Question

If ${\text{e}}^{\tan x}+(\log x{)}^{\tan x}$ ,then find $\frac{dy}{dx}$ .

Answer 1

$e^{\tan x}+(\log x{)}^{\tan x}$

Put $u=e^{\tan x}$ and $v=(\log x{)}^{\tan x}$

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$u=e^{\tan x}$

Take log both sides

$\log x=\tan x-\log e=\tan x$

$\frac{1}{u}\frac{du}{dx}={\sec }^{2}x$---------$ii$

$v=(\log x{)}^{\tan x}$

Take log both sides

$\log v=\tan x.\log \left(\log x\right)$

$\frac{1}{v}\frac{dv}{dx}=\tan x.\frac{1}{\log x}.\frac{1}{x}+\log \left(\log x\right){\sec }^{2}x$

$\frac{dv}{dx}=v[\frac{\tan x}{x\log x}+\log (\log x\left){\sec }^{2}x\right]$

$(\log x{)}^{\tan x}[\frac{\tan x}{x\log x}+\log (\log x\left){\sec }^{2}x\right]$---------$iii$

From $i,ii$ and $iii$ we get

$\frac{dy}{dx}=e^{\tan x}{\sec }^{2}x+(\log x{)}^{\tan x}[\frac{\tan x}{2\log x}+\log (\log x\left){\sec }^{2}x\right]$