cos x)^{3}} d x
$$ | ## Solution
$$
\int \frac{2 \cos x+3 \sin x}{(2 \sin x-3 \cos x)^{3}} d x=\int \frac{d(2 \sin x-3 \cos x)}{(2 \sin x-3 \cos x)^{3}}=-\frac{1}{2 \cdot(2 \sin x-3 \cos x)^{2}}+C
$$
Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\... | -\frac{1}{2\cdot(2\sinx-3\cosx)^{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,183 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{8 x-\operatorname{arctg} 2 x}{1+4 x^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{8 x - \operatorname{arctg} 2 x}{1 + 4 x^{2}} d x = \int \frac{8 x}{1 + 4 x^{2}} d x - \int \frac{\operatorname{arctg} 2 x}{1 + 4 x^{2}} d x = \\
& = \int \frac{d\left(1 + 4 x^{2}\right)}{1 + 4 x^{2}} - \frac{1}{2} \cdot \int \operatorname{arctg} 2 x \cdot d(\operatorname{arc... | \ln|1+4x^{2}|-\frac{1}{4}\cdot\operatorname{arctg}^{2}2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,184 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{\frac{1}{2 \sqrt{x}}+1}{(\sqrt{x}+x)^{2}} d x
$$ | ## Solution
$$
\int \frac{\frac{1}{2 \sqrt{x}}+1}{(\sqrt{x}+x)^{2}} d x=\int \frac{d(\sqrt{x}+x)}{(\sqrt{x}+x)^{2}}=-\frac{1}{\sqrt{x}+x}+C
$$
Source — "http://pluspi.org/wiki/index.php/?%?%?%?%?%?%?%?%?%?%? \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \m... | -\frac{1}{\sqrt{x}+x}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,185 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x+\frac{1}{x}}{\sqrt{x^{2}+1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x+\frac{1}{x}}{\sqrt{x^{2}+1}} d x=\int \frac{x}{\sqrt{x^{2}+1}} d x+\int \frac{1}{x \sqrt{x^{2}+1}} d x= \\
& =\frac{1}{2} \cdot \int \frac{d\left(x^{2}+1\right)}{\sqrt{x^{2}+1}}+\int \frac{1}{x \sqrt{x^{2}+1}} d x=\frac{1}{2} \cdot 2 \cdot \sqrt{x^{2}+1}+\int \frac{1}{x^{2... | \sqrt{x^{2}+1}-\ln|\frac{1+\sqrt{x^{2}+1}}{x}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,187 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x-\frac{1}{x}}{\sqrt{x^{2}+1}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x-\frac{1}{x}}{\sqrt{x^{2}+1}} d x=\int \frac{x}{\sqrt{x^{2}+1}} d x-\int \frac{1}{x \sqrt{x^{2}+1}} d x= \\
& =\frac{1}{2} \cdot \int \frac{d\left(x^{2}+1\right)}{\sqrt{x^{2}+1}}-\int \frac{1}{x \sqrt{x^{2}+1}} d x=\frac{1}{2} \cdot 2 \cdot \sqrt{x^{2}+1}-\int \frac{1}{x^{2... | \sqrt{x^{2}+1}+\ln|\frac{1+\sqrt{x^{2}+1}}{x}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,188 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{\operatorname{arctg} x + x}{1 + x^{2}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{\operatorname{arctg} x + x}{1 + x^2} \, dx = \int \frac{\operatorname{arctg} x}{1 + x^2} \, dx + \int \frac{x}{1 + x^2} \, dx = \\
& = \int \operatorname{arctg} x \, d(\operatorname{arctg} x) + \frac{1}{2} \int \frac{d(1 + x^2)}{1 + x^2} = \frac{1}{2} \left( \operatorname{ar... | \frac{1}{2}(\operatorname{arctg}^2x+\ln|1+x^2|)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,189 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x
$$
Note: In the context of this problem, $\operatorname{arctg} x$ is the same as $\arctan x$. | ## Solution
$$
\begin{aligned}
& \int \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x=\int \frac{x d x}{1+x^{2}}-\int \frac{(\operatorname{arctg} x)^{4}}{1+x^{2}}= \\
& =\frac{1}{2} \int \frac{d\left(1+x^{2}\right)}{1+x^{2}}-\int(\operatorname{arctg} x)^{4} d(\operatorname{arctg} x)=\frac{1}{2} \ln \left(1+x^{2}\ri... | \frac{1}{2}\ln(1+x^{2})-\frac{(\operatorname{arctg}x)^{5}}{5}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,190 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{x^{3}}{x^{2}+1} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{x^{3}}{x^{2}+1} d x=\int \frac{x\left(x^{2}+1\right)-x}{x^{2}+1} d x=\int x d x-\int \frac{x}{x^{2}+1}= \\
& =\frac{1}{2} x^{2}-\frac{1}{2} \int \frac{d\left(x^{2}+1\right)}{x^{2}+1}=\frac{1}{2}\left(x^{2}-\ln \left(x^{2}+1\right)\right)+C
