## Task Condition

Find the derivative.

$y=\left(x^{2}+1\right)^{\cos x}$
## Solution

$y=\left(x^{2}+1\right)^{\cos x}$

$\ln y=\ln \left(x^{2}+1\right)^{\cos x}=\cos x \cdot \ln \left(x^{2}+1\right)$

$\frac{y^{\prime}}{y}=-\sin x \cdot \ln \left(x^{2}+1\right)+\cos x \cdot \frac{1}{x^{2}+1} \cdot 2 x=$

$=\frac{2 x \cdot \cos x}{x^{2}+1}-\sin x \cdot \ln \left(x^{2}+1\right)$

$$
\begin{aligned}
& y^{\prime}=y \cdot\left(\frac{2 x \cdot \cos x}{x^{2}+1}-\sin x \cdot \ln \left(x^{2}+1\right)\right)= \\
& =\left(x^{2}+1\right)^{\cos x} \cdot\left(\frac{2 x \cdot \cos x}{x^{2}+1}-\sin x \cdot \ln \left(x^{2}+1\right)\right)
\end{aligned}
$$

## Problem Kuznetsov Differentiation 12-18