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August 30, 2014

Posted by **Sean** on Tuesday, January 22, 2008 at 9:08pm.

f(x) = cos(arcsin(2x))

What is f'(x)?

- Calculus III -
**Damon**, Tuesday, January 22, 2008 at 9:18pmI did this down below, after the one drwls answered

- Calculus III -
**Damon**, Tuesday, January 22, 2008 at 9:19pmcopied:

It is so similar that you should be able to do it the same way. Draw the triangle and you have sqrt(1-4x^2)

f(x) = (1-4x^2)^.5

f'(x) = .5 [(1-4x^2)^-.5] (-8x)

= -4x/sqrt(1-4x^2)

- Calculus III -
- Calculus III -
**Anonymous**, Tuesday, January 22, 2008 at 9:30pmOh.

But the text book I'm using gives me the answer:

Sqrt(1-4x^2)

No denominators or anything.

- Calculus III -
**Damon**, Tuesday, January 22, 2008 at 9:59pmThat is just the first step.

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