# Thread: Derivative of inverse function

1. ## Derivative of inverse function

Hi everybody,

I got the following:

f'(x)= xarcsinx - sqrt(1-x^2)
I use the sum rule and my answer is f(x)=1/sqrt(1-x^2)+1, but I need your approval, because I'm not quite sure. Please give me the whole solution step-by-step so that I can get it. Many thanks in advance!

2. Originally Posted by lookr
Hi everybody,

I got the following:

f'(x)= xarcsinx - sqrt(1-x^2)
I use the sum rule and my answer is f(x)=1/sqrt(1-x^2)+1, but I need your approval, because I'm not quite sure. Please give me the whole solution step-by-step so that I can get it. Many thanks in advance!
$f'(x) = x. \frac{1}{\sqrt{1-x^2}} + 1.\sin^{-1} x - \frac{-2x}{2 \sqrt{1-x^2}}$

$f'(x) = \frac{x}{\sqrt{1-x^2}} + \sin^{-1} x + \frac{x}{\sqrt{1-x^2}}$

$f'(x) = \frac{2x}{\sqrt{1-x^2}} + \sin^{-1} x$

Now you solve further.