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September 16, 2014

September 16, 2014

Posted by **Sean** on Tuesday, January 22, 2008 at 4:26pm.

f(x) = sin(arccos(4x))

What is f'(x)?

- Calculus III -
**drwls**, Tuesday, January 22, 2008 at 4:47pmf(x)= sin(arccos(4x)) = sqrt(1 - 16x^2)

(To prove that, draw yourself a triangle with cos A = 4x and figure out the sin of A)

let u = 1 - 16x^2

f(u) = sqrt u

f'(x) = df/dx = df/du du/dx

= (1/2)(u)^-1/2 * -32 x

= -16 x/sqrt[1 - 16x^2]

- Calculus III -
**Sean**, Tuesday, January 22, 2008 at 8:38pmThanks. The next question I have is pretty similar.

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

Find f'(x)

- Calculus III -
**Damon**, Tuesday, January 22, 2008 at 9:17pmIt 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)

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