Question

if y=tan(x+y) ,find dy/dx​

Answer 1

Given,

$\longrightarrow y=\tan(x+y)$

Taking tan inverse on both sides,

$\longrightarrow \tan^{-1}y=x+y$

$\longrightarrow x=\tan^{-1}y-y$

Differentiating with respect to $y,$

$\longrightarrow \dfrac{dx}{dy}=\dfrac{d}{dy}\left[\tan^{-1}y-y\right]$

We have,

• $\dfrac{d}{dy}\big(\tan^{-1}y\big)=\dfrac{1}{1+y^2}$

$\longrightarrow \dfrac{d}{dy}\big(y\big)=1$

Then,

$\longrightarrow \dfrac{dx}{dy}=\dfrac{1}{1+y^2}-1$

$\longrightarrow \dfrac{dx}{dy}=\dfrac{1-(1+y^2)}{1+y^2}$

$\longrightarrow \dfrac{dx}{dy}=\dfrac{1-1-y^2}{1+y^2}$

$\longrightarrow \dfrac{dx}{dy}=-\dfrac{y^2}{1+y^2}$

Taking the reciprocal we get,

$\longrightarrow \dfrac{dy}{dx}=-\dfrac{1+y^2}{y^2}$

$\longrightarrow \underline{\underline{\dfrac{dy}{dx}=-1-\dfrac{1}{y^2}}}$

This is the derivative.