Question

Find the derivative of the following:

${\left(\sin x\right)}^x+\log x+{\left(\log x\right)}^{\tan x}$

Answer 1

$\frac{d}{dx}(sinx{)}^x+logx+(logx{)}^{tanx}$

$=\frac{d}{dx}(sinx{)}^x+\frac{d}{dx}(logx)+\frac{d}{dx}(logx{)}^{tanx}$

$=\left(sinx{)}^{2}logx\frac{d}{dx}\right(sinx)+\frac{1}{x}+(logx{)}^{tanx}log\left(tanx\right)+\frac{d}{dx}\left(logx\right)$

$=\left(cosx\right)(sinx{)}^{x}logx+\frac{1}{x}+(logx{)}^{tanx}log\left(tanx\right)+\frac{1}{x}$