Question

diffrentiate (x+y)=tan inverse y​

Answer 1

See the attachement below.

Answer 2

Answer:

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

Step-by-step explanation:

Formula used:

$\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^2}$

I have applied chain rule to differentiate the given function

Now,

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

Differentiate with respect to x

$1+\frac{dy}{dx}=\frac{1}{1+y^2}(\frac{dy}{dx})$

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

$1=[\frac{1}{1+y^2}-1]\frac{dy}{dx}$

$1=[\frac{1-1-y^2}{1+y^2}]\frac{dy}{dx}$

$1=[\frac{-y^2}{1+y^2}]\frac{dy}{dx}$

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