\end{aligned}
$$
Source — "http://... | \frac{1}{2}(x^{2}-\ln(x^{2}+1))+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,191 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{(\arcsin x)^{2}+1}{\sqrt{1-x^{2}}} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{(\arcsin x)^{2}+1}{\sqrt{1-x^{2}}} d x=\int \frac{\arcsin ^{2} x}{\sqrt{1-x^{2}}} d x+\int \frac{d x}{\sqrt{1-x^{2}}}= \\
& =\int(\arcsin x)^{2} d(\arcsin x)+\arcsin x=\frac{1}{3} \cdot(\arcsin x)^{3}+\arcsin x+C
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/\... | \frac{1}{3}\cdot(\arcsinx)^{3}+\arcsinx+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,192 |
## Problem Statement
Calculate the indefinite integral:
$$
\int \frac{1-\sqrt{x}}{\sqrt{x} \cdot(x+1)} d x
$$ | ## Solution
$$
\begin{aligned}
& \int \frac{1-\sqrt{x}}{\sqrt{x} \cdot(x+1)} d x=\int \frac{1}{\sqrt{x} \cdot(x+1)} d x-\int \frac{1}{x+1} d x= \\
& =2 \cdot \int \frac{1}{x+1} d(\sqrt{x})-\ln |x+1|=2 \arctan \sqrt{x}-\ln |x+1|+C
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/%D0%97%D0%B0%D0%B4%D0%B0%D1%... | 2\arctan\sqrt{x}-\ln|x+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,193 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{8 ; 1 ; 12\}$
$p=\{1 ; 2 ;-1\}$
$q=\{3 ; 0 ; 2\}$
$r=\{-1 ; 1 ; 1\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\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+4q+3r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,194 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{8 ; 3 ;-1\}$
$b=\{4 ; 1 ; 3\}$
$c_{1}=2 a-b$
$c_{2}=2 b-4 a$ | ## 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.
It is not hard to notice that $c_{2}=-2(2 a-b)=-2 c_{1}$ for any $a$ and $b$.
That is, $c_{1}=-2 \cdot c_{2}$, which means the vectors $c_{1}... | c_{1}=-2\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,195 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(3 ; 3 ;-1), B(5 ; 1 ;-2), C(4 ; 1 ; 1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(5-3 ; 1-3 ;-2-(-1))=(2 ;-2 ;-1) \\
& \overrightarrow{A C}=(4-3 ; 1-3 ; 1-(-1))=(1 ;-2 ; 2)
\end{aligned}
$$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\over... | \frac{4}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,196 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=7 p-2 q$
$b=p+3 q$
$|p|=\frac{1}{2}$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$ | ## 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=(7 p-2 q) \times(p+3 q)=7 \cdot p \times p+7 \cdot 3 \cdot p \times q-2 \cdot q \... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,197 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 1 ;-1\}$
$b=\{1 ; 0 ;-1\}$
$c=\{8 ; 3 ;-2\}$ | ## 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.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 1 & -1 \\
1 & 0 & -1 \\
8 & 3 & -2
\end{array}\right|= \\
& =3 \cdot\left|\... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,198 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(-1 ; 2 ;-3) \)
\( A_{2}(4 ;-1 ; 0) \)
\( A_{3}(2 ; 1 ;-2) \)
\( A_{4}(3 ; 4 ; 5) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{4-(-1) ;-1-2 ; 0-(-3)\}=\{5 ;-3 ; 3\} \\
& \overrightarrow{A_{1} A_{3}}=\{2-(-1) ; 1-2 ;-2-(-3)\}=\{3 ;-1 ; 1\} \\
& \vec{A}_{1} A_{4}=\{3-(-1) ; 4-2 ; 5-(-3)\}=\{4 ; 2 ; 8\}
\end{aligned}
$$
According to the geomet... | 6\frac{2}{3},5\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,199 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(7 ; 2 ; 4)$
$M_{2}(7 ;-1 ;-2)$
$M_{3}(-5 ;-2 ;-1)$
$M_{0}(10 ; 1 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
7-7 & -1-2 & -2-4 \\
-5-7 & -2-2 & -1-4
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
0 & -3 & -6 \\
-12 &... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,200 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(2 ; 1 ; 7)$
