Question

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

Answer 1

Let $y=({\tan }^{-1}x{)}^y+y^{\cot x}-1$

Now let $u=({\tan }^{-1}x{)}^y$ and $v=y^{\cot x}$

starting with $u$, we get,

$u=({\tan }^{-1}x{)}^y$

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

$\frac{1}{u}.\frac{du}{dx}=y\frac{d}{dx}\log \left({\tan }^{-1}\right)+\log \left({\tan }^{-1}\right)\frac{dy}{dx}$

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

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

$\Rightarrow$ $\frac{du}{dx}=\frac{y.({\tan }^{-1}x{)}^{y-1}}{1-x^2}+({\tan }^{-1}x{)}^y.log({\tan }^{-1}x)\frac{dy}{dx}$ ------ ( 1 )

And now $v=y^{\cot x}$

Taking log we get,

$\log v=\cot x\log y$

On differentiating we get,

$\frac{1}{v}\frac{dv}{dx}=\cot x\frac{d}{dx}\log y+\log y\frac{d}{dx}\cot x$

$\Rightarrow$ $\frac{1}{v}\frac{dv}{dx}=\frac{\cot x}{y}\frac{dy}{dx}-cose{c}^{2}x.\log y$

$\Rightarrow$ $\frac{dv}{dx}=y^{\cot x}.\frac{\cot x}{y}.\frac{dy}{dx}-y^{\cot x}.cose{c}^{2}x.\log y$

$\Rightarrow$ $\frac{dv}{dx}=y^{\cot x-1}.\cot x.\frac{dy}{dx}-y^{\cot x}.cose{c}^{2}x.\log y$ ----- ( 2 )

Now, $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}-0$

$\Rightarrow$ $\frac{dy}{dx}=\frac{y.({\tan }^{-1}x{)}^{y-1}}{1+x^2}+({\tan }^{-1}x{)}^y.\log ({\tan }^{-1}x)\frac{dy}{dx}+y^{\cot x-1}.\cot x.\frac{dy}{dx}-y^{\cot x}.cose{c}^{2}x.\log y$

$\Rightarrow$ $\frac{dy}{dx}-({\tan }^{-1}{)}^y.\log ({\tan }^{-1}x)\frac{dy}{dx}-y^{\cot x-1}.\cot x.\frac{dy}{dx}=\frac{y({\tan }^{-1}x{)}^{y-1}}{1+x^2}-y^{\cot x}.cose{c}^{2}x.\log y$

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

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