Question

If $y=x^{\cos x}+{\left(\tan x\right)}^{\cot x},find\frac{dy}{dx}$

Answer 1

We have,

$y=x^{\cos x}+{\left(\tan x\right)}^{\cot x}$

Taking log both side and we get,

$\log y=\log {\left(x\right)}^{\cos x}+\log {\left(\tan x\right)}^{\cot x}$

$\log y=\cos x\log x+\cot x\log \tan x$

On differentiating with respect to x and we get,

$\frac{d}{dx}\log y=\cos x\frac{d}{dx}\log x+\log x\frac{d}{dx}\cos x+\cot x\frac{d}{dx}\log \tan x+\mathrm{logtan}x\frac{d}{dx}\cot x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos x\frac{1}{x}+\log x(-\sin x)+\cot x\frac{1}{\tan x}\left({\sec }^{2}x\right)+\log \tan x(-{\csc }^{2}x)$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\cos x}{x}-\sin x\log x-\frac{\cot x}{\tan x}{\sec }^{2}x-{\csc }^{2}x\cos x\log \tan x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{\cos x}{x}-{\csc }^{2}x-\sin x\log x-{\csc }^{2}x\cos x\log \tan x$

$\Rightarrow \frac{dy}{dx}=y[\frac{\cos x}{x}-{\csc }^{2}x-\sin x\log x-{\csc }^{2}x\cos x\log \tan x]$

$\Rightarrow \frac{dy}{dx}=x^{\cos x}+{\left(\tan x\right)}^{\cot x}[\frac{\cos x}{x}-{\csc }^{2}x-\sin x\log x-{\csc }^{2}x\cos x\log \tan x]$

Hence, this is the answer.