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## Problem Statement
Find the derivative.
$y=\arcsin \left(e^{-4 x}\right)+\ln \left(e^{4 x}+\sqrt{e^{8 x}-1}\right)$ | ## Solution
$y^{\prime}=\left(\arcsin \left(e^{-4 x}\right)+\ln \left(e^{4 x}+\sqrt{e^{8 x}-1}\right)\right)^{\prime}=$
$=\frac{1}{\sqrt{1-\left(e^{-4 x}\right)^{2}}} \cdot e^{-4 x} \cdot(-4)+\frac{1}{e^{4 x}+\sqrt{e^{8 x}-1}} \cdot\left(e^{4 x}+\sqrt{e^{8 x}-1}\right)^{\prime}=$
$=-\frac{4}{e^{4 x} \sqrt{1-e^{-8 x}... | 4\sqrt{\frac{e^{4x}-1}{e^{4x}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,783 |
## Task Condition
Find the derivative.
$y=\sqrt{(3-x)(2+x)}+5 \arcsin \sqrt{\frac{x+2}{5}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{(3-x)(2+x)}+5 \arcsin \sqrt{\frac{x+2}{5}}\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{(3-x)(2+x)}} \cdot(-1 \cdot(2+x)+(3-x) \cdot 1)+5 \cdot \frac{1}{\sqrt{1-\left(\sqrt{\frac{x+2}{5}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{\frac{x+2}{5}}} \cdot \frac{1}{5}=
\end{alig... | \sqrt{\frac{3-x}{2+x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,784 |
## Task Condition
Find the derivative.
$$
y=\frac{5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)}{9+\ln ^{2} 5}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)}{9+\ln ^{2} 5}\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 5} \cdot\left(5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 5} \cdot\left(5^{x} \cdot \ln 5 \cdot(\sin 3 x \cdot \ln 5-3 \cos 3 x)... | 5^{x}\cdot\sin3x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,785 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\ln \left(1-t^{2}\right) \\
y=\arcsin \sqrt{1-t^{2}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\ln \left(1-t^{2}\right)\right)^{\prime}=\frac{1}{1-t^{2}} \cdot(-2 t)=\frac{-2 t}{1-t^{2}}$
$$
\begin{aligned}
& y_{t}^{\prime}=\left(\arcsin \sqrt{1-t^{2}}\right)^{\prime}=\frac{1}{\sqrt{1-\left(\sqrt{1-t^{2}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)= \\
& =\frac{1... | \frac{\sqrt{1-^{2}}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,786 |
Condition of the problem
To find the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$.
$\left\{\begin{array}{l}x=3 \cos t \\ y=4 \sin t\end{array}\right.$
$t_{0}=\frac{\pi}{4}$ | ## Solution
Since $t_{0}=\frac{\pi}{4}$, then
$x_{0}=3 \cos \frac{\pi}{4}=\frac{3 \sqrt{2}}{2}$
$y_{0}=4 \sin \frac{\pi}{4}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}$
Let's find the derivatives:
$x_{t}^{\prime}=(3 \cos t)^{\prime}=-3 \sin t$
$y_{t}^{\prime}=(4 \sin t)^{\prime}=4 \cos t$
$y_{x}^{\prime}=\frac{y_{t}^{\prime... | -\frac{4}{3}\cdotx+4\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,787 |
## Task Condition
Find the $n$-th order derivative.
$y=\sqrt{e^{3 x+1}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sqrt{e^{3 x+1}}\right)^{\prime}=\left(\left(e^{3 x+1}\right)^{\frac{1}{2}}\right)^{\prime}=\left(e^{\frac{3 x+1}{2}}\right)^{\prime}=e^{\frac{3 x+1}{2}} \cdot \frac{3}{2} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(e^{\frac{3 x+1}{2}} \cdot \frac{3}{... | (\frac{3}{2})^{n}\cdot\sqrt{e^{3x+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,788 |
## Task Condition
Find the derivative of the specified order.
$y=\frac{\ln x}{x^{5}}, y^{\prime \prime \prime}=?$ | ## Solution
$y^{\prime}=\left(\frac{\ln x}{x^{5}}\right)^{\prime}=\frac{\frac{1}{x} \cdot x^{5}-\ln x \cdot 5 x^{4}}{x^{10}}=\frac{1-5 \ln x}{x^{6}}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1-5 \ln x}{x^{6}}\right)^{\prime}=\frac{-5 \cdot \frac{1}{x} \cdot x^{6}-(1-5 \ln x) \cdot 6 x^{5}}{x^{1... | \frac{107-210\lnx}{x^{8}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,789 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$\left\{\begin{array}{l}x=e^{t} \\ y=\arcsin t\end{array}\right.$ | ## Solution
$x_{t}^{\prime}=\left(e^{t}\right)^{\prime}=e^{t}$
$y_{t}^{\prime}=(\arcsin t)^{\prime}=\frac{1}{\sqrt{1-t^{2}}}$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1}{\sqrt{1-t^{2}}}\right) / e^{t}=\frac{1}{e^{t} \cdot \sqrt{1-t^{2}}} \\
& \left(y_{x}^{\pri... | \frac{^{2}+-1}{e^{2}\cdot\sqrt{(1-^{2})^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,790 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
$y=\frac{2 x}{x^{3}+1}+\frac{1}{x}$
$x\left(x^{3}+1\right) y^{\prime}+\left(2 x^{3}-1\right) y=\frac{x^{3}-2}{x}$ | ## Solution
$y^{\prime}=\left(\frac{2 x}{x^{3}+1}+\frac{1}{x}\right)^{\prime}=\frac{2 \cdot\left(x^{3}+1\right)-2 x \cdot 3 x^{2}}{\left(x^{3}+1\right)^{2}}-\frac{1}{x^{2}}=$
$=\frac{2-4 x^{3}}{\left(x^{3}+1\right)^{2}}-\frac{1}{x^{2}}$
Substitute into equation (1):
$x\left(x^{3}+1\right) \cdot\left(\frac{2-4 x^{3}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 45,791 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{3 ; 3 ;-1\}$
$p=\{3 ; 1 ; 0\}$
$q=\{-1 ; 2 ; 1\}$
$r=\{-1 ; 0 ; 2\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\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+q-r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,792 |
## problem statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a_{\text {and }} b$, collinear?
$$
\begin{aligned}
& a=\{-2 ;-3 ;-2\} \\
& b=\{1 ; 0 ; 5\} \\
& c_{1}=3 a+9 b \\
& c_{2}=-a-3 b
\end{aligned}
$$ | ## 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 $c_{1}=-3(-a-3 b)=-3 c_{2}$ for any $a$ and $b$.
