# Derivative of an inverse trig

ttttrigg3r

## Homework Statement

tan^-1(x/(1-x^2)^1/2) find the derivative

the problem comes from 3g from MIT's PDF I found
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset5sol.pdf [Broken]

here is the solution key
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset5prb.pdf [Broken]

## Homework Equations

The solution key says if y = x/(1-x^2)^1/2 then y' = (1-x^2)^-3/2 When I do it, y' comes out differently. This is how I attempt to solve it.

## The Attempt at a Solution

Original problem: tan^-1((x/(1-x^2)^1/2)) what is the derivative with respect to x?

let y=x/(1-x^2)^1/2 y=x*(1-x^2)^-1/2 using the product rule I get:
y'=(1)(1-x^2)^-1/2 + x(-1/2)((1-x^2)^-3/2)(-2)
y'=(1-x^2)^-1/2 + (x^2)(1-x^2)^-3/2
This looks a lot different than the y' stated in the solution manual: y' = (1-x^2)^-3/2

Am I on the right track? Is the solution manual wrong? or am I missing a step? Thank you ahead.

Last edited by a moderator:

## Answers and Replies

Homework Helper
You need to use the chain rule, you have:
$$f(x)=\tan^{-1}\left[ \frac{x}{\sqrt{1-x^{2}}} \right]$$
write (as you did) y=x/(1-x^2)^1/2 and use the chain rule:
$$\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}$$
I believe you have already computed the second factor.

ttttrigg3r
Thank you for the reply. However my question is why is my computed derivative of the inside term different than the answer key?

Homework Helper
It's not different at all, take out a factor of $1/\sqrt{1-x^{2}}$ and you will see (with a little work) why they are the same.

ttttrigg3r
Oh nice. I was wondering what I could do. Ty.