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## Problem Statement
Calculate the definite integral:
$$
\int_{\frac{\pi}{2}}^{2 \operatorname{arctan} 2} \frac{d x}{\sin x(1+\sin x)}
$$ | ## Solution
## Method 1
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg} \frac{\pi}{4}=1 \\... | \ln2-\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,945 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{d x}{(1+\cos x+\sin x)^{2}}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow... | 1-\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,946 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x \, dx}{2+\sin x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left... | \frac{\pi}{2}-\frac{2\pi}{3\sqrt{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,947 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{4}} \frac{d x}{\cos x(1+\cos x)}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0
\end{aligned}
$$
Since $\operat... | \frac{\sqrt{2}-2}{\sqrt{2}}-\ln(\sqrt{2}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,948 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin x d x}{5+3 \sin x}
$$ | ## Solution
We will use the universal substitution:
$$
t=\operatorname{tg} \frac{x}{2}
$$
From which:
$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left... | \frac{\pi-5\operatorname{arctg}2+5\operatorname{arctg}\frac{3}{4}}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,949 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{13 ; 2 ; 7\}$
$p=\{5 ; 1 ; 0\}$
$q=\{2 ;-1 ; 3\}$
$r=\{1 ; 0 ;-1\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 3p+q-4r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,950 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{1 ; 4 ;-2\}$
$b=\{1 ; 1 ;-1\}$
$c_{1}=a+b$
$c_{2}=4 a+2 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=a+b=\{1+1 ; 4+1 ;-2+(-1)\}=\{2 ; 5 ;-3\} \\
& c_{2}=4 a+2 b=\{4 \cdot 1+2 \cdot 1 ; 4 \cdot 4+2 \cdot 1 ;... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,951 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(5 ; 3 ;-1), B(5 ; 2 ; 0), C(6 ; 4 ;-1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(5-5 ; 2-3 ; 0-(-1))=(0 ;-1 ; 1)$
$\overrightarrow{A C}=(6-5 ; 4-3 ;-1-(-1))=(1 ; 1 ; 0)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightar... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,952 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=p+3q$
$b=p-2q$
$|p|=2$
$|q|=3$
$(\widehat{p, q})=\frac{\pi}{3}$ | ## 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+3 q) \times(p-2 q)=p \times p-2 \cdot p \times q+3 \cdot q \times p+3 \cdot(-2... | 15\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,953 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 1 ;-1\}$
$b=\{-2 ;-1 ; 0\}$
$c=\{5 ; 2 ;-1\}$ | ## 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.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 1 & -1 \\
-2 & -1 & 0 \\
5 & 2 & -1
\end{array}\right|= \\
& =3 \cdot\left|... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,954 |
## problem statement
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4}$ and its height dropped from vertex $A_{4}$ to the face $A_{1} A_{2} A_{3}$.
$A_{1}(0 ;-1 ;-1)$
$A_{2}(-2 ; 3 ; 5)$
$A_{3}(1 ;-5 ;-9)$
$A_{4}(-1 ;-6 ; 3)$ | ## Solution
From vertex $A_{1}$, we will draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-2-0 ; 3-(-1) ; 5-(-1)\}=\{-2 ; 4 ; 6\} \\
& \overrightarrow{A_{1} A_{3}}=\{1-0 ;-5-(-1) ;-9-(-1)\}=\{1 ;-4 ;-8\} \\
& \overrightarrow{A_{1} A_{4}}=\{-1-0 ;-6-(-1) ; 3-(-1)\}=\{-1 ;-5 ; 4\}
\end{aligned}
$$
Acc... | \frac{37}{3\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,955 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(1 ; 0 ; 2)$
$M_{2}(1 ; 2 ;-1)$
$M_{3}(2 ;-2 ; 1)$
$M_{0}(-5 ;-9 ; 1)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-1 & y-0 & z-2 \\
1-1 & 2-0 & -1-2 \\
2-1 & -2-0 & 1-2
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y & z-2 \\
0 & 2 & -3 \\
1 & -2 & -1... | \sqrt{77} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,956 |
## Task Condition
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-3 ; 5 ;-2)$
$B(-4 ; 0 ; 3)$
$C(-3 ; 2 ; 5)$ | ## Solution
Let's find the vector $\overrightarrow{B C}:$
$\overrightarrow{B C}=\{-3-(-4) ; 2-0 ; 5-3\}=\{1 ; 2 ; 2\}$
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:
$$
\begin{aligned}
& (x-(-3))+2 \c... | x+2y+2z-3=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,957 |
## Task Condition
Find the angle between the planes
$x-y \sqrt{2}+z-1=0$
$x+y \sqrt{2}-z+3=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{1 ;-\sqrt{2} ; 1\}$
$\overrightarrow{n_{2}}=\{1 ; \sqrt{2} ;-1\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{align... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,958 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(-5 ;-5 ; 6)$
$C(-7 ; 6 ; 2)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$$
\begin{aligned}
& A B=\sqrt{(-5-0)^{2}+(-5-0)^{2}+(6-z)^{2}}=\sqrt{25+25+36-12 z+z^{2}}=\sqrt{z^{2}-12 z+86} \\
& A C=\sqrt{(-7-0)^{2}+(6-0)^{2}+(2-z)^{2}}=\sqrt{49+36+4-4 z+z^{2}}=\sqrt{z^{2}-4 z+89}
\end{aligned}
$$
Since by the condition of the problem $A B... | A(0;0;-\frac{3}{8}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,959 |
## Task Condition
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 $a$?
$A\left(\frac{1}{2} ; \frac{1}{3} ; 1\right)$
$a: 2 x-3 y+3 z-2=0$
$k=1.5$ | ## 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 $a$:
$a^{\prime}: 2 x-3 y+3 z-3=0$
Substitute the coordinates of point $A$ into the equ... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,960 |
Condition of the problem
Write the canonical equations of the line.