$B(9 ; 0 ; 2)$
$C(9 ; 2 ; 3)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{9-9 ; 2-0 ; 3-2\}=\{0 ; 2 ; 1\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$0 \cdot(x-2)+2 \cdot(y-1)+(z-7)=0$
$2 y... | 2y+z-9=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,201 |
## Problem Statement
Find the angle between the planes:
\[
\begin{aligned}
& 3 x-2 y+3 z+23=0 \\
& y+z+5=0
\end{aligned}
\] | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{3 ;-2 ; 3\}$
$\overrightarrow{n_{2}}=\{0 ; 1 ; 1\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \ph... | 8119^{\}45^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,202 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(0 ; 5 ;-9)$
$C(-1 ; 0 ; 5)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(0-0)^{2}+(5-y)^{2}+(-9-0)^{2}}=\sqrt{25-10 y+y^{2}+81}=\sqrt{y^{2}-10 y+106} \\
& A C=\sqrt{(-1-0)^{2}+(0-y)^{2}+(5-0)^{2}}=\sqrt{1+y^{2}+25}=\sqrt{y^{2}+26}
\end{aligned}
$$
Since according to the problem $A B=A C$, then
$\sqrt{y^... | A(0;8;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,203 |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A\left(\frac{1}{3} ; 1 ; 1\right)$
$a: 3 x-y+5 z-6=0$
$k=\frac{5}{6}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 3 x-y+5 z-5=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 47,204 |
## Task Condition
Write the canonical equations of the line.
$5 x+y+2 z+4=0$
$x-y-3 z+2=0$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vect... | \frac{x+1}{-1}=\frac{y-1}{17}=\frac{z}{-6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,205 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z+3}{2}$
$3 x+4 y+7 z-16=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-3}{2}=\frac{y+1}{3}=\frac{z+3}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3+2 t \\
y=-1+3 t \\
z=-3+2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$3(3+2 t)+4(-1+3 t)+7(-3+2 t)-16=0$
... | (5;2;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 47,206 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(x)$:
$f(x)=\left\{\begin{array}{c}\sin \left(e^{x^{2} \sin \frac{5}{x}}-1\right)+x, x \neq 0 \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$$
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 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,208 |
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$$
y=\frac{1+\sqrt{x}}{1-\sqrt{x}}, x_{\bar{u}}=4
$$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)^{\prime}=\frac{(1+\sqrt{x})^{\prime}(1-\sqrt{x})-(1+\sqrt{x})(1-\sqrt{x})^{\prime}}{(1-\sqrt{x})^{2}}= \\
& =\frac{\frac{1}{2 \sqrt{x}} \cdot(1-\sqrt{x})-(1+\sqrt{x})\left(-\frac{1}{2 \sqrt{x}}\right)}{(1-\... | -2x+5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,209 |
Condition of the problem
Find the differential $d y$.
$y=\operatorname{arctg}(\operatorname{sh} x)+(\operatorname{sh} x) \ln (\operatorname{ch} x)$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=(\operatorname{arctg}(\operatorname{sh} x)+(\operatorname{sh} x) \ln (\operatorname{ch} x))^{\prime} d x= \\
& =\left(\frac{1}{1+\operatorname{sh}^{2} x} \cdot \operatorname{ch} x+\operatorname{ch} x \cdot \ln (\operatorname{ch} x)+\operatorname{sh} x \cdot \fr... | \operatorname{ch}x\cdot(1+\ln(\operatorname{ch}x)) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,210 |
## Task Condition
Calculate approximately using the differential.
$$
y=\sqrt[3]{x}, x=26.46
$$ | ## Solution
If the increment $\Delta x = x - x_{\text{argument}} x$ is small in absolute value, then $f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x^{x}$
Choose:
$x_{\bar{u}} = 2 \bar{i}$
Then:
$\Delta x = -0.54$
Calculate:
$$
\begin{aligned}
& y(2... | 2.98 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,211 |
Condition of the problem
Find the derivative.
