# Derivative of an inverse trig

• ttttrigg3r
In summary, the conversation is about finding the derivative of a function given in the form of tan^-1(x/(1-x^2)^1/2). The solution key provides a different answer than the one obtained by the speaker, leading to a question about the discrepancy. The speaker is using the chain rule to solve the problem and the final answer turns out to be the same as the one provided in the solution key after simplification.

## 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:
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.

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

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.

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

## 1. What is the derivative of an inverse trig function?

The derivative of an inverse trig function is the inverse of the derivative of the corresponding trig function. For example, the derivative of arctan(x) is 1/(1+x^2) and the derivative of arccos(x) is -1/sqrt(1-x^2).

## 2. How do you find the derivative of an inverse trig function?

To find the derivative of an inverse trig function, use the chain rule. First, take the derivative of the inside function (the trig function) and then multiply it by the derivative of the outside function (the inverse trig function).

## 3. What is the derivative of inverse sine?

The derivative of inverse sine is 1/sqrt(1-x^2).

## 4. Can the derivative of an inverse trig function be negative?

Yes, the derivative of an inverse trig function can be negative. This depends on the value of the input x and the specific inverse trig function being used.

## 5. Why is the derivative of inverse tangent equal to 1/(1+x^2)?

The derivative of inverse tangent is equal to 1/(1+x^2) because this is the derivative of the tangent function. The inverse of tangent is arctan, so the derivative of arctan must be the inverse of the derivative of tangent, which is 1/(1+x^2).