That is, $c_{1}=-3 \cdot c_{2}$, which means the vectors $c_{1... | c_{1}=-3\cdotc_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,793 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(0 ; 1 ;-2), B(3 ; 1 ; 2), C(4 ; 1 ; 1)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(3-0 ; 1-1 ; 2-(-2))=(3 ; 0 ; 4)$
$\overrightarrow{A C}=(4-0 ; 1-1 ; 1-(-2))=(4 ; 0 ; 3)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightar... | 0.96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,794 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
\[
\begin{aligned}
& a=p-4 q \\
& b=3 p+q \\
& |p|=1 \\
& |q|=2 \\
& (\widehat{p, q})=\frac{\pi}{6}
\end{aligned}
\] | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(p-4 q) \times(3 p+q)=3 \cdot p \times p+p \times q-4 \cdot 3 \cdot q \times p-4 ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,795 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 2 ; 1\}$
$b=\{1 ;-3 ;-7\}$
$c=\{1 ; 2 ; 3\}$ | ## 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 & 2 & 1 \\
1 & -3 & -7 \\
1 & 2 & 3
\end{array}\right|= \\
& =3 \cdot\left|\b... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,796 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(-2 ; 0 ;-4) \)
\( A_{2}(-1 ; 7 ; 1) \)
\( A_{3}(4 ;-8 ;-4) \)
\( A_{4}(1 ;-4 ; 6) \) | ## Solution
From vertex $A_{1 \text {, we draw vectors: }}$
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-1-(-2) ; 7-0 ; 1-(-4)\}=\{1 ; 7 ; 5\} \\
& A_{1} A_{3}=\{4-(-2) ;-8-0 ;-4-(-4)\}=\{6 ;-8 ; 0\} \\
& \overrightarrow{A_{1} A_{4}}=\{1-(-2) ;-4-0 ; 6-(-4)\}=\{3 ;-4 ; 10\}
\end{aligned}
$$
According to the ... | 83\frac{1}{3},5\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,797 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(-1 ; 2 ; 4)$
$M_{2}(-1 ;-2 ;-4)$
$M_{3}(3 ; 0 ;-1)$
$M_{0}(-2 ; 3 ; 5)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-(-1) & y-2 & z-4 \\ -1-(-1) & -2-2 & -4-4 \\ 3-(-1) & 0-2 & -1-4\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y-2 & z-4 \\
0 & -4 & -8 \\
4 &... | \frac{5}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,798 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(1 ; 9 ;-4)$
$B(5 ; 7 ; 1)$
$C(3 ; 5 ; 0)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{3-5 ; 5-7 ; 0-1\}=\{-2 ;-2 ;-1\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$-2 \cdot(x-1)-2 \cdot(y-9)-(z-(-4))=0$
... | -2x-2y-z+16=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,799 |
## Task Condition
Find the angle between the planes:
$x+2 y+2 z-3=0$
$16 x+12 y-15 z-1=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{1 ; 2 ; 2\}$
$\overrightarrow{n_{2}}=\{16 ; 12 ;-15\}$
The angle $\phi_{\text {between the planes is determined by the formula: }}$
$\cos \phi=\frac{\l... | \arccos\frac{2}{15}\approx82^{0}20^{\}16^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,800 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(0 ; 0 ; z)$
$B(-6 ; 7 ; 5)$
$C(8 ;-4 ; 3)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(-6-0)^{2}+(7-0)^{2}+(5-z)^{2}}=\sqrt{36+49+25-10 z+z^{2}}=\sqrt{z^{2}-10 z+110}$
$A C=\sqrt{(8-0)^{2}+(-4-0)^{2}+(3-z)^{2}}=\sqrt{64+16+9-6 z+z^{2}}=\sqrt{z^{2}-6 z+89}$
Since by the condition of the problem $A B=A C$, then
$$
\begin{aligned}
& \sqr... | A(0;0;5.25) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,801 |
## 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 $a$?
$A(2; -5; 4)$
$a: 5x + 2y - z + 3 = 0$
$k = \frac{4}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 5 x+2 y-z+4=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,802 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x+2}{-1}=\frac{y-1}{1}=\frac{z+4}{-1}$
$2 x-y+3 z+23=0$ | ## Solution
Let's write the parametric equations of the line.
$\frac{x+2}{-1}=\frac{y-1}{1}=\frac{z+4}{-1}=t \Rightarrow$
$\left\{\begin{array}{l}x=-2-t \\ y=1+t \\ z=-4-t\end{array}\right.$
Substitute into the equation of the plane:
$2(-2-t)-(1+t)+3(-4-t)+23=0$
$-4-2 t-1-t-12-3 t+23=0$
$-6 t+6=0$
$t=1$
Find t... | (-3,2,-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,804 |
Condition of the problem
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the line.
$$
\begin{aligned}
& M(0 ; 2 ; 1) \\
& \frac{x-1.5}{2}=\frac{y}{-1}=\frac{z-2}{1}
\end{aligned}
$$ | ## 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:
$\vec{n}=\vec{s}=\{2 ;-1 ; 1\}$
Then the equation of the desired plane is:
$2 ... | M^{\}(-1;0;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,805 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
2 x^{2}+x^{2} \cos \frac{1}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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 | 45,806 |
## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa \( x_{0} \).
$$
y=\frac{-2\left(x^{8}+2\right)}{3\left(x^{4}+1\right)}, x_{0}=1
$$ | ## Solution
Let's find $y^{\prime}:$
$$
\begin{aligned}
& y^{\prime}=\left(\frac{-2\left(x^{8}+2\right)}{3\left(x^{4}+1\right)}\right)^{\prime}=\frac{-2\left(x^{8}+2\right)^{\prime} \cdot 3\left(x^{4}+1\right)-(-2) \cdot\left(x^{8}+2\right) \cdot 3\left(x^{4}+1\right)^{\prime}}{9\left(x^{4}+1\right)^{2}}= \\
& =\frac... | -\frac{2}{3}\cdotx-\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,807 |
## Condition of the problem
Find the differential $d y$.
$y=\sqrt{\operatorname{ctg} x}-\frac{\sqrt{\operatorname{tg}^{3} x}}{3}$ | ## Solution
$d y=y^{\prime} \cdot d x=\left(\sqrt{\operatorname{ctg} x}-\frac{\sqrt{\operatorname{tg}^{3} x}}{3}\right)^{\prime} d x=\left(\frac{1}{2 \sqrt{\operatorname{ctg} x}} \cdot \frac{-1}{\sin ^{2} x}-\frac{1}{3} \cdot \frac{1}{2 \sqrt{\operatorname{tg}^{3} x}} \cdot 3 \operatorname{tg}^{2} x \cdot \frac{1}{\co... | -\frac{\sqrt{2}}{\cosx\cdot\sqrt{\sin^{3}2x}}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,808 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt[3]{x}, x=7.64$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 8$
Then:
$\Delta x = -0.36$
Calculate:
$y(8) = \sqrt[3]{8} = 2$
$y^{\prime}... | 1.97 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,809 |
## Task Condition
Find the derivative.
$y=\frac{x^{6}+8 x^{3}-128}{\sqrt{8-x^{3}}}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{6}+8 x^{3}-128}{\sqrt{8-x^{3}}}\right)^{\prime}=\frac{\left(6 x^{5}+24 x^{2}\right) \cdot \sqrt{8-x^{3}}-\left(x^{6}+8 x^{3}-128\right) \cdot \frac{1}{2 \sqrt{8-x^{3}}} \cdot\left(-3 x^{2}\right)}{8-x^{3}}= \\
& =\frac{2\left(6 x^{5}+24 x^{2}\right) \cdot\left... | \frac{9x^{5}}{2\sqrt{8-x^{3}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,810 |
## Task Condition
Find the derivative.
