## Task Condition

Find the derivative of the specified order.

$y=x \cdot \cos x^{2}, y^{\prime \prime \prime}=?$
## Solution

$y^{\prime}=\left(x \cdot \cos x^{2}\right)^{\prime}=\cos x^{2}+x \cdot\left(-\sin x^{2}\right) \cdot 2 x=\cos x^{2}-2 x^{2} \cdot \sin x^{2}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\cos x^{2}-2 x^{2} \cdot \sin x^{2}\right)^{\prime}=-\sin x^{2} \cdot 2 x-\left(4 x \cdot \sin x^{2}+2 x^{2} \cdot \cos x^{2} \cdot 2 x\right)=$

$=-2 x \cdot \sin x^{2}-4 x \cdot \sin x^{2}-4 x^{3} \cdot \cos x^{2}=-6 x \cdot \sin x^{2}-4 x^{3} \cdot \cos x^{2}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-6 x \cdot \sin x^{2}-4 x^{3} \cdot \cos x^{2}\right)^{\prime}=$

$=-6 \cdot \sin x^{2}-12 x^{2} \cdot \cos x^{2}-12 x^{2} \cdot \cos x^{2}+8 x^{4} \cdot \sin x^{2}=$

$=-24 x^{2} \cdot \cos x^{2}+\left(8 x^{4}-6\right) \cdot \sin x^{2}$

## Problem Kuznetsov Differentiation $19-3$

Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.

$\left\{\begin{array}{l}x=e^{t} \cos t \\ y=e^{t} \sin t\end{array}\right.$

## Solution

$$
\begin{aligned}
& x_{t}^{\prime}=\left(e^{t} \cos t\right)^{\prime}=e^{t} \cos t-e^{t} \sin t=e^{t} \cdot(\cos t-\sin t) \\
& y_{t}^{\prime}=\left(e^{t} \sin t\right)^{\prime}=e^{t} \sin t+e^{t} \cos t=e^{t} \cdot(\sin t+\cos t)
\end{aligned}
$$

We get:

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{e^{t} \cdot(\sin t+\cos t)}{e^{t} \cdot(\cos t-\sin t)}=\frac{\sin t+\cos t}{\cos t-\sin t}$

$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{\sin t+\cos t}{\cos t-\sin t}\right)^{\prime}=$

$=\frac{(\cos t-\sin t) \cdot(\cos t-\sin t)-(\sin t+\cos t) \cdot(-\sin t-\cos t)}{(\cos t-\sin t)^{2}}=$

$=\frac{\cos ^{2} t-\cos t \cdot \sin t-\sin \cdot \cos t+\sin ^{2} t+\sin ^{2} t+\sin t \cdot \cos t+\cos t \cdot \sin t+\cos ^{2} t}{(\cos t-\sin t)^{2}}=$

$=\frac{2}{(\cos t-\sin t)^{2}}$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{2}{(\cos t-\sin t)^{2}}\right) /\left(e^{t} \cdot(\cos t-\sin t)\right)=\frac{2}{e^{t} \cdot(\cos t-\sin t)^{3}}$

## Problem Kuznetsov Differentiation 20-3