$y=\frac{1}{2} \cdot \ln \left(\epsilon^{2 x}+1\right)-2 \operatorname{arctg} e^{x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot \ln \left(\epsilon^{2 x}+1\right)-2 \operatorname{arctg} e^{x}\right)^{\prime}=\frac{1}{2} \cdot \frac{1}{\varepsilon^{2 x}+1} \cdot 2 \epsilon^{2 x}-2 \cdot \frac{1}{1+\epsilon^{2 x}} \cdot e^{x}= \\
& =\frac{\epsilon^{2 x}}{1+\epsilon^{2 x}}-\frac{2... | \frac{\epsilon^{2x}-\epsilon^{x}}{1+\epsilon^{2x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,213 |
Condition of the problem
Find the derivative.
$y=\ln ^{2}(x+\cos x)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln ^{2}(x+\cos x)\right)^{\prime}=2 \ln (x+\cos x) \cdot \frac{1}{x+\cos x} \cdot(1-\sin x)= \\
& =\frac{1-\sin x}{x+\cos x} \cdot 2 \ln (x+\cos x)
\end{aligned}
$$
## Problem Kuznetsov Differentiation 8-7 | \frac{1-\sinx}{x+\cosx}\cdot2\ln(x+\cosx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,214 |
Condition of the problem
Find the derivative.
$$
y=\frac{\cos \ln 7 \cdot \sin ^{2} 7 x}{7 \cos 14 x}
$$ | ## Solution
$y^{\prime}=\left(\frac{\cos \ln 7 \cdot \sin ^{2} 7 x}{7 \cos 14 x}\right)^{\prime}=\frac{\cos \ln 7}{7} \cdot\left(\frac{\sin ^{2} 7 x}{\cos 14 x}\right)^{\prime}=$
$=\frac{\cos \ln 7}{7} \cdot \frac{2 \sin 7 x \cdot \cos 7 x \cdot 7 \cdot \cos 14 x-\sin ^{2} 7 x \cdot(-\sin 14 x) \cdot 14}{\cos ^{2} 14... | \frac{\cos\ln7\cdot\operatorname{tg}14x}{\cos14x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,215 |
## Problem Statement
Find the derivative.
$y=\frac{1}{4} \cdot \ln \frac{x-1}{x+1}-\frac{1}{2} \cdot \operatorname{arctg} x$ | ## Solution
$y^{\prime}=\left(\frac{1}{4} \cdot \ln \frac{x-1}{x+1}-\frac{1}{2} \cdot \operatorname{arctg} x\right)^{\prime}=$
$=\frac{1}{4} \cdot \frac{x+1}{x-1} \cdot\left(\frac{x-1}{x+1}\right)^{\prime}-\frac{1}{2} \cdot \frac{1}{1+x^{2}}=$
$=\frac{1}{4} \cdot \frac{x+1}{x-1} \cdot \frac{1 \cdot(x+1)-(x-1) \cdot ... | \frac{1}{x^{4}-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,216 |
## Problem Statement
Find the derivative.
$$
y=\frac{1}{2 a \sqrt{1+a^{2}}} \ln \frac{a+\sqrt{1+a^{2}} \tanh x}{a-\sqrt{1+a^{2}} \tanh x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2 a \sqrt{1+a^{2}}} \ln \frac{a+\sqrt{1+a^{2}} \operatorname{th} x}{a-\sqrt{1+a^{2}} \operatorname{th} x}\right)^{\prime}= \\
& =\frac{1}{2 a \sqrt{1+a^{2}}} \cdot \frac{a-\sqrt{1+a^{2}} \operatorname{th} x}{a+\sqrt{1+a^{2}} \operatorname{th} x} \cdot\left(\fr... | \frac{1}{^{2}\cdot\operatorname{ch}^{2}x+(1+^{2})\cdot\operatorname{sh}^{2}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,217 |
Problem condition
Find the derivative.