$y=x+\frac{8}{1+e^{x / 4}}$ | ## Solution
$y^{\prime}=\left(x+\frac{8}{1+e^{x / 4}}\right)^{\prime}=1-\frac{8}{\left(1+e^{x / 4}\right)^{2}} \cdot e^{\frac{x}{4}} \cdot \frac{1}{4}=$
$=\frac{\left(1+e^{x / 4}\right)^{2}}{\left(1+e^{x / 4}\right)^{2}}-\frac{2 e^{x / 4}}{\left(1+e^{x / 4}\right)^{2}}=\frac{1+2 e^{x / 4}+e^{x / 2}-2 e^{x / 4}}{\left... | \frac{1+e^{x/2}}{(1+e^{x/4})^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,811 |
## Task Condition
Find the derivative.
$$
y=\frac{x(\cos (\ln x)+\sin (\ln x))}{2}
$$ | ## Solution
$y^{\prime}=\left(\frac{x(\cos (\ln x)+\sin (\ln x))}{2}\right)^{\prime}=$
$=\frac{1}{2} \cdot\left(\cos (\ln x)+\sin (\ln x)+x \cdot\left(-\sin (\ln x) \cdot \frac{1}{x}+\cos (\ln x) \cdot \frac{1}{x}\right)\right)=$
$=\frac{1}{2} \cdot(\cos (\ln x)+\sin (\ln x)-\sin (\ln x)+\cos (\ln x))=$
$=\frac{1}{... | \cos(\lnx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,812 |
## Task Condition
Find the derivative.
$$
y=\frac{\sin \left(\tan \frac{1}{7}\right) \cdot \cos ^{2} 16 x}{32 \sin 32 x}
$$ | ## Solution
$y^{\prime}=\left(\frac{\sin \left(\tan \frac{1}{7}\right) \cdot \cos ^{2} 16 x}{32 \sin 32 x}\right)^{\prime}=$
$=\sin \left(\tan \frac{1}{7}\right) \cdot\left(\frac{\cos ^{2} 16 x}{64 \sin 16 x \cdot \cos 16 x}\right)^{\prime}=$
$=\frac{\sin \left(\tan \frac{1}{7}\right)}{64} \cdot\left(\frac{\cos 16 x... | -\frac{\sin(\tan\frac{1}{7})}{4\sin^{2}16x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,813 |
Problem condition
Find the derivative.
$y=6 \arcsin \frac{\sqrt{x}}{2}-\frac{6+x}{2} \cdot \sqrt{x(4-x)}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(6 \arcsin \frac{\sqrt{x}}{2}-\frac{6+x}{2} \cdot \sqrt{x(4-x)}\right)^{\prime}= \\
& =6 \cdot \frac{1}{\sqrt{1-\left(\frac{\sqrt{x}}{2}\right)^{2}}} \cdot \frac{1}{4 \sqrt{x}}-\left(\frac{1}{2} \cdot \sqrt{x(4-x)}+\frac{6+x}{2} \cdot \frac{1}{2 \sqrt{x(4-x)}} \cdot(4-2... | \frac{x^{2}-3}{\sqrt{x(4-x)}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,814 |
## Task Condition
Find the derivative.
$y=-\frac{12 \operatorname{sh}^{2} x+1}{3 \operatorname{sh}^{2} x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(-\frac{12 \operatorname{sh}^{2} x+1}{3 \operatorname{sh}^{2} x}\right)^{\prime}=\left(-4+\frac{1}{3 \operatorname{sh}^{2} x}\right)^{\prime}= \\
& =0+(-2) \cdot \frac{1}{3 \operatorname{sh}^{3} x} \cdot \operatorname{ch} x=-\frac{2 \operatorname{ch} x}{3 \operatorname{... | -\frac{2\operatorname{ch}x}{3\operatorname{sh}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,815 |
## Task Condition
Find the derivative.
$$
y=\left(x^{4}+5\right)^{\cot x}
$$ | ## Solution
$y=\left(x^{4}+5\right)^{\cot x}$
$\ln y=\ln \left(x^{4}+5\right)^{\cot x}=\cot x \cdot \ln \left(x^{4}+5\right)$
$\frac{y'}{y}=\frac{-1}{\sin ^{2} x} \cdot \ln \left(x^{4}+5\right)+\cot x \cdot \frac{1}{x^{4}+5} \cdot 4 x^{3}=$
$=\frac{4 x^{3} \cdot \cot x}{x^{4}+5}-\frac{\ln \left(x^{4}+5\right)}{\sin ... | (x^{4}+5)^{\cotx}\cdot(\frac{4x^{3}\cdot\cotx}{x^{4}+5}-\frac{\ln(x^{4}+5)}{\sin^{2}x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,816 |
## Task Condition
Find the derivative.
$$
y=\left(3 x^{2}-4 x+2\right) \sqrt{9 x^{2}-12 x+3}+(3 x-2)^{4} \cdot \arcsin \frac{1}{3 x-2}, 3 x-2>0
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\left(3 x^{2}-4 x+2\right) \sqrt{9 x^{2}-12 x+3}+(3 x-2)^{4} \cdot \arcsin \frac{1}{3 x-2}\right)^{\prime}= \\
& =(6 x-4) \sqrt{9 x^{2}-12 x+3}+\left(3 x^{2}-4 x+2\right) \cdot \frac{1}{2 \sqrt{9 x^{2}-12 x+3}} \cdot(18 x-12)+ \\
& +4(3 x-2)^{3} \cdot 3 \cdot \arcsin \... | 12(3x-2)^{3}\cdot\arcsin\frac{1}{3x-2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,817 |
## Task Condition
Find the derivative.
$y=(2+3 x) \sqrt{x-1}-\frac{3}{2} \operatorname{arctg} \sqrt{x-1}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left((2+3 x) \sqrt{x-1}-\frac{3}{2} \operatorname{arctg} \sqrt{x-1}\right)^{\prime}= \\
& =3 \cdot \sqrt{x-1}+(2+3 x) \cdot \frac{1}{2 \sqrt{x-1}}-\frac{3}{2} \cdot \frac{1}{1+(\sqrt{x-1})^{2}} \cdot \frac{1}{2 \sqrt{x-1}}= \\
& =\frac{3(x-1)}{\sqrt{x-1}}+\frac{2+3 x}{2 \sq... | \frac{18x^{2}-8x-3}{4x\sqrt{x-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,818 |
## Task Condition
Find the derivative.
$$
y=\operatorname{arctg} \frac{\sqrt{2 \operatorname{tg} x}}{1-\operatorname{tg} x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{\sqrt{2 \operatorname{tg} x}}{1-\operatorname{tg} x}\right)^{\prime}=\frac{1}{1+\left(\frac{\sqrt{2 \operatorname{tg} x}}{1-\operatorname{tg} x}\right)^{2}} \cdot\left(\frac{\sqrt{2 \operatorname{tg} x}}{1-\operatorname{tg} x}\right)^{\prime}... | \frac{1-\operatorname{tg}x+\sqrt{2\operatorname{tg}x}}{\sqrt{2\operatorname{tg}x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,819 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\ln \frac{1-t}{1+t} \\
y=\sqrt{1-t^{2}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\ln \frac{1-t}{1+t}\right)^{\prime}=\frac{1+t}{1-t} \cdot \frac{-1 \cdot(1+t)-(1-t) \cdot 1}{(1+t)^{2}}=$
$=\frac{-1-t-1+t}{1-t^{2}}=-\frac{2}{1-t^{2}}$
$y_{t}^{\prime}=\left(\sqrt{1-t^{2}}\right)^{\prime}=\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)=-\frac{t}{\sqrt{1-t^{2}}}$
We obtain:... | \frac{\cdot\sqrt{1-^{2}}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,820 |
## Problem Statement
Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=a \cdot \sin ^{3} t \\
y=a \cdot \cos ^{3} t
\end{array}\right.
\]
$t_{0}=\frac{\pi}{6}$ | ## Solution
Since $t_{0}=\frac{\pi}{6}$, then
$x_{0}=a \sin ^{3} \frac{\pi}{6}=a \cdot\left(\frac{1}{2}\right)^{3}=\frac{a}{8}$
$y_{0}=a \cos ^{3} \frac{\pi}{6}=a \cdot\left(\frac{\sqrt{3}}{2}\right)=\frac{3 \sqrt{3} \cdot a}{8}$
Let's find the derivatives:
$x_{t}^{\prime}=\left(a \sin ^{3} t\right)^{\prime}=a \cd... | \begin{aligned}& | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,821 |
## Task Condition
Find the $n$-th order derivative.