\[
\begin{aligned}
& 3 x+y-z-6=0 \\
& 3 x-y+2 z=0
\end{aligned}
\] | ## 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-1}{1}=\frac{y-3}{-9}=\frac{z}{-6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,961 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x+1}{-3}=\frac{y+2}{2}=\frac{z-3}{-2}$
$x+3 y-5 z+9=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x+1}{-3}=\frac{y+2}{2}=\frac{z-3}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-1-3 t \\
y=-2+2 t \\
z=3-2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(-1-3 t)+3(-2+2 t)-5(3-2 t)+9=0$
... | (-4;0;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,962 |
## Task Condition
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$M(2 ; 1 ; 0)$
$\frac{x-2}{0}=\frac{y+1.5}{-1}=\frac{z+0.5}{1}$ | ## 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:
$\vec{n}=\vec{s}=\{0 ;-1 ; 1\}$
Then the equation of the desired plane is:
$0 ... | M^{\}(2;-2;-3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,963 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{1+3 n}{6-n}, a=-3$ | ## Solution
By the definition of the limit:
$$
\begin{aligned}
& \forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right|<\varepsilon
\end{aligned}
$$
$$
\left|\frac{1+3 n+18-3 n}{6-n}\right|
$$
$$
\left|\frac{19}{n-6}\right|
$$
$$
\frac{19}{n-6}
$$
$$
n-... | N(\varepsilon)=7+[\frac{19}{\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 46,964 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n+6}-\sqrt{n^{2}-5}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n+6}-\sqrt{n^{2}-5}\right)}{\frac{1}{n}\left(\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{3}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \fra... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,966 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \sqrt[3]{n}\left(\sqrt[3]{n^{2}}-\sqrt[3]{n(n-1)}\right)
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \sqrt[3]{n}\left(\sqrt[3]{n^{2}}-\sqrt[3]{n(n-1)}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n}\left(\sqrt[3]{n^{2}}-\sqrt[3]{n(n-1)}\right)\left(\sqrt[3]{n^{4}}+\sqrt[3]{n^{2}} \cdot \sqrt[3]{n(n-1)}+\sqrt[3]{n^{2}(n-1)^{2}}\right)}{\sqrt[3]... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,967 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^{n}+7^{n}}{2^{n}-7^{n-1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{7^{n}}\left(2^{n}+7^{n}\right)}{\frac{1}{7^{n}}\left(2^{n}-7^{n-1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\frac{2}{7}\right)^{n}+1}{\left(\frac{2}{7}\right)^{n}-\... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,968 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty}\left(\frac{n^{3}+n+1}{n^{3}+2}\right)^{2 n^{2}}
$$ | ## Solution
$\lim _{n \rightarrow \infty}\left(\frac{n^{3}+n+1}{n^{3}+2}\right)^{2 n^{2}}=\lim _{n \rightarrow \infty}\left(\frac{n^{3}+2+n-1}{n^{3}+2}\right)^{2 n^{2}}=$
$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(1+\frac{n-1}{n^{3}+2}\right)^{2 n^{2}}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,969 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow \frac{1}{2}} \frac{2 x^{2}-5 x+2}{x-\frac{1}{2}}=-3$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | \delta(\varepsilon)=\frac{\varepsilon}{2} | Calculus | proof | Yes | Yes | olympiads | false | 46,970 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{)}$:
$f(x)=-2 x^{2}+9, x_{0}=4$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$$
\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, such that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\righ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 46,971 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{3}-2 x-1}{x^{4}+2 x+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{3}-2 x-1}{x^{4}+2 x+1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)\left(x^{2}-x-1\right)}{(x+1)\left(x^{3}-x^{2}+x+1\right)}= \\
& =\lim _{x \rightarrow-1} \frac{x^{2}-x-1}{x^{3}-x^{2}+x+1}=\frac{(-1)^{2}-(-1)-1}{(-1)^{3}-(-1)^{2}+(-1... | -\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,972 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt{x}-4)^{2}}}=\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{\sqrt[3]{(\sqrt[4]{x}-2)^{2}(\sqrt[4]{x}+2)^{2}}}= \\
& =\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{(\sqrt[4]{x}-2)^{\frac{2}{3}} \sqrt[3]{(\sqrt[4]{x}+2)^{2}}}=\l... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,973 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\operatorname{tg} x-\sin x}{x(1-\cos 2 x)}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
\(\sin x \sim x\), as \(x \rightarrow 0\)
\(1 - \cos x \sim \frac{x^2}{2}\), as \(x \rightarrow 0\)
\(\tan x \sim x\), as \(x \rightarrow 0\)
We get:
\[
\begin{aligned}
& \lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x(1 - \cos 2x)} = \left\{\... | \frac{1}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,974 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1} \frac{1-x^{2}}{\sin \pi x}$ | ## Solution
Substitution: u
$x=y+1 \Rightarrow y=x-1$
$x \rightarrow 1 \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow 1} \frac{1-x^{2}}{\sin \pi x}=\lim _{y \rightarrow 0} \frac{1-(y+1)^{2}}{\sin \pi(y+1)}=$
$=\lim _{y \rightarrow 0} \frac{1-\left(y^{2}+2 y+1\right)}{\sin (\pi y+\pi)}=\lim _{y \rightar... | \frac{2}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,975 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow a} \frac{a^{\left(x^{2}-a^{2}\right)}-1}{\tan \ln \left(\frac{x}{a}\right)}$ | ## Solution
Substitution:
$x=y+a \Rightarrow y=x-a$
$x \rightarrow a \Rightarrow y \rightarrow 0$
We get:
$\lim _{x \rightarrow a} \frac{a^{\left(x^{2}-a^{2}\right)}-1}{\tan \ln \left(\frac{x}{a}\right)}=\lim _{y \rightarrow 0} \frac{a^{\left((y+a)^{2}-a^{2}\right)}-1}{\tan \ln \left(\frac{y+a}{a}\right)}=$
$=\li... | 2^{2}\cdot\ln | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,976 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{3^{5 x}-2^{-7 x}}{2 x-\tan x}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{3^{5 x}-2^{-7 x}}{2 x-\operatorname{tg} x}=\lim _{x \rightarrow 0} \frac{\left(243^{x}-1\right)-\left(128^{-x}-1\right)}{2 x-\operatorname{tg} x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 243}\right)^{x}-1\right)-\left(\left(e^{\ln 128}\right... | \ln(3^{5}\cdot2^{7}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,977 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{\sqrt{\cos x}-1}{\sin ^{2} 2 x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{\cos x}-1}{\sin ^{2} 2 x}=\lim _{x \rightarrow 0} \frac{-(1-\sqrt{\cos x})(1+\sqrt{\cos x})}{\sin ^{2} 2 x(1+\sqrt{\cos x})}= \\
& =\lim _{x \rightarrow 0} \frac{-(1-\cos x)}{\sin ^{2} 2 x(1+\sqrt{\cos x})}=
\end{aligned}
$$
