Question

differentiation of tan inverse x

Answer 1

d/dx(tan inverse x) = 1/1+(x)square

Answer 2

Answer:

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

Step-by-step explanation:

Let

$y=tan^{-1}x$

Take

$x=tany$

$\frac{dx}{dy}=sec^2y$

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

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

Now,

$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$

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

That is,

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