$y=7^{5 x}$ | ## Solution
$y^{\prime}=5 \ln (7) \cdot 7^{5 x}$
$y^{\prime \prime}=5^{2} \ln (7)^{2} \cdot 7^{5 x}$
$y^{\prime \prime \prime}=5^{3} \ln (7)^{3} \cdot 7^{5 x}$
Assume that $y^{(n)}=5^{n} \ln (7)^{n} \cdot 7^{5 x}$.
We will prove this by mathematical induction.
1.
$$
n=1 \Rightarrow y^{(n)}=5 \ln (7) 7^{5 x}
$$
... | y^{(n)}=5^{n}\ln(7)^{n}\cdot7^{5x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,822 |
## Task Condition
Find the derivative of the specified order.
$$
y=\left(x^{2}+3\right) \ln (x-3), y^{IV}=?
$$ | ## Solution
$y^{\prime}=\left(\left(x^{2}+3\right) \ln (x-3)\right)^{\prime}=2 x \cdot \ln (x-3)+\left(x^{2}+3\right) \cdot \frac{1}{x-3}=$
$=2 x \cdot \ln (x-3)+\frac{x^{2}+3}{x-3}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2 x \cdot \ln (x-3)+\frac{x^{2}+3}{x-3}\right)^{\prime}=$
$=2 \ln (x-3)+2 x... | \frac{-2x^{2}+24x-126}{(x-3)^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,823 |
## Task Condition
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\cos ^{2} t \\
y=\operatorname{tg}^{2} t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\cos ^{2} t\right)^{\prime}=2 \cos t \cdot(-\sin t)=-2 \cos t \cdot \sin t$
$y_{t}^{\prime}=\left(\operatorname{tg}^{2} t\right)^{\prime}=2 \operatorname{tg} t \cdot \frac{1}{\cos ^{2} t}=\frac{2 \sin t}{\cos ^{3} t}$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}... | \frac{2}{\cos^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,824 |
## Task Condition
Show that the function $y_{\text {satisfies equation (1). }}$.
$$
\begin{aligned}
& y=\operatorname{tg}(\ln 3 x) \\
& \left(1+y^{2}\right) d x=x \cdot d y
\end{aligned}
$$ | ## Solution
$y^{\prime}=(\tan(\ln 3 x))^{\prime}=\frac{1}{\cos ^{2}(\ln 3 x)} \cdot \frac{1}{3 x} \cdot 3=\frac{1}{x \cdot \cos ^{2}(\ln 3 x)}$
Equation (1):
$\left(1+y^{2}\right) d x=x \cdot d y$
$1+y^{2}=x \cdot \frac{d y}{d x}$
$1+y^{2}=x \cdot y^{\prime}$
Substitute $y_{\text {and }} y^{\prime}:$
$1+\tan^{2}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,825 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
\operatorname{arctg}\left(x \cos \left(\frac{1}{5 x}\right)\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## 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:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | f^{\}(0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,826 |
Condition of the problem
To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.
$y=x-x^{3}, x_{0}=-1$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=\left(x-x^{3}\right)^{\prime}=1-3 x^{2}$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=1-3 x_{0}^{2}=1-3 \cdot(-1)^{2}=1-3=-2$
Since $y^{\prime}\left(x_{0}\right) \neq 0$, the equation of the normal line is:
$y-y_{0}=-\frac{1}{y_{0}^{\prime}} \cdot\left(x-x_{0... | \frac{x}{2}+\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,827 |
## Problem Statement
Find the differential $d y$.
$y=\sqrt{1+2 x}-\ln |x+\sqrt{1+2 x}|$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=(\sqrt{1+2 x}-\ln |x+\sqrt{1+2 x}|)^{\prime} d x= \\
& =\left(\frac{1}{2 \sqrt{1+2 x}} \cdot 2-\frac{1}{x+\sqrt{1+2 x}} \cdot\left(1+\frac{1}{2 \sqrt{1+2 x}} \cdot 2\right)\right) d x=
\end{aligned}
$$
$$
\begin{aligned}
& =\left(\frac{1}{\sqrt{1+2 x}}-\frac{1... | \frac{x-1}{(x+\sqrt{1+2x})\sqrt{1+2x}}\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,828 |
## Task Condition
Calculate approximately using the differential.
$y=\frac{x+\sqrt{5-x^{2}}}{2}, x=0.98$ | ## Solution
If the increment $\Delta x = x - x_{0 \text{ of the argument }} x$ is small in absolute value, then $f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 1$
Then:
$\Delta x = -0.02$
Calculate:
\[
\begin{aligned}
& y(1) = \fr... | 1.495 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,829 |
## Task Condition
Find the derivative.
$y=\frac{x^{4}-8 x^{2}}{2\left(x^{2}-4\right)}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x^{4}-8 x^{2}}{2\left(x^{2}-4\right)}\right)^{\prime}=\frac{\left(4 x^{3}-16 x\right) \cdot\left(x^{2}-4\right)-\left(x^{4}-8 x^{2}\right) \cdot 2 x}{2\left(x^{2}-4\right)^{2}}= \\
& =\frac{4 x^{5}-16 x^{3}-16 x^{3}+64 x-2 x^{5}+16 x^{3}}{2\left(x^{2}-4\right)^{2... | x+\frac{16x}{(x^{2}-4)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,830 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{2} \cdot \operatorname{arctan} \frac{e^{x}-3}{2}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot \operatorname{arctg} \frac{e^{x}-3}{2}\right)^{\prime}=\frac{1}{2} \cdot \frac{1}{1+\left(\frac{e^{x}-3}{2}\right)^{2}} \cdot \frac{e^{x}}{2}= \\
& =\frac{e^{x}}{4} \cdot \frac{4}{4+\left(e^{x}-3\right)^{2}}=\frac{e^{x}}{4+e^{2 x}-6 e^{x}+9}=\frac{e^{... | \frac{e^{x}}{e^{2x}-6e^{x}+13} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,831 |
Condition of the problem
Find the derivative.
$y=2 \sqrt{x}-4 \ln (2+\sqrt{x})$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=(2 \sqrt{x}-4 \ln (2+\sqrt{x}))^{\prime}=2 \cdot \frac{1}{2 \sqrt{x}}-4 \cdot \frac{1}{2+\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}= \\
& =\frac{2+\sqrt{x}}{\sqrt{x}(2+\sqrt{x})}-\frac{2}{(2+\sqrt{x}) \sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x}(2+\sqrt{x})}
\end{aligned}
$$
## Problem Kuz... | \frac{\sqrt{x}}{\sqrt{x}(2+\sqrt{x})} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,832 |
## Problem Statement
Find the derivative.
$y=\frac{2 x-1}{4} \cdot \sqrt{2+x-x^{2}}+\frac{9}{8} \cdot \arcsin \frac{2 x-1}{3}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x-1}{4} \cdot \sqrt{2+x-x^{2}}+\frac{9}{8} \cdot \arcsin \frac{2 x-1}{3}\right)^{\prime}= \\
& =\frac{2}{4} \cdot \sqrt{2+x-x^{2}}+\frac{2 x-1}{4} \cdot \frac{1}{2 \sqrt{2+x-x^{2}}} \cdot(1-2 x)+\frac{9}{8} \cdot \frac{1}{\sqrt{1-\left(\frac{2 x-1}{3}\right)^{2... | \sqrt{2+x-x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,834 |
## Problem Statement
Find the derivative.
