## Task Condition

Find the derivative.

$y=\left(x^{3}+4\right)^{\tan x}$
## Solution

$y=\left(x^{3}+4\right)^{\tan x}$

$\ln y=\ln \left(x^{3}+4\right)^{\tan x}=\tan x \cdot \ln \left(x^{3}+4\right)$

$\frac{y'}{y}=\frac{1}{1+x^{2}} \cdot \ln \left(x^{3}+4\right)+\tan x \cdot \frac{1}{x^{3}+4} \cdot 3 x^{2}=$

$=\frac{\ln \left(x^{3}+4\right)}{1+x^{2}}+\frac{3 x^{2} \cdot \tan x}{x^{3}+4}$

$y'=y \cdot\left(\frac{\ln \left(x^{3}+4\right)}{1+x^{2}}+\frac{3 x^{2} \cdot \tan x}{x^{3}+4}\right)=\left(x^{3}+4\right)^{\tan x} \cdot\left(\frac{\ln \left(x^{3}+4\right)}{1+x^{2}}+\frac{3 x^{2} \cdot \tan x}{x^{3}+4}\right)$

## Problem Kuznetsov Differentiation 12-13