Using the substitution o... | -\frac{1}{16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,978 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{1+x \cdot 3^{x}}{1+x \cdot 7^{x}}\right)^{\frac{1}{x^2}}
$$ | ## Solution
$\lim _{x \rightarrow 0}\left(\frac{1+x \cdot 3^{x}}{1+x \cdot 7^{x}}\right)^{\frac{1}{\tan^{2} x}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(\left(1+x \cdot 3^{x}\right) /\left(1+x \cdot 7^{x}\right)\right)}\right)^{\frac{1}{\tan^{2} x}}=$
$=\lim _{x \rightarrow 0} e^{\frac{1}{\tan^{2} x} \ln \left(... | \frac{3}{7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,979 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^{\left(e^{x}-1\right) / x}=\left(\lim _{x \rightarrow 0} \tan\left(\frac{\pi}{4}-x\right)\right)^{\lim _{x \rightarrow 0}\left(e^{x}-1\right) / x}= \\
& =\left(\tan\left(\frac{\pi}{4}-0\right)\right)^{\lim _{x \righta... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,980 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 1}\left(\frac{x+1}{2 x}\right)^{\frac{\ln (x+2)}{\ln (2-x)}}$ | ## Solution
$\lim _{x \rightarrow 1}\left(\frac{x+1}{2 x}\right)^{\frac{\ln (x+2)}{\ln (2-x)}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{x+1}{2 x}\right)}\right)^{\frac{\ln (x+2)}{\ln (2-x)}}=$
$=\lim _{x \rightarrow 1} e^{\frac{\ln (x+2)}{\ln (2-x)} \cdot \ln \left(\frac{x+1}{2 x}\right)}=\exp \left\{\lim _{x ... | \sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,981 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}$ | ## Solution
$\lim _{x \rightarrow \frac{\pi}{2}}(\cos x+1)^{\sin x}=\left(\cos \frac{\pi}{2}+1\right)^{\sin \frac{\pi}{2}}=(0+1)^{1}=1^{1}=1$
## Problem Kuznetsov Limits 20-27 | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,982 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{2+\cos x \cdot \sin \frac{2}{2 x-\pi}}{3+2 x \sin x}$ | ## Solution
Since $\sin \frac{2}{2 x-\pi}$ is bounded, and $\cos x \rightarrow 0$ as $x \rightarrow \frac{\pi}{2}$, then
$\cos x \cdot \sin \frac{2}{2 x-\pi} \rightarrow 0$ as $x \rightarrow \frac{\pi}{2}$
Then:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{2+\cos x \cdot \sin \frac{2}{2 x-\pi}}{3+2 x \sin x}=\frac{2+... | \frac{2}{3+\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,983 |
## Task Condition
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{5 ; 15 ; 0\}$
$p=\{1 ; 0 ; 5\}$
$q=\{-1 ; 3 ; 2\}$
$r=\{0 ;-1 ; 1\}$ | ## 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:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 4p-q-18r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,984 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a_{\text {and }} b$, collinear?
$a=\{-2 ; 7 ;-1\}$
$b=\{-3 ; 5 ; 2\}$
$c_{1}=2 a+3 b$
$c_{2}=3 a+2 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
c_{1}=2 a+3 b=\{2 \cdot(-2)+3 \cdot(-3) ; 2 \cdot 7+3 \cdot 5 ; 2 \cdot(-1)+3 \cdot 2\}=\{-13 ; 29 ; 4\}
$$
$$
c_{2}=3 a+2 b=\{3... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,985 |
## problem statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(6 ; 2 ;-3), B(6 ; 3 ;-2), C(7 ; 3 ;-3)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(6-6 ; 3-2 ;-2-(-3))=(0 ; 1 ; 1)$
$\overrightarrow{A C}=(7-6 ; 3-2 ;-3-(-3))=(1 ; 1 ; 0)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overri... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,986 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p+3 q$
$b=p-2 q$
$|p|=6$
$|q|=7$
$(\widehat{p, q})=\frac{\pi}{3}$ | ## 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=(2 p+3 q) \times(p-2 q)=2 \cdot p \times p+2 \cdot(-2) \cdot p \times q+3 \cdot q... | 147\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,987 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{7 ; 3 ; 4\}$
$b=\{-1 ;-2 ;-1\}$
$c=\{4 ; 2 ; 4\}$ | ## 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.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
7 & 3 & 4 \\
-1 & -2 & -1 \\
4 & 2 & 4
\end{array}\right|= \\
& =7 \cdot\left|\... | -18\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,988 |
## problem statement
Calculate the volume of the tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4_{\text {and }}}$ its height dropped from vertex $A_{4 \text { to the face }} A_{1} A_{2} A_{3}$.
$A_{1}(1 ; 1 ; 2)$
$A_{2}(-1 ; 1 ; 3)$
$A_{3}(2 ;-2 ; 4)$
$A_{4}(-1 ; 0 ;-2)$ | ## Solution
From vertex $A_{1}$, we draw vectors:
$\overrightarrow{A_{1} A_{2}}=\{-1-1 ; 1-1 ; 3-2\}=\{-2 ; 0 ; 1\}$
$\overrightarrow{A_{1} A_{3}}=\{2-1 ;-2-1 ; 4-2\}=\{1 ;-3 ; 2\}$
$\overrightarrow{A_{1} A_{4}}=\{-1-1 ; 0-1 ;-2-2\}=\{-2 ;-1 ;-4\}$
According to the geometric meaning of the scalar triple product, w... | 5\frac{5}{6},\sqrt{\frac{35}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,989 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-3 ;-5 ; 6)$
$M_{2}(2 ; 1 ;-4)$
$M_{3}(0 ;-3 ;-1)$
$M_{0}(3 ; 6 ; 68)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{ccc}
x-(-3) & y-(-5) & z-6 \\
2-(-3) & 1-(-5) & -4-6 \\
0-(-3) & -3-(-5) & -1-6
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+3 & y+5 & z-6 \\
5 & ... | \sqrt{573} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,990 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-3 ; 7 ; 2)$
$B(3 ; 5 ; 1)$
$C(4 ; 5 ; 3)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{4-3 ; 5-5 ; 3-1\}=\{1 ; 0 ; 2\}$
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:
$1 \cdot(x+3)+0 \cdot(y-7)+2 \cdot(z-2)=0... | x+2z-1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,991 |
## Task Condition
Find the angle between the planes:
$3 x-2 y-2 z-16=0$
$x+y-3 z-7=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:
$\overrightarrow{n_{1}}=\{3 ;-2 ;-2\}$
$\overrightarrow{n_{2}}=\{1 ; 1 ;-3\}$
The angle $\phi$ between the planes is determined by the formula:
$$
\begin{aligned}
& \cos \ph... | \phi=\arccos\frac{7}{\sqrt{187}}\approx5912^{\}37^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,992 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; y ; 0)$
$B(1 ; 6 ; 4)$
$C(5 ; 7 ; 1)$ | ## Solution
Let's find the distances $A B$ and $A C$:
\[
\begin{aligned}
& A B=\sqrt{(1-0)^{2}+(6-y)^{2}+(4-0)^{2}}=\sqrt{1+36-12 y+y^{2}+16}=\sqrt{y^{2}-12 y+53} \\
& A C=\sqrt{(5-0)^{2}+(7-y)^{2}+(1-0)^{2}}=\sqrt{25+49-14 y+y^{2}+1}=\sqrt{y^{2}-14 y+75}
\end{aligned}
\]
Since by the condition of the problem $A B=A... | A(0;11;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,993 |
## 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 $a$?
$A(0 ; 1 ;-1)$
$a: 6 x-5 y+3 z-4=0$
$k=-\frac{3}{4}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0$ and the coefficient $k$, the plane transitions to
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 6 x-5 y+3 z+3=0$
Substitute the coordinates of point $A$ into the equati... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,994 |
## Task Condition
Write the canonical equations of the line.