$y=\frac{1}{2} \cdot \ln \frac{1+\sqrt{\tanh x}}{1-\sqrt{\tanh x}}-\arctan \sqrt{\tanh x}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot \ln \frac{1+\sqrt{\tanh x}}{1-\sqrt{\tanh x}}-\arctan \sqrt{\tanh x}\right)^{\prime}= \\
& =\frac{1}{2} \cdot \frac{1-\sqrt{\tanh x}}{1+\sqrt{\tanh x}} \cdot \frac{\frac{1}{2 \sqrt{\tanh x}} \cdot \frac{1}{\cosh^{2} x} \cdot(1-\sqrt{\tanh x})-(1+\sqrt... | \sqrt{\tanhx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,835 |
## Task Condition
Find the derivative.
$y=(\sin x)^{5 e^{x}}$ | Solution
$y=(\sin x)^{5 e^{x}}$
$\ln y=5 e^{x} \cdot \ln (\sin x)$
$\frac{y^{\prime}}{y}=\left(5 e^{x} \cdot \ln (\sin x)\right)^{\prime}=5 e^{x} \cdot \ln (\sin x)+5 e^{x} \cdot \frac{1}{\sin x} \cdot \cos x=$
$=5 e^{x} \cdot \ln (\sin x)+5 e^{x} \cdot \operatorname{ctg} x=5 e^{x} \cdot(\ln (\sin x)+\operatorname{... | 5e^{x}\cdot(\sinx)^{5e^{x}}\cdot(\ln(\sinx)+\operatorname{ctg}x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,836 |
## Task Condition
Find the derivative.
$y=2 x-\ln \left(1+\sqrt{1-e^{4 x}}\right)-e^{-2 x} \cdot \arcsin \left(e^{2 x}\right)$ | ## Solution
$y^{\prime}=\left(2 x-\ln \left(1+\sqrt{1-e^{4 x}}\right)-e^{-2 x} \cdot \arcsin \left(e^{2 x}\right)\right)^{\prime}=$
$=2-\frac{1}{1+\sqrt{1-e^{4 x}}} \cdot \frac{1}{2 \sqrt{1-e^{4 x}}} \cdot\left(-4 e^{4 x}\right)-$
$-\left(-2 e^{-2 x} \cdot \arcsin \left(e^{2 x}\right)+e^{-2 x} \cdot \frac{1}{\sqrt{1... | 2e^{-2x}\cdot\arcsin(e^{2x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,837 |
## Task Condition
Find the derivative.
$$
y=x\left(2 x^{2}+5\right) \sqrt{x^{2}+1}+3 \ln \left(x+\sqrt{x^{2}+1}\right)
$$ | ## Solution
$y^{\prime}=\left(x\left(2 x^{2}+5\right) \sqrt{x^{2}+1}+3 \ln \left(x+\sqrt{x^{2}+1}\right)\right)^{\prime}=$
$=\left(6 x^{2}+5\right) \sqrt{x^{2}+1}+\left(2 x^{3}+5 x\right) \cdot \frac{1}{2 \sqrt{x^{2}+1}} \cdot 2 x+$
$+3 \cdot \frac{1}{x+\sqrt{x^{2}+1}} \cdot\left(1+\frac{1}{2 \sqrt{x^{2}+1}} \cdot 2... | 8\sqrt{(x^{2}+1)^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,838 |
Problem condition
Find the derivative.
$y=\frac{1}{2 \sqrt{2}}(\sin (\ln x)-(\sqrt{2}-1) \cdot \cos (\ln x)) x^{\sqrt{2}+1}$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2 \sqrt{2}}(\sin (\ln x)-(\sqrt{2}-1) \cdot \cos (\ln x)) x^{\sqrt{2}+1}\right)^{\prime}= \\
& =\frac{1}{2 \sqrt{2}}\left(\cos (\ln x) \cdot \frac{1}{x}+(\sqrt{2}-1) \cdot \sin (\ln x) \cdot \frac{1}{x}\right) x^{\sqrt{2}+1}+ \\
& +\frac{1}{2 \sqrt{2}}(\sin (\... | \frac{x^{\sqrt{2}}}{2\sqrt{2}}\cdot(\sqrt{2}\cdot\sin(\lnx)-\sqrt{2}\cdot\cos(\lnx)+2\cos(\lnx)) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,839 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$.
$$
\left\{\begin{array}{l}
x=\sqrt{2 t-t^{2}} \\
y=\frac{1}{\sqrt[3]{(1-t)^{2}}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\sqrt{2 t-t^{2}}\right)^{\prime}=\frac{1}{2 \sqrt{2 t-t^{2}}} \cdot(2-2 t)=\frac{1-t}{\sqrt{2 t-t^{2}}}$
$y_{t}^{\prime}=\left(\frac{1}{\sqrt[3]{(1-t)^{2}}}\right)^{\prime}=\left((1-t)^{-\frac{2}{3}}\right)^{\prime}=-\frac{2}{3} \cdot(1-t)^{-\frac{5}{3}} \cdot(-1)=\frac{2}{3 \sqrt[3]... | \frac{2\sqrt{2-^{2}}}{3(1-)^{2}\cdot\sqrt[3]{(1-)^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,840 |
## Problem Statement
Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{
\begin{array}{l}
x=a(t-\sin t) \\
y=a(1-\cos t)
\end{array}
\right.
\]
$t_{0}=\frac{\pi}{3}$ | ## Solution
Since $t_{0}=\frac{\pi}{3}$, then
$x_{0}=a\left(\frac{\pi}{3}-\sin \frac{\pi}{3}\right)=a \cdot\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)$
$y_{0}=a\left(1-\cos \frac{\pi}{3}\right)=a \cdot\left(1-\frac{1}{2}\right)=\frac{a}{2}$
Find the derivatives:
$x_{t}^{\prime}=(a(t-\sin t))^{\prime}=a(1-\cos t)... | \sqrt{3}\cdotx-(\frac{\pi}{\sqrt{3}}+2)\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,841 |
## Task Condition
Find the $n$-th order derivative.
$y=\sqrt[5]{e^{7 x-1}}$ | ## Solution
$y=\sqrt[5]{e^{7 x-1}}=e^{\frac{7 x-1}{5}}$
$y^{\prime}=\left(e^{\frac{7 x-1}{5}}\right)^{\prime}=e^{\frac{7 x-1}{5}} \cdot \frac{7}{5}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(e^{\frac{7 x-1}{5}} \cdot \frac{7}{5}\right)^{\prime}=e^{\frac{7 x-1}{5}} \cdot\left(\frac{7}{5}\right)^{2}$
... | (\frac{7}{5})^{n}\cdot\sqrt[5]{e^{7x-1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,842 |
## Task Condition
Find the derivative of the specified order.
$y=x \cdot \cos x^{2}, y^{\prime \prime \prime}=?$ | ## Solution
$y^{\prime}=\left(x \cdot \cos x^{2}\right)^{\prime}=\cos x^{2}+x \cdot\left(-\sin x^{2}\right) \cdot 2 x=\cos x^{2}-2 x^{2} \cdot \sin x^{2}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\cos x^{2}-2 x^{2} \cdot \sin x^{2}\right)^{\prime}=-\sin x^{2} \cdot 2 x-\left(4 x \cdot \sin x^{2}+2 x^... | y^{\\\}=-24x^{2}\cdot\cosx^{2}+(8x^{4}-6)\cdot\sinx^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,843 |
## Problem Statement
Show that the function $y_{\text {satisfies equation (1). }}$.