$y=\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{3 x-1}{\sqrt{2}}+\frac{1}{3} \cdot \frac{3 x-1}{3 x^{2}-2 x+1}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sqrt{2}} \cdot \operatorname{arctg} \frac{3 x-1}{\sqrt{2}}+\frac{1}{3} \cdot \frac{3 x-1}{3 x^{2}-2 x+1}\right)^{\prime}= \\
& =\frac{1}{\sqrt{2}} \cdot \frac{1}{1+\left(\frac{3 x-1}{\sqrt{2}}\right)^{2}} \cdot \frac{3}{\sqrt{2}}+\frac{1}{3} \cdot \frac{3\lef... | \frac{4}{3(3x^{2}-2x+1)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,219 |
## Task Condition
Find the derivative.
$$
y=2 \arcsin \frac{2}{3 x+4}+\sqrt{9 x^{2}+24 x+12}, 3 x+4>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(2 \arcsin \frac{2}{3 x+4}+\sqrt{9 x^{2}+24 x+12}\right)^{\prime}= \\
& =2 \cdot \frac{1}{\sqrt{1-\left(\frac{2}{3 x+4}\right)^{2}}}+\frac{1}{2 \sqrt{9 x^{2}+24 x+12}} \cdot(18 x+24)= \\
& =2 \cdot \frac{3 x+4}{\sqrt{(3 x+4)^{2}-4}}+\frac{6(3 x+4)}{2 \sqrt{9 x^{2}+24 x+... | \frac{8(3x+4)}{\sqrt{9x^{2}+24x+12}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,220 |
## Task Condition
Find the derivative.
$y=\frac{7^{x}(3 \sin 3 x+\cos 3 x \cdot \ln 7)}{9+\ln ^{2} 7}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{7^{x}(3 \sin 3 x+\cos 3 x \cdot \ln 7)}{9+\ln ^{2} 7}\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 7} \cdot\left(7^{x}(3 \sin 3 x+\cos 3 x \cdot \ln 7)\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 7} \cdot\left(7^{x} \cdot \ln 7 \cdot(3 \sin 3 x+\cos 3 x \cdot \ln 7)... | 7^{x}\cdot\cos3x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,221 |
## Condition of the problem
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\operatorname{ctg}\left(2 e^{t}\right) \\
y=\ln \left(\operatorname{tg} e^{t}\right)
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\operatorname{ctg}\left(2 e^{t}\right)\right)^{\prime}=\frac{-1}{\sin ^{2}\left(2 e^{t}\right)} \cdot 2 e^{t}=\frac{-2 e^{t}}{\sin ^{2}\left(2 e^{t}\right)} \\
& y_{t}^{\prime}=\left(\ln \left(\operatorname{tg} e^{t}\right)\right)^{\prime}=\frac{1}{\operatorname{tg... | -\sin(2e^{}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,222 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=t(t \cdot \cos t-2 \sin t) \\
y=t(t \cdot \sin t+2 \cos t)
\end{array}\right. \\
& t_{0}=\frac{\pi}{4}
\end{aligned}
$$ | ## Solution
Since $t_{0}=\frac{\pi}{4}$, then
$x_{0}=\frac{\pi}{4} \cdot\left(\frac{\pi}{4} \cdot \cos \frac{\pi}{4}-2 \sin \frac{\pi}{4}\right)=\frac{\pi}{4} \cdot\left(\frac{\pi}{4} \cdot \frac{\sqrt{2}}{2}-2 \cdot \frac{\sqrt{2}}{2}\right)=$
$=\frac{\pi}{4} \cdot\left(\frac{\pi \cdot \sqrt{2}}{8}-\sqrt{2}\right)$... | -x+\frac{\pi^{2}\cdot\sqrt{2}}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,223 |
## Task Condition
Find the $n$-th order derivative.
$y=\frac{x}{2(3 x+2)}$ | ## Solution
$$
\begin{aligned}
& y=\frac{x}{2(3 x+2)} \\
& y^{\prime}=\left(\frac{x}{2(3 x+2)}\right)^{\prime}=\frac{1 \cdot(3 x+2)-x \cdot 3}{2(3 x+2)^{2}}=\frac{3 x+2-3 x}{2(3 x+2)^{2}}= \\
& =\frac{2}{2(3 x+2)^{2}}=\frac{1}{(3 x+2)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{(3 x+2)^... | y^{(n)}=\frac{(-1)^{n-1}\cdotn!\cdot3^{n-1}}{(3x+2)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,224 |
## Task Condition
Find the derivative of the specified order.
