Derivative of an inverse trig

  • Thread starter ttttrigg3r
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  • #1
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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.
 
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Answers and Replies

  • #2
hunt_mat
Homework Helper
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You need to use the chain rule, you have:
[tex]
f(x)=\tan^{-1}\left[ \frac{x}{\sqrt{1-x^{2}}} \right]
[/tex]
write (as you did) y=x/(1-x^2)^1/2 and use the chain rule:
[tex]
\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}
[/tex]
I believe you have already computed the second factor.
 
  • #3
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Thank you for the reply. However my question is why is my computed derivative of the inside term different than the answer key?
 
  • #4
hunt_mat
Homework Helper
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26
It's not different at all, take out a factor of [itex]1/\sqrt{1-x^{2}}[/itex] and you will see (with a little work) why they are the same.
 
  • #5
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Oh nice. I was wondering what I could do. Ty.
 

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