$6 x-7 y-4 z-2=0$
$x+7 y-z-5=0$ | ## 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}{35}=\frac{y-\frac{4}{7}}{2}=\frac{z}{49} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,995 |
## Problem Statement
Find the point of intersection of the line and the plane.
$$
\begin{aligned}
& \frac{x+2}{-1}=\frac{y-1}{1}=\frac{z+3}{2} \\
& x+2 y-z-2=0
\end{aligned}
$$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x+2}{-1}=\frac{y-1}{1}=\frac{z+3}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-2-t \\
y=1+t \\
z=-3+2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$(-2-t)+2(1+t)-(-3+2 t)-2=0$
$-2-t+2+2 ... | (-3,2,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 46,996 |
## Condition of the problem
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$$
\begin{aligned}
& M(2 ;-2 ;-3) \\
& \frac{x-1}{-1}=\frac{y+0.5}{0}=\frac{z+1.5}{0}
\end{aligned}
$$ | ## 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:
$\vec{n}=\vec{s}=\{-1 ; 0 ; 0\}$
Then the equation of the desired plane is:
$-... | M^{\}(2;1;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 46,997 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{9 x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,998 |
## Condition of the problem
Compose the equation of the tangent to the given curve at the point with abscissa $x_{0}$.
$$
y=\frac{x^{29}+6}{x^{4}+1}, x_{0}=1
$$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{29}+6}{x^{4}+1}\right)^{\prime}=\frac{\left(x^{29}+6\right)^{\prime}\left(x^{4}+1\right)-\left(x^{29}+6\right)\left(x^{4}+1\right)^{\prime}}{\left(x^{4}+1\right)^{2}}= \\
& =\frac{29 x^{28}\left(x^{4}+1\right)-\left(x^{29}+6\right) \c... | 7.5x-4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 46,999 |
## Condition of the problem
Find the differential $d y$
$y=\ln \left(\tan \frac{x}{2}\right)-\frac{x}{\sin x}$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\ln \left(\tan \frac{x}{2}\right)-\frac{x}{\sin x}\right)^{\prime} d x= \\
& =\left(\frac{1}{\tan \frac{x}{2}} \cdot\left(\tan \frac{x}{2}\right)^{\prime}-\frac{x^{\prime} \cdot \sin x-x \cdot(\sin x)^{\prime}}{\sin ^{2} x}\right) d x= \\
& =\left(\frac{1... | \frac{x\cdot\cosx}{\sin^{2}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,000 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt[3]{x}, x=8,24$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 8$
Then:
$\Delta x = 0.24$
Calculate:
$y(8) = \sqrt[3]{8} = 2$
$y^{\prime} ... | 2.02 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,001 |
## Task Condition
Find the derivative.
$y=\frac{\sqrt{x-1}(3 x+2)}{4 x^{2}}$ | Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\sqrt{x-1}(3 x+2)}{4 x^{2}}\right)^{\prime}=\frac{\left(\frac{1}{2 \sqrt{x-1}} \cdot(3 x+2)+\sqrt{x-1} \cdot 3\right) \cdot x^{2}-\sqrt{x-1}(3 x+2) \cdot 2 x}{4 x^{4}}= \\
& =\frac{((3 x+2)+6(x-1)) \cdot x-4 \cdot(x-1)(3 x+2)}{8 x^{3} \sqrt{x-1}}= \\
& =\frac{(9 x-4... | \frac{-3x^{2}+8}{8x^{3}\sqrt{x-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,002 |
## Task Condition
Find the derivative.
$y=x+\frac{1}{1+e^{x}}-\ln \left(1+e^{x}\right)$ | ## Solution
$y^{\prime}=\left(x+\frac{1}{1+e^{x}}-\ln \left(1+e^{x}\right)\right)^{\prime}=1-\frac{1}{\left(1+e^{x}\right)^{2}} \cdot e^{x}-\frac{1}{1+e^{x}} \cdot e^{x}=$
$=\frac{1+2 e^{x}+e^{2 x}-e^{x}-e^{x}-e^{2 x}}{\left(1+e^{x}\right)^{2}}=\frac{1}{\left(1+e^{x}\right)^{2}}$
## Problem Kuznetsov Differentiation... | \frac{1}{(1+e^{x})^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,003 |
## Task Condition
Find the derivative.
$y=\log _{16} \log _{5} \operatorname{tg} x$ | ## Solution
$y^{\prime}=\left(\log _{16} \log _{5} \operatorname{tg} x\right)^{\prime}=\frac{1}{\log _{5} \operatorname{tg} x \cdot \ln 16} \cdot\left(\log _{5} \operatorname{tg} x\right)^{\prime}=$
$=\frac{1}{\log _{5} \operatorname{tg} x \cdot \ln 16} \cdot \frac{1}{\operatorname{tg} x \cdot \ln 5} \cdot \frac{1}{\... | \frac{1}{\sin2x\cdot\ln4\cdot\ln\operatorname{tg}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,004 |
## Task Condition
Find the derivative.
