Calculus III
posted by Sean .
Derivative of Inverse Trigonometric Functