Question

If ${\left({\tan }^{-1}x\right)}^y+y^{\cot x}$, then find $\frac{dy}{dx}$

Answer 1

Let $u={\left({\tan }^{-1}x\right)}^y$ and $v=y^{\cot x}$

$\Rightarrow \log u=y\log \left({\tan }^{-1}x\right)$ and $\log v=\cot x\log y$

or $\frac{1}{u}\frac{du}{dx}=\frac{y}{{\tan }^{-1}x}\frac{1}{1+x^2}+\log {\tan }^{-1}x\frac{dy}{dx}$

$\frac{du}{dx}={\left({\tan }^{-1}x\right)}^y[\frac{y}{{\tan }^{-1}x(1+x^2)}+\log {\tan }^{-1}x\frac{dy}{dx}]$

$\frac{1}{v}\frac{dv}{dx}=\frac{\cot x}{y}\frac{dy}{dx}-{\csc }^{2}x\log y$

or $\frac{dv}{dx}=y^{\cot x}[\frac{\cot x}{y}\frac{dy}{dx}-{\csc }^{2}x\log y]$

Given:$u+v=0$

$\Rightarrow \frac{du}{dx}+\frac{dv}{dx}=0$

$\Rightarrow \frac{y{\left({\tan }^{-1}x\right)}^{y-1}}{1+x^2}+{\left({\tan }^{-1}x\right)}^y\log {\tan }^{-1}x\frac{dy}{dx}+\cot x.y^{\cot x-1}\frac{dy}{dx}-y^{\cot x}{\csc }^{2}x\log y=0$

$\therefore \frac{dy}{dx}=\frac{y^{\cot x}{\csc }^{2}x\log y-\frac{y{\left({\tan }^{-1}x\right)}^{y-1}}{1+x^2}}{{\left({\tan }^{-1}x\right)}^y\log {\tan }^{-1}x+\cot x.y^{\cot x-1}}$