$y=\frac{\cos (\operatorname{ctg} 3) \cdot \cos ^{2} 14 x}{28 \sin 28 x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\cos (\operatorname{ctg} 3) \cdot \cos ^{2} 14 x}{28 \sin 28 x}\right)^{\prime}=\frac{\cos (\operatorname{ctg} 3)}{28} \cdot\left(\frac{\cos ^{2} 14 x}{\sin 28 x}\right)^{\prime}= \\
& =\frac{\cos (\operatorname{ctg} 3)}{28} \cdot\left(\frac{\cos ^{2} 14 x}{2 \si... | -\frac{\cos(\operatorname{ctg}3)}{4\sin^{2}14x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,005 |
## Task Condition
Find the derivative.
$$
y=\arcsin \sqrt{\frac{x}{x+1}}+\operatorname{arctg} \sqrt{x}
$$ | ## Solution
$y^{\prime}=\left(\arcsin \sqrt{\frac{x}{x+1}}+\operatorname{arctg} \sqrt{x}\right)^{\prime}=$
$=\frac{1}{\sqrt{1-\left(\sqrt{\frac{x}{x+1}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{\frac{x}{x+1}}} \cdot \frac{1 \cdot(x+1)-x \cdot 1}{(x+1)^{2}}+\frac{1}{1+(\sqrt{x})^{2}} \cdot \frac{1}{2 \sqrt{x}}=$
$=\frac{\... | \frac{1}{\sqrt{x}(x+1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,006 |
## Task Condition
Find the derivative.
$$
y=\frac{\sinh 3 x}{\sqrt{\cosh 6 x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\operatorname{sh} 3 x}{\sqrt{\operatorname{ch} 6 x}}\right)^{\prime}=\frac{3 \operatorname{ch} 3 x \cdot \sqrt{\operatorname{ch} 6 x}-\operatorname{sh} 3 x \cdot \frac{1}{2 \sqrt{\operatorname{ch} 6 x}} \cdot \operatorname{sh} 6 x \cdot 6}{\operatorname{ch} 6 x}=... | \frac{3\operatorname{ch}3x}{\operatorname{ch}6x\cdot\sqrt{\operatorname{ch}6x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,007 |
## Task Condition
Find the derivative.
$y=x^{\sin x^{3}}$ | ## Solution
$y=x^{\sin x^{3}}$
$\ln y=\ln x^{\sin x^{3}}=\sin x^{3} \cdot \ln x$
$\frac{y^{\prime}}{y}=\cos x^{3} \cdot 3 x^{2} \cdot \ln x+\sin x^{3} \cdot \frac{1}{x}=3 x^{2} \cdot \ln x \cdot \cos x^{3}+\frac{\sin x^{3}}{x}$
$y^{\prime}=y \cdot\left(3 x^{2} \cdot \ln x \cdot \cos x^{3}+\frac{\sin x^{3}}{x}\right... | x^{\sinx^{3}}\cdot(3x^{2}\cdot\lnx\cdot\cosx^{3}+\frac{\sinx^{3}}{x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,008 |
## Task Condition
Find the derivative.
$$
y=\sqrt{x^{2}-8 x+17} \cdot \operatorname{arctg}(x-4)-\ln \left(x-4+\sqrt{x^{2}-8 x+17}\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{x^{2}-8 x+17} \cdot \operatorname{arctg}(x-4)-\ln \left(x-4+\sqrt{x^{2}-8 x+17}\right)\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{x^{2}-8 x+17}} \cdot(2 x-8) \cdot \operatorname{arctg}(x-4)+\sqrt{x^{2}-8 x+17} \cdot \frac{1}{1+(x-4)^{2}}- \\
& -\frac{1}{x-4+\sqrt{x^... | \frac{x-4}{\sqrt{x^{2}-8x+17}}\cdot\operatorname{arctg}(x-4) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,009 |
## Task Condition
Find the derivative.
$$
y=4 \arcsin \frac{4}{2 x+3}+\sqrt{4 x^{2}+12 x-7}, 2 x+3>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(4 \arcsin \frac{4}{2 x+3}+\sqrt{4 x^{2}+12 x-7}\right)^{\prime}= \\
& =4 \cdot \frac{1}{\sqrt{1-\left(\frac{4}{2 x+3}\right)^{2}}} \cdot\left(-\frac{4}{(2 x+3)^{2}} \cdot 2\right)+\frac{1}{2 \sqrt{4 x^{2}+12 x-7}} \cdot(8 x+12)= \\
& =-\frac{4(2 x+3)}{\sqrt{(2 x+3)^{2}... | \frac{2\sqrt{4x^{2}+12x-7}}{2x+3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,010 |
## Task Condition
Find the derivative.
$$
y=\operatorname{arctg} \frac{\sqrt{\sqrt{x^{4}+1}-x^{2}}}{x}, x>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{\sqrt{\sqrt{x^{4}+1}-x^{2}}}{x}\right)^{\prime}=\frac{1}{1+\left(\frac{\sqrt{\sqrt{x^{4}+1}-x^{2}}}{x}\right)^{2}} \cdot\left(\frac{\sqrt{\sqrt{x^{4}+1}-x^{2}}}{x}\right)^{\prime}=
\end{aligned}
$$
\cdot\sqrt{\sqrt{x^{4}+1}-x^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,011 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\frac{t}{\sqrt{1-t^{2}}} \\
y=\ln \frac{1+\sqrt{1-t^{2}}}{t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\frac{t}{\sqrt{1-t^{2}}}\right)^{\prime}=\frac{1 \cdot \sqrt{1-t^{2}}-t \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)}{1-t^{2}}=$
$$
\begin{aligned}
& =\frac{1-t^{2}+t \cdot t}{\sqrt{\left(1-t^{2}\right)^{3}}}=\frac{1}{\sqrt{\left(1-t^{2}\right)^{3}}} \\
& y_{t}^{\prime}=\left(\ln \fr... | \frac{^{2}-1}{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,012 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=\frac{1+\ln t}{t^{2}} \\
y=\frac{3+2 \ln t}{t}
\end{array}
\right.
\]
$t_{0}=1$ | ## Solution
Since $t_{0}=1$, then
$x_{0}=\frac{1+\ln 1}{1^{2}}=1$
$y_{0}=\frac{3+2 \ln 1}{1}=3$
Let's find the derivatives:
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\frac{1+\ln t}{t^{2}}\right)^{\prime}=\frac{\frac{1}{t} \cdot t^{2}-(1+\ln t) \cdot 2 t}{t^{4}}= \\
& =\frac{1-2-2 \ln t}{t^{3}}=\frac{-1-2 \ln t}{t^... | x+2-x+4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,013 |
## Task Condition
Find the $n$-th order derivative.
$$
y=\frac{4+15 x}{5 x+1}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{4+15 x}{5 x+1}\right)^{\prime}=\frac{15 \cdot(5 x+1)-(4+15 x) \cdot 5}{(5 x+1)^{2}}=\frac{75 x+15-20-75 x}{(5 x+1)^{2}}= \\
& =-\frac{5}{(5 x+1)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{5}{(5 x+1)^{2}}\right)^{\prime}=-\frac{5 \cd... | y^{(n)}=\frac{(-1)^{n}\cdotn!\cdot5^{n}}{(5x+1)^{n+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,014 |
## Task Condition
Find the derivative of the specified order.
$$
y=\frac{\ln (3+x)}{3+x}, y^{\prime \prime \prime}=?