$y=5 e^{-2 x}+\frac{e^{x}}{3}$
$y^{\prime}+2 y=e^{x}$ | ## Solution
$y^{\prime}=\left(5 e^{-2 x}+\frac{e^{x}}{3}\right)^{\prime}=-10 e^{-2 x}+\frac{e^{x}}{3}$
Substitute into equation (1):
$-10 e^{-2 x}+\frac{e^{x}}{3}+2\left(5 e^{-2 x}+\frac{e^{x}}{3}\right)=e^{x}$
Simplify:
$-10 e^{-2 x}+\frac{e^{x}}{3}+10 e^{-2 x}+\frac{2 e^{x}}{3}=e^{x}$
$e^{x}=e^{x}$
$0=0$
The ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,844 |
## Problem Statement
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{8 ;-7 ;-13\}$
$p=\{0 ; 1 ; 5\}$
$q=\{3 ;-1 ; 2\}$
$r=\{-1 ; 0 ; 1\}$ | ## 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... | -4p+3q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,845 |
## problem statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{4 ; 2 ;-7\}$
$b=\{5 ; 0 ;-3\}$
$c_{1}=a-3 b$
$c_{2}=6 b-2 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(a-3 b)=-2 c_{1}$ for any $a$ and $b$.
Thus, ${ }^{c_{1}}=-\frac{1}{2} \cdot c_{2}$, which means the v... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,846 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(0 ; 1 ; 0), B(0 ; 2 ; 1), C(1 ; 2 ; 0)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(0-0 ; 2-1 ; 1-0)=(0 ; 1 ; 1)$
$\overrightarrow{A C}=(1-0 ; 2-1 ; 0-0)=(1 ; 1 ; 0)$
We find the cosine of the angle $\phi_{\text {between vectors }} \overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightarro... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,847 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=2 p+3 q$
$b=p-2 q$
$|p|=2$
$|q|=1$
$(\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... | 7\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,848 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{-3 ; 3 ; 3\}$
$b=\{-4 ; 7 ; 6\}$
$c=\{3 ; 0 ;-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 & 3 & 3 \\
-4 & 7 & 6 \\
3 & 0 & -1
\end{array}\right|= \\
& =-3 \cdot\left|... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,849 |
## 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 \( A_{1} A_{2} A_{3} \).
\( A_{1}(-3 ; -5 ; 6) \)
\( A_{2}(2 ; 1 ; -4) \)
\( A_{3}(0 ; -3 ; -1) \)
\( A_{4}(-5 ; 2 ; -8) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{2-(-3) ; 1-(-5) ;-4-6\}=\{5 ; 6 ;-10\} \\
& \overrightarrow{A_{1} A_{3}}=\{0-(-3) ;-3-(-5) ;-1-6\}=\{3 ; 2 ;-7\} \\
& \overrightarrow{A_{1} A_{4}}=\{-5-(-3) ; 2-(-5) ;-8-6\}=\{-2 ; 7 ;-14\}
\end{aligned}
$$
Accordin... | \frac{191}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,850 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(2 ; 3 ; 1)$
$M_{2}(4 ; 1 ;-2)$
$M_{3}(6 ; 3 ; 7)$
$M_{0}(-5 ;-4 ; 8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$\left|\begin{array}{ccc}x-2 & y-3 & z-1 \\ 4-2 & 1-3 & -2-1 \\ 6-2 & 3-3 & 7-1\end{array}\right|=0$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-3 & z-1 \\
2 & -2 & -3 \\
4 & 0 & 6
\end{... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,851 |
## problem statement
Write the equation of the plane passing through point $A$ and perpendicular to the vector $\overrightarrow{B C}$.
$A(7 ; -5 ; 0)$
$B(8 ; 3 ; -1)$
$C(8 ; 5 ; 1)$ | ## Solution
Let's find the vector $\overrightarrow{B C}$:
$\overrightarrow{B C}=\{8-8 ; 5-3 ; 1-(-1)\}=\{0 ; 2 ; 2\}$
Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$0 \cdot(x-7)+2 \cdot(y-(-5))+2 \cd... | y+z+5=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,852 |
## Problem Statement
Find the angle between the planes:
$$
\begin{aligned}
& 3 x-y-5=0 \\
& 2 x+y-3=0
\end{aligned}
$$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{3 ;-1 ; 0\}$
$\overrightarrow{n_{2}}=\{2 ; 1 ; 0\}$
The angle $\phi_{\text{between the planes is determined by the formula: }}$
$\cos \phi=\frac{\left(... | \frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,853 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(4 ; 6 ; 8)$
$C(2 ; 4 ; 6)$ | ## Solution
Let's find the distances $A B$ and $A C$:
$A B=\sqrt{(4-x)^{2}+(6-0)^{2}+(8-0)^{2}}=\sqrt{16-8 x+x^{2}+36+64}=\sqrt{x^{2}-8 x+116}$
$A C=\sqrt{(2-x)^{2}+(4-0)^{2}+(6-0)^{2}}=\sqrt{4-4 x+x^{2}+16+36}=\sqrt{x^{2}-4 x+56}$
Since by the condition of the problem $A B=A C$, then
$$
\begin{aligned}
& \sqrt{x^... | A(15;0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,854 |
## 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 $a$?
$A(3 ; 2 ; 4)$
$a: 2x - 3y + z - 6 = 0$
$k = \frac{2}{3}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 2 x-3 y+z-4=0$
Substitute the coordinates of point $A$ into the equat... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,855 |
## Task Condition
Write the canonical equations of the line.
$$
\begin{aligned}
& 3 x+3 y+z-1=0 \\
& 2 x-3 y-2 z+6=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- are coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction ... | \frac{x+1}{-3}=\frac{y-\frac{4}{3}}{8}=\frac{z}{-15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,856 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-1}{1}=\frac{y-3}{0}=\frac{z+2}{-2}$
$3 x-7 y-2 z+7=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-1}{1}=\frac{y-3}{0}=\frac{z+2}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+t \\
y=3 \\
z=-2-2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$3(1+t)-7 \cdot 3-2(-2-2 t)+7=0$
$3+3 t-21... | (2;3;-4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,857 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$f(x)=\left\{\begin{array}{c}x^{2} \cos \left(\frac{4}{3 x}\right)+\frac{x^{2}}{2}, x \neq 0 \\ 0, x=0\end{array}\right.$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,859 |
## Condition of the problem
To compose the equation of the normal to the given curve at the point with abscissa $\tau_{0}$.
$y=8 \sqrt[1]{x}-70, x_{0}=16$ | ## Solution
Let's find $y^{\prime}:$
$$
y^{\prime}=(8 \sqrt[1]{x}-70)^{\prime}=\left(8 \cdot x^{\frac{1}{4}}-70\right)^{\prime}=8 \cdot \frac{1}{4} \cdot x^{-\frac{3}{4}}=\frac{2}{\sqrt[4]{x^{3}}}
$$
Then:
$$
y_{\overline{0}}^{\prime}=y^{\prime}\left(x_{\overline{0}}\right)=\frac{2}{\sqrt[2]{x_{\overline{0}}^{3}}}=... | -4x+10 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,860 |
## Task Condition
Find the differential $d y$.
$$
y=\arccos \left(\frac{x^{2}-1}{x^{2} \sqrt{2}}\right)
$$ | ## Solution
$d y=y^{\prime} \cdot d x=\left(\arccos \left(\frac{x^{2}-1}{x^{2} \sqrt{2}}\right)\right)^{\prime} d x=\frac{-1}{\sqrt{1-\left(\frac{x^{2}-1}{x^{2} \sqrt{2}}\right)^{2}}} \cdot\left(\frac{x^{2}-1}{x^{-2} \sqrt{2}}\right)^{\prime} d x=$
$=\frac{-1}{\sqrt{\frac{2 x^{4}}{2 x^{4}}-\frac{\left(x^{2}-1\right)^{... | y(1.97)\approx2.975 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,861 |
## Task Condition
Find the derivative.
$$
y=\frac{\left(x^{2}-8\right) \sqrt{x^{2}-8}}{6 x^{3}}
$$ | ## Solution
$y^{\prime}=\left(\frac{\left(x^{2}-8\right) \sqrt{x^{2}-8}}{6 x^{3}}\right)^{\prime}=\left(\frac{\left(x^{2}-8\right)^{\frac{3}{2}}}{6 x^{3}}\right)^{\prime}=\frac{\frac{3}{2} \cdot\left(x^{2}-8\right)^{\frac{1}{2}} \cdot 2 x \cdot x^{-5}-\left(x^{2}-8\right) \sqrt{x^{2}-8} \cdot 3 x^{2}}{6 x^{6}}=$
$=\f... | \frac{\\sqrt{x^{2}-8}}{x^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,862 |
## Task Condition
Find the derivative.