$y=x^{2} \cdot \sin (5 x-3), y^{\prime \prime \prime}=?$ | ## Solution
$y^{\prime}=\left(x^{2} \cdot \sin (5 x-3)\right)^{\prime}=2 x \cdot \sin (5 x-3)+x^{2} \cdot \cos (5 x-3) \cdot 5=$ $=2 x \cdot \sin (5 x-3)+5 x^{2} \cdot \cos (5 x-3)$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2 x \cdot \sin (5 x-3)+5 x^{2} \cdot \cos (5 x-3)\right)^{\prime}=$
$=2 \sin ... | -150x\cdot\sin(5x-3)+(30-125x^{2})\cdot\cos(5x-3) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,225 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sqrt{t} \\
y=\frac{1}{\sqrt{1-t}}
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
& x_{t}^{\prime}=(\sqrt{t})^{\prime}=\frac{1}{2 \sqrt{t}} \\
& y_{t}^{\prime}=\left(\frac{1}{\sqrt{1-t}}\right)^{\prime}=\left((1-t)^{-\frac{1}{2}}\right)^{\prime}=-\frac{1}{2} \cdot(1-t)^{-\frac{3}{2}} \cdot(-1)= \\
& =\frac{1}{2 \sqrt{(1-t)^{3}}}
\end{aligned}
$$
We obtain:
$$
\begin... | (1+2)\sqrt{1-} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,226 |
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
\frac{x^{2}}{9}+y^{2}=1, z=y, z=0(y \geq 0)
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=1 \\
& y=0 \text { when } x=3
\end{aligned}
$$
That is, $x \in[-3,3], y \in[0,1]$
Consider the surface $z=y:$
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=2 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$
That is, $x \in[-\sqrt{3}, \sqrt{3}], y \in[0,2]$
Consider the surface $z=y \sqrt{3}$:
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=4 \\
& y=0 \text { when } x=\sqrt{3}
\end{aligned}
$$
That is, $x$ belongs to the interval $[-\sqrt{3}, \sqrt{3}]$, and $y \in [0,4]$
Consider the surface $z=y \sqrt{3}$:
$$
V_{z}=\int_{0}^{4} z d y=\i... | 32 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,243 |
## Task Condition
Calculate the volumes of bodies bounded by surfaces.
$$
z=x^{2}+5 y^{2}, z=5
$$ | ## Solution
In the section of the given figure by the plane $z=$ const, there is an ellipse:
$$
x^{2}+5 y^{2}=z
$$
The area of the ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\pi \cdot a \cdot b$
Let's find the radii of the ellipse:
$$ | ## Solution
The base of the considered area is a semi-ellipse, in which
$$
\begin{aligned}
& x=0 \text { when } y=5 \\
& y=0 \text { when } x=\sqrt{27}=3 \sqrt{3}
\end{aligned}
$$
That is,
$x \in[-3 \sqrt{3}, 3 \sqrt{3}], y \in[0,5]$
Consider the surface $z=\frac{y}{\sqrt{3}}:$
![](https://cdn.mathpix.com/cropped... | 1250 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,248 |
## Task Condition
Calculate the volumes of bodies bounded by surfaces.
$$
z=4 x^{2}+9 y^{2}, z=6
$$ | ## Solution
In the section of the given figure by the plane $z=$ const, there is an ellipse:
$$
4 x^{2}+9 y^{2}=z
$$
The area of the ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\pi \cdot a \cdot b$
Let's find the radii of the ellipse:
![](https://cdn.mathpix.com/cropped/2024_0... | 3\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,249 |
## Problem Statement
Calculate the volumes of the bodies bounded by the surfaces.
$$
x^{2}+\frac{y^{2}}{4}-z^{2}=1, z=0, z=3
$$ | ## Solution
In the section of the given figure by the plane $z=$ const, there is an ellipse:
$$
x^{2}+\frac{y^{2}}{4}=z^{2}+1
$$
The area of the ellipse described by the formula:
$$
\begin{aligned}
& \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { is } \\
& \pi \cdot a \cdot b
\end{aligned}
$$
Let's find the rad... | 24\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,250 |
## Problem Statement
Calculate the volumes of bodies bounded by the surfaces.
$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}-\frac{z^{2}}{100}=-1, z=20
$$ | ## Solution
In the section of the given figure by the plane $z=$ const, there is an ellipse:
$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}=\frac{z^{2}}{100}-1
$$
The area of an ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
is $\pi \cdot a \cdot b$
Let's find the radii of the ellipse:
![](htt... | 200\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,251 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.