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\ln (3+x)}{3+x}\right)^{\prime}=\frac{\frac{1}{3+x} \cdot(3+x)-\ln (3+x) \cdot 1}{(3+x)^{2}}=\frac{1-\ln (3+x)}{(3+x)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1-\ln (3+x)}{(3+x)^{2}}\right)^{\prime}=\frac{-\frac{1}{3+x} \cdot(3+x)^... | \frac{11-6\ln(3+x)}{(3+x)^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,015 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\operatorname{sh} t \\
y=\operatorname{th}^{2} t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\operatorname{sh} t)^{\prime}=\operatorname{ch} t$
$y_{t}^{\prime}=\left(\operatorname{th}^{2} t\right)^{\prime}=2 \operatorname{th} t \cdot \frac{1}{\operatorname{ch}^{2} t}=\frac{2 \operatorname{sh} t}{\operatorname{ch}^{3} t}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t... | \frac{2-6\operatorname{sh}^{2}}{\operatorname{ch}^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,016 |
## Problem Statement
Show that the function $y$ satisfies equation (1).
$$
\begin{aligned}
& y=\sqrt[3]{2+3 x-3 x^{2}} \\
& y \cdot y^{\prime}=\frac{1-2 x}{y}
\end{aligned}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt[3]{2+3 x-3 x^{2}}\right)^{\prime}=\frac{1}{3 \sqrt[3]{\left(2+3 x-3 x^{2}\right)^{2}}} \cdot(3-6 x)= \\
& =\frac{1-2 x}{\sqrt[3]{\left(2+3 x-3 x^{2}\right)^{2}}}
\end{aligned}
$$
Substitute into equation (1):
$$
\sqrt[3]{2+3 x-3 x^{2}} \cdot \frac{1-2 x}{\sqrt[... | proof | Algebra | proof | Yes | Yes | olympiads | false | 47,017 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0):$
$f(x)=\left\{\begin{array}{c}\tan\left(x^{3}+x^{2} \sin \left(\frac{2}{x}\right)\right), x \neq 0 \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\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 | 47,018 |
## Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=\frac{4 x-x^{2}}{4}, x_{0}=2$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=\left(\frac{4 x-x^{2}}{4}\right)^{\prime}=\frac{4-2 x}{4}=\frac{2-x}{2}
$$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{2-x_{0}}{2}=\frac{2-2}{2}=0$
Since $y^{\prime}\left(x_{0}\right)=0$, the equation of the normal line is:
$x=x_{0}$
$x=2$
Thus, th... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,019 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt[3]{x}, x=7.76$ | ## 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:
$x_{0} = 8$
Then:
$\Delta x = -0.24$
Calculate:
$y(8) = \sqrt[3]{8} = 2$
$y^{... | 1.98 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,020 |
## Task Condition
Find the derivative.
$$
y=\frac{2\left(3 x^{3}+4 x^{2}-x-2\right)}{15 \sqrt{1+x}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2\left(3 x^{3}+4 x^{2}-x-2\right)}{15 \sqrt{1+x}}\right)^{\prime}=\frac{2}{15} \cdot \frac{\left(9 x^{2}+8 x-1\right) \cdot \sqrt{1+x}-\left(3 x^{3}+4 x^{2}-x-2\right) \cdot \frac{1}{2 \sqrt{1+x}}}{1+x}= \\
& =\frac{1}{15} \cdot \frac{2\left(9 x^{2}+8 x-1\right) ... | x\sqrt{1+x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,021 |
## Task Condition
Find the derivative.
$$
y=x-\ln \left(2+e^{x}+2 \sqrt{e^{2 x}+e^{x}+1}\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x-\ln \left(2+e^{x}+2 \sqrt{e^{2 x}+e^{x}+1}\right)\right)^{\prime}= \\
& =1-\frac{1}{2+e^{x}+2 \sqrt{e^{2 x}+e^{x}+1}} \cdot\left(e^{x}+2 \cdot \frac{1}{2 \sqrt{e^{2 x}+e^{x}+1}} \cdot\left(2 e^{2 x}+e^{x}\right)\right)= \\
& =\frac{\left(2+e^{x}+2 \sqrt{e^{2 x}+e^{x}... | \frac{1}{\sqrt{e^{2x}+e^{x}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,022 |
Condition of the problem
Find the derivative.
$$
y=\sqrt{x} \ln (\sqrt{x}+\sqrt{x+a})-\sqrt{x+a}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=(\sqrt{x} \ln (\sqrt{x}+\sqrt{x+a})-\sqrt{x+a})^{\prime}= \\
& =\frac{1}{2 \sqrt{x}} \cdot \ln (\sqrt{x}+\sqrt{x+a})+\sqrt{x} \cdot \frac{1}{\sqrt{x}+\sqrt{x+a}} \cdot\left(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{x+a}}\right)-\frac{1}{2 \sqrt{x+a}}= \\
& =\frac{1}{2 \sqrt{x}} ... | \frac{1}{2\sqrt{x}}\cdot\ln(\sqrt{x}+\sqrt{x+}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,023 |
## Task Condition
Find the derivative.
$$
y=\sin \sqrt{3}+\frac{1}{3} \cdot \frac{\sin ^{2} 3 x}{\cos 6 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sin \sqrt{3}+\frac{1}{3} \cdot \frac{\sin ^{2} 3 x}{\cos 6 x}\right)^{\prime}=\frac{1}{3} \cdot\left(\frac{\sin ^{2} 3 x}{\cos 6 x}\right)^{\prime}= \\
& =\frac{1}{3} \cdot \frac{2 \sin 3 x \cdot \cos 3 x \cdot 3 \cdot \cos 6 x-\sin ^{2} 3 x \cdot(-\sin 6 x \cdot 6)}{... | \frac{\sin6x}{\cos^{2}6x}=\frac{\tan6x}{\cos6x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,024 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{4 \sqrt{5}} \ln \frac{2+\sqrt{5} \tanh x}{2-\sqrt{5} \cdot \tanh x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{4 \sqrt{5}} \ln \frac{2+\sqrt{5} \tanh x}{2-\sqrt{5} \cdot \tanh x}\right)^{\prime}=\frac{1}{4 \sqrt{5}} \cdot \frac{2-\sqrt{5} \tanh x}{2+\sqrt{5} \cdot \tanh x} \cdot\left(\frac{2+\sqrt{5} \tanh x}{2-\sqrt{5} \cdot \tanh x}\right)^{\prime}= \\
& =\frac{1}{4 ... | \frac{1}{4-\sinh^{2}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,025 |
Condition of the problem
Find the derivative.