$y=\ln ^{3}(1+\cos x)$ | ## Solution
$y^{\prime}=\left(\ln ^{3}(1+\cos x)\right)^{\prime}=3 \ln ^{2}(1+\cos x) \cdot \frac{1}{1+\cos x} \cdot(-\sin x)=-\frac{3 \cdot \sin x \cdot \ln ^{2}(1+\cos x)}{1+\cos x}$
## Problem Kuznetsov Differentiation 8-8 | -\frac{3\cdot\sinx\cdot\ln^{2}(1+\cosx)}{1+\cosx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,864 |
## Condition of the problem
Find the derivative.
$$
y=\cos (\operatorname{ctg} 2)-\frac{1}{16} \cdot \frac{\cos ^{2} 8 x}{\sin 16 x}
$$ | ## Solution
$y^{\prime}=\left(\cos (\operatorname{ctg} 2)-\frac{1}{16} \cdot \frac{\cos ^{2} 8 x}{\sin 16 x}\right)^{\prime}=-\frac{1}{16} \cdot\left(\frac{\cos ^{2} 8 x}{\sin 16 x}\right)^{\prime}=-\frac{1}{16} \cdot\left(\frac{\cos ^{2} 8 x}{2 \sin 8 x \cdot \cos 8 x}\right)^{\prime}=$ $=-\frac{1}{32} \cdot\left(\fr... | \frac{1}{4\sin^{2}8x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,865 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{2} \cdot(x-4) \sqrt{8 x-x^{2}-7}-9 \arccos \sqrt{\frac{x-1}{6}}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{2} \cdot(x-4) \sqrt{8 x-x^{2}-7}-9 \arccos \sqrt{\frac{x-1}{6}}\right)^{\prime}= \\
& =\frac{1}{2} \cdot\left(\sqrt{8 x-x^{2}-7}+(x-4) \cdot \frac{1}{2 \sqrt{8 x-x^{2}-7}} \cdot(8-2 x)\right)-9 \cdot \frac{-1}{\sqrt{1-\frac{x-1}{6}}} \cdot \frac{1}{2 \sqrt{\fr... | \sqrt{8x-x^{2}-7} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,866 |
## Task Condition
Find the derivative.
$$
y=\frac{1}{18 \sqrt{2}} \ln \frac{1+\sqrt{2} \operatorname{coth} x}{1-\sqrt{2} \operatorname{coth} x}
$$ | ## Solution
$y^{\prime}=\left(\frac{1}{18 \sqrt{2}} \ln \frac{1+\sqrt{2} \operatorname{cth} x}{1-\sqrt{2} \operatorname{cth} x}\right)^{\prime}=\frac{1}{18 \sqrt{2}} \cdot \frac{1-\sqrt{2} \operatorname{cth} x}{1+\sqrt{2} \operatorname{cth} x} \cdot\left(\frac{1+\sqrt{2} \operatorname{cth} x}{1-\sqrt{2} \operatorname{... | \frac{1}{9\cdot(1+\operatorname{ch}^{2}x)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,867 |
## Task Condition
Find the derivative.
$$
y=3 x-\ln \left(1+\sqrt{1-e^{6 x}}\right)-e^{-3 x} \cdot \arcsin \left(e^{3 x}\right)
$$ | ## Solution
$y^{\prime}=\left(3 x-\ln \left(1+\sqrt{1-e^{6 x}}\right)-e^{-3 x} \cdot \arcsin \left(e^{3 x}\right)\right)^{\prime}$
$=3-\frac{1}{1+\sqrt{1-e^{6 x}}} \cdot\left(1+\sqrt{1-e^{6 x}}\right)^{\prime}-\left(-3 e^{-3 x} \cdot \arcsin \left(e^{3 x}\right)+e^{-3 x} \cdot \frac{1}{\sqrt{1-e^{6 x}}} \cdot e^{3 x}... | 3e^{-3x}\cdot\arcsin(e^{3x}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,869 |
## Problem Statement
Find the derivative.
$$
y=x\left(2 x^{2}+1\right) \sqrt{x^{2}+1}-\ln \left(x+\sqrt{x^{2}+1}\right)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(x\left(2 x^{2}+1\right) \sqrt{x^{2}+1}-\ln \left(x+\sqrt{x^{2}+1}\right)^{\prime}=\left(\left(2 x^{3}+x\right) \sqrt{x^{2}+1}-\ln \left(x+\sqrt{x^{2}+1}\right)\right)^{\prime}=\right. \\
& =\left(6 x^{2}+1\right) \sqrt{x^{2}+1}+\left(2 x^{2}+x\right) \frac{1}{2 \sqrt{x... | 8x^{2}\sqrt{x^{2}+1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,870 |
## Task Condition
Find the derivative.
$$
y=\ln \frac{\sin x}{\cos x+\sqrt{\cos 2 x}}
$$ | ## Solution
$y^{\prime}=\left(\ln \frac{\sin x}{\cos x+\sqrt{\cos 2 x}}\right)^{\prime}=\frac{\cos x+\sqrt{\cos 2 x}}{\sin x} \cdot\left(\frac{\sin x}{\cos x+\sqrt{\cos 2 x}}\right)^{\prime}=$
$=\frac{\cos x+\sqrt{\cos 2 x}}{\sin x} \cdot \frac{\cos x \cdot(\cos x+\sqrt{\cos 2 x})-\sin x \cdot\left(-\sin x+\frac{1}{2... | \frac{1}{\sinx\cdot\sqrt{\cos2x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,871 |
## Task Condition
Find the derivative $y_{x}^{\prime}$
$$
\left\{\begin{array}{l}
x=\ln (\operatorname{ctg} t) \\
y=\frac{1}{\cos ^{2} t}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\ln (\operatorname{ctg} t))^{\prime}=\frac{1}{\operatorname{ctg} t} \cdot \frac{-1}{\sin ^{2} t}=-\frac{1}{\cos t \cdot \sin t}$
$y_{t}^{\prime}=\left(\frac{1}{\cos ^{2} t}\right)^{\prime}=\frac{-2}{\cos ^{3} t} \cdot(-\sin t)=\frac{2 \sin t}{\cos ^{3} t}$
We obtain:
$$
y_{x}^{\prime}=\... | -2\operatorname{tg}^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,872 |
## Problem Statement
Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$.
\[
\left\{\begin{array}{l}
x=\frac{3 a t}{1+t^{2}} \\
y=\frac{3 a t^{2}}{1+t^{2}}
\end{array}\right.