$$
y=\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}, x>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}\right)^{\prime}= \\
& =\frac{1}{24} \cdot 2 x \cdot \sqrt{x^{2}-4}+\frac{1}{24}\left(x^{2}+8\right) \cdot \frac{1}{2 \sqrt{x^{2}-4}} \cdot 2 x+\frac{2 x}{16} \cdot \arcsin \frac{2}{x}+... | \frac{x^{3}-x}{8\sqrt{x^{2}-4}}+\frac{x}{8}\cdot\arcsin\frac{2}{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,026 |
## Problem Statement
Find the derivative.
$$
y=\frac{x \cdot \arcsin x}{\sqrt{1-x^{2}}}+\ln \sqrt{1-x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x \cdot \arcsin x}{\sqrt{1-x^{2}}}+\ln \sqrt{1-x^{2}}\right)^{\prime}= \\
& =\frac{\left(\arcsin x+x \cdot \frac{1}{\sqrt{1-x^{2}}}\right) \cdot \sqrt{1-x^{2}}-x \cdot \arcsin x \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2 x)}{1-x^{2}}+ \\
& +\frac{1}{\sqrt{1-x^{2}}... | \frac{\arcsinx}{\sqrt{(1-x^2)^3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,027 |
## Task Condition
Find the derivative.
$y=\frac{1}{\sin \alpha} \ln (\tan x+\cot \alpha)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{\sin \alpha} \ln (\operatorname{tg} x+\operatorname{ctg} \alpha)\right)^{\prime}=\frac{1}{\sin \alpha} \cdot \frac{1}{\operatorname{tg} x+\operatorname{ctg} \alpha} \cdot \frac{1}{\cos ^{2} x}= \\
& =\frac{1}{\frac{\sin \alpha \cdot \sin x \cdot \cos ^{2} x}{\... | \frac{1}{\cosx\cdot\cos(\alpha-x)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,028 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\frac{3 t^{2}+1}{3 t^{3}} \\
y=\sin \left(\frac{t^{3}}{3}+t\right)
\end{array}\right.
$$ | ## Solution
$$
\begin{aligned}
x_{t}^{\prime} & =\left(\frac{3 t^{2}+1}{3 t^{3}}\right)^{\prime}=\left(\frac{1}{t}+\frac{1}{3 t^{3}}\right)^{\prime}=-\frac{1}{t^{2}}-\frac{1}{t^{4}}=-\frac{t^{2}+1}{t^{4}} \\
y_{t}^{\prime} & =\left(\sin \left(\frac{t^{3}}{3}+t\right)\right)^{\prime}=\cos \left(\frac{t^{3}}{3}+t\right)... | -^{4}\cdot\cos(\frac{^{3}}{3}+) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,029 |
## Problem Statement
Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=a \sin ^{3} t \\
y=a \cos ^{3} t
\end{array}\right.
\]
$t_{0}=\frac{\pi}{3}$ | ## Solution
Since $t_{0}=\frac{\pi}{3}$, then
$x_{0}=a \sin ^{3} \frac{\pi}{3}=a \cdot\left(\frac{\sqrt{3}}{2}\right)^{3}=\frac{3 \sqrt{3} \cdot a}{8}$
$y_{0}=a \cos ^{3} \frac{\pi}{3}=a \cdot\left(\frac{1}{2}\right)=\frac{a}{8}$
Let's find the derivatives:
$x_{t}^{\prime}=\left(a \sin ^{3} t\right)^{\prime}=a \cd... | -\frac{x}{\sqrt{3}}+\frac{}{2}\sqrt{3}\cdotx- | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,030 |
## Task Condition
Find the $n$-th order derivative.
$y=x \cdot e^{a x}$ | ## Solution
$y=x \cdot e^{a x}$
$y^{\prime}=\left(x \cdot e^{a x}\right)^{\prime}=e^{a x}+x \cdot e^{a x} \cdot a=(1+a \cdot x) \cdot e^{a x}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left((1+a \cdot x) \cdot e^{a x}\right)^{\prime}=a \cdot e^{a x}+(1+a \cdot x) \cdot e^{a x} \cdot a=$
$=(2+a \cdot x) \... | y^{(n)}=(n+\cdotx)\cdote^{}\cdot^{n-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,031 |
## Task Condition
Find the derivative of the specified order.
$y=\left(2 x^{2}-7\right) \ln (x-1), y^{V}=?$ | ## Solution
$y^{\prime}=\left(\left(2 x^{2}-7\right) \ln (x-1)\right)^{\prime}=4 x \cdot \ln (x-1)+\left(2 x^{2}-7\right) \cdot \frac{1}{x-1}=$
$=4 x \cdot \ln (x-1)+\frac{2 x^{2}-7}{x-1}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(4 x \cdot \ln (x-1)+\frac{2 x^{2}-7}{x-1}\right)^{\prime}=$
$=4 \ln (... | \frac{8(x^{2}-5x-11)}{(x-1)^{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,032 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\cos 2 t \\
y=2 \sec ^{2} t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\cos 2 t)^{\prime}=-2 \sin 2 t$
$y_{t}^{\prime}=\left(2 \sec ^{2} t\right)^{\prime}=\left(\frac{2}{\cos ^{2} t}\right)^{\prime}=-\frac{4}{\cos ^{3} t} \cdot(-\sin t)=\frac{4 \sin t}{\cos ^{3} t}$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{4 \sin t}{\cos... | \frac{1}{\cos^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,033 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
$y=x \cdot e^{-\frac{x^{2}}{2}}$
$x \cdot y^{\prime}=\left(1-x^{2}\right) y$ | ## Solution
$y^{\prime}=\left(x \cdot e^{-\frac{x^{2}}{2}}\right)^{\prime}=e^{-\frac{x^{2}}{2}}+x \cdot e^{-\frac{x^{2}}{2}} \cdot(-x)=e^{-\frac{x^{2}}{2}} \cdot\left(1-x^{2}\right)$
Substitute into equation (1):
$x \cdot e^{-\frac{x^{2}}{2}} \cdot\left(1-x^{2}\right)=\left(1-x^{2}\right) \cdot x \cdot e^{-\frac{x^{... | proof | Calculus | proof | Yes | Yes | olympiads | false | 47,034 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
3^{x^{2} \sin \frac{2}{x}}-1+2 x, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,035 |
## Condition of the problem
To derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=\frac{x}{x^{2}+1}, x_{0}=-2$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x}{x^{2}+1}\right)^{\prime}=\frac{x^{\prime}\left(x^{2}+1\right)-x\left(x^{2}+1\right)^{\prime}}{\left(x^{2}+1\right)^{2}}= \\
& =\frac{x^{2}+1-x \cdot 2 x}{\left(x^{2}+1\right)^{2}}=\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}}
\end{aligned}... | -\frac{3}{25}\cdotx-\frac{16}{25} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,036 |
## problem statement
Find the differential $d y$
$$
y=\operatorname{arctg}\left(\operatorname{tg} \frac{x}{2}+1\right)
$$ | ## Solution
$d y=y^{\prime} \cdot d x=\left(\operatorname{arctg}\left(\operatorname{tg} \frac{x}{2}+1\right)\right)^{\prime} d x=$
$=\frac{1}{1+\left(\operatorname{tg} \frac{x}{2}+1\right)^{2}} \cdot \frac{1}{\cos ^{2} \frac{x}{2}} \cdot \frac{1}{2} \cdot d x=\frac{1}{1+\operatorname{tg}^{2} \frac{x}{2}+2 \operatorna... | \frac{}{3+2\sinx+\cosx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,037 |
## Task Condition
Approximately calculate using the differential.
$y=x^{7}, x=2,002$ | ## 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:
$x_{0} = 2$
Then:
$\Delta x = 0.002$
Calculate:
$y(2) = 2^{7} = 128$
$y^{\prim... | 128.896 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,038 |
## Task Condition
Find the derivative.
$$
y=\frac{(2 x+1) \sqrt{x^{2}-x}}{x^{2}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{(2 x+1) \sqrt{x^{2}-x}}{x^{2}}\right)^{\prime}= \\
& =\frac{\left(2 \sqrt{x^{2}-x}+(2 x+1) \frac{1}{2 \sqrt{x^{2}-x}} \cdot(2 x-1)\right) \cdot x^{2}-(2 x+1) \sqrt{x^{2}-x} \cdot 2 x}{x^{4}}= \\
& =\frac{\left(4\left(x^{2}-x\right)+\left(4 x^{2}-1\right)\right) \... | \frac{3}{x^{2}\sqrt{x^{2}-x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,039 |
## Problem Statement
Find the derivative.
$$
y=\frac{1}{m \sqrt{a b}} \operatorname{arctan}\left(e^{m x} \cdot \sqrt{\frac{a}{b}}\right)
$$ | ## Solution
$y^{\prime}=\left(\frac{1}{m \sqrt{a b}} \operatorname{arctg}\left(e^{m x} \cdot \sqrt{\frac{a}{b}}\right)\right)^{\prime}=\frac{1}{m \sqrt{a b}} \cdot \frac{1}{1+\left(e^{m x} \cdot \sqrt{\frac{a}{b}}\right)^{2}} \cdot e^{m x} \cdot \sqrt{\frac{a}{b}} \cdot m=$
$=\frac{e^{m x}}{b+a \cdot e^{2 m x}}$
## ... | \frac{e^{x}}{b+\cdote^{2x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,040 |
## Task Condition
Find the derivative.
$$
y=\ln \left(\arcsin \sqrt{1-e^{2 x}}\right)
$$ | ## Solution
$y^{\prime}=\left(\ln \left(\arcsin \sqrt{1-e^{2 x}}\right)\right)^{\prime}=\frac{1}{\arcsin \sqrt{1-e^{2 x}}} \cdot\left(\arcsin \sqrt{1-e^{2 x}}\right)^{\prime}=$
$$
=\frac{1}{\arcsin \sqrt{1-e^{2 x}}} \cdot \frac{1}{\sqrt{1-\left(\sqrt{1-e^{2 x}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{1-e^{2 x}}} \cdot\le... | -\frac{e^{x}}{\sqrt{1-e^{2x}}\cdot\arcsin\sqrt{1-e^{2x}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,041 |
## Task Condition
Find the derivative.
$$
y=\sqrt{\tan 4}+\frac{\sin ^{2} 21 x}{21 \cos 42 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{\tan 4}+\frac{\sin ^{2} 21 x}{21 \cos 42 x}\right)^{\prime}=0+\left(\frac{\sin ^{2} 21 x}{21 \cos 42 x}\right)^{\prime}= \\
& =\frac{2 \sin 21 x \cdot \cos 21 x \cdot 21 \cdot \cos 42 x-\sin ^{2} 21 x \cdot(-\sin 42) \cdot 42}{21 \cos ^{2} 42 x}= \\
& =\frac{\sin... | \frac{\sin42x}{\cos^{2}42x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,042 |
## Task Condition
Find the derivative.
$y=\operatorname{arctg} x+\frac{5}{6} \cdot \ln \frac{x^{2}+1}{x^{2}+4}$ | $$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} x+\frac{5}{6} \cdot \ln \frac{x^{2}+1}{x^{2}+4}\right)^{\prime}= \\
& =\frac{1}{1+x^{2}}+\frac{5}{6} \cdot \frac{x^{2}+4}{x^{2}+1} \cdot \frac{2 x \cdot\left(x^{2}+4\right)-\left(x^{2}+1\right) \cdot 2 x}{\left(x^{2}+4\right)^{2}}= \\
& =\frac{1}{1+x^{2}}+\frac... | \frac{x^{2}+9}{(1+x^{2})\cdot(x^{2}+4)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,043 |
## Task Condition
Find the derivative.
$y=-\frac{1}{4} \arcsin \frac{5+3 \operatorname{ch} x}{3+5 \operatorname{ch} x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(-\frac{1}{4} \arcsin \frac{5+3 \operatorname{ch} x}{3+5 \operatorname{ch} x}\right)^{\prime}= \\
& =-\frac{1}{4} \cdot \frac{1}{\sqrt{1-\left(\frac{5+3 \operatorname{ch} x}{3+5 \operatorname{ch} x}\right)^{2}}} \cdot \frac{3 \operatorname{sh} x \cdot(3+5 \operatorname{... | \frac{1}{3+5\operatorname{ch}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,044 |
## Task Condition
Find the derivative.
$y=(\sin \sqrt{x})^{e^{1 / x}}$ | ## Solution
$y=(\sin \sqrt{x})^{e^{1 / x}}$
$\ln y=e^{\frac{1}{x}} \cdot \ln (\sin \sqrt{x})$
$\frac{y^{\prime}}{y}=\left(e^{\frac{1}{x}} \cdot \ln (\sin \sqrt{x})\right)^{\prime}=e^{\frac{1}{x}} \cdot \frac{1}{x} \cdot \ln (\sin \sqrt{x})+e^{\frac{1}{x}} \cdot \frac{1}{\sin \sqrt{x}} \cdot \cos \sqrt{x} \cdot \frac... | \frac{(\sin\sqrt{x})^{e^{1}{x}}\cdote^{\frac{1}{x}}\cdot(\frac{\ln(\sin\sqrt{x})}{x}+\frac{\tan\sqrt{x}}{2\sqrt{x}})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 47,045 |
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