\]
$t_{0}=2$ | ## Solution
Since $t_{0}=2$, then
$x_{0}=\frac{3 a \cdot 2}{1+2^{2}}=\frac{6 a}{5}$
$y_{0}=\frac{3 a \cdot 2^{2}}{1+2^{2}}=\frac{12 a}{5}$
Let's find the derivatives:
$$
\begin{aligned}
& x_{t}^{\prime}=\left(\frac{3 a t}{1+t^{2}}\right)^{\prime}=\frac{3 a\left(1+t^{2}\right)-3 a t \cdot 2 t}{\left(1+t^{2}\right)^... | -\frac{4}{3}\cdotx-4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,873 |
## Task Condition
Find the $n$-th order derivative.
$y=\lg (x+4)$ | ## Solution
$y=\lg (x+4)$
$y^{\prime}=(\lg (x+4))^{\prime}=\frac{1}{(x+4) \ln 10}=\frac{1}{\ln 10} \cdot(x+4)^{-1}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{\ln 10} \cdot(x+4)^{-1}\right)^{\prime}=-\frac{1}{\ln 10} \cdot(x+4)^{-2}$
$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{... | y^{(n)}=\frac{(-1)^{n-1}\cdot(n-1)!}{\ln10\cdot(x+4)^{n}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,874 |
## Task Condition
Find the derivative of the specified order.
$y=\frac{\ln x}{x^{2}}, y^{IV}=?$ | ## Solution
$y^{\prime}=\left(\frac{\ln x}{x^{2}}\right)^{\prime}=\frac{\frac{1}{x} \cdot x^{2}-\ln x \cdot 2 x}{x^{4}}=\frac{1-2 \ln x}{x^{3}}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1-2 \ln x}{x^{3}}\right)^{\prime}=\frac{\frac{-2}{x} \cdot x^{3}-(1-2 \ln x) 3 x^{2}}{x^{6}}=$
$=\frac{-2-3(1... | \frac{-154+120\lnx}{x^{6}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,875 |
Condition of the problem
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sin t \\
y=\sec t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$
$y_{t}^{\prime}=(\sec t)^{\prime}=\left(\frac{1}{\cos t}\right)^{\prime}=-\frac{1}{\cos ^{2} t} \cdot(-\sin t)=\frac{\sin t}{\cos ^{2} t}$
We obtain:
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{\sin t}{\cos ^{2} t}\right) / \cos t=\frac{\si... | \frac{1+2\sin^{2}}{\cos^{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,876 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{2 n-5}{3 n+1}, a=\frac{2}{3}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left|\frac{6 n-15-6 n-2}{3(3 n+1)}\right| \\
& \left|\frac{-17}{3(3 n+1)}\right| \\
& \frac{17}{3(3 n+1)} \\
& 3 n+1>\frac{17}{3 \varepsilon} ;=> \\
& ... | N(\varepsilon)=[\frac{17+6\varepsilon}{9\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 45,878 |
## Condition of the problem
$$
\lim _{n \rightarrow x}\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}-4\right)}-\sqrt{n^{4}-9}\right)
$$ | ## Solution
$$
\lim _{n \rightarrow x}\left(\sqrt{\left(n n^{2}+1\right)\left(n n^{2}-4\right)}-\sqrt{n^{4}-9}\right)=
$$
$$
\begin{aligned}
& =\lim _{n \rightarrow x} \frac{\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}-4\right)}-\sqrt{n^{4}-9}\right)\left(\sqrt{\left(n^{2}+1\right)\left(n^{2}-4\right)}-\sqrt{n^{4}-9}\... | -\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,881 |
## Problem Statement
Calculate the limit of the numerical sequence:
$$
\lim _{n \rightarrow \infty} \frac{2^{n+1}+3^{n+1}}{2^{n}+3^{n}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2^{n+1}+3^{n+1}}{2^{n}+3^{n}}=\lim _{n \rightarrow \infty} \frac{2 \cdot 2^{n}+3 \cdot 3^{n}}{2^{n}+3^{n}}= \\
& =\lim _{n \rightarrow \infty} \frac{2 \cdot 2^{n}+2 \cdot 3^{n}+3^{n}}{2^{n}+3^{n}}=\lim _{n \rightarrow \infty} \frac{2\left(2^{n}+3^{n}\... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,882 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ :
$\lim _{x \rightarrow 3} \frac{4 x^{2}-14 x+6}{x-3}=10$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function \( f: M \subset \mathbb{R} \rightarrow \mathbb{R} \) and \( a \in M' \) is a limit point of the set \( M \). A number \( A \in \mathbb{R} \) is called the limit of the function \( f \) as \( x \) approaches \( a (x \rightarrow... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,884 |
## problem statement
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0 \text { (find }} \delta(\varepsilon)_{\text {) }}$ :
$f(x)=2 x^{2}-4, x_{0}=3$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0$ : $\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon_{\text {when }}$ $\left|x-x_{0}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,885 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1} \frac{\left(2 x^{2}-x-1\right)^{2}}{x^{3}+2 x^{2}-x-2}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\left(2 x^{2}-x-1\right)^{2}}{x^{3}+2 x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 1} \frac{(2 x+1)^{2}(x-1)^{2}}{\left(x^{2}+3 x+2\right)(x-1)}= \\
& =\lim _{x \rightarrow 1} \frac{(2 x+1)^{2}(x-1)}{x^{2}+3 x+2}=\frac{(2 \cdot 1+1)^{2}(1-1)}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,886 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{\cos 7 x-\cos 3 x}$ | ## Solution
We will use the substitution of equivalent infinitesimals:
$$
\begin{aligned}
& 1-\cos 2 x \sim \frac{(2 x)^{2}}{2}, \text { as } x \rightarrow 0(2 x \rightarrow 0) \\
& \sin 5 x \sim 5 x, \text { as } x \rightarrow 0(5 x \rightarrow 0) \\
& \sin 2 x \sim 2 x, \text { as } x \rightarrow 0(2 x \rightarrow ... | -\frac{1}{10} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,888 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sin 2 x}{(\pi-4 x)^{2}}$ | ## Solution
Substitution:
$x=y+\frac{\pi}{4} \Rightarrow y=x-\frac{\pi}{4}$
$x \rightarrow \frac{\pi}{4} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sin 2 x}{(\pi-4 x)^{2}}=\lim _{y \rightarrow 0} \frac{1-\sin 2\left(y+\frac{\pi}{4}\right)}{\left(\pi-4\lef... | \frac{1}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,889 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2} \frac{\tan x - \tan 2}{\sin (\ln (x-1))}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\tan x - \tan 2}{\sin (\ln (x-1))} = \lim _{x \rightarrow 2} \frac{\frac{\sin (x-2)}{\cos x \cdot \cos 2}}{\sin (\ln (x-1))} = \\
& = \lim _{x \rightarrow 2} \frac{\sin (x-2)}{\cos x \cdot \cos 2 \cdot \sin (\ln (x-1))} =
\end{aligned}
$$
Substitution:
$... | \frac{1}{\cos^2(2)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,890 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{3 x}}{\sin 2 x-\sin x}$ | ## Solution
$\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{3 x}}{\sin 2 x-\sin x}=\lim _{x \rightarrow 0} \frac{\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)}{\sin 2 x-\sin x}=$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{5 x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}(\sin 2 x-\sin x)}=$
$=\frac{\... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,891 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow a} \frac{\tan x - \tan a}{\ln x - \ln a}$ | ## Solution
$\lim _{x \rightarrow a} \frac{\tan x-\tan a}{\ln x-\ln a}=\lim _{x \rightarrow a} \frac{\left(\frac{\sin (x-a)}{\cos x \cdot \cos a}\right)}{\ln \frac{x}{a}}=$
Substitution:
$x=y+a \Rightarrow y=x-a$
$x \rightarrow a \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \... | \frac{}{\cos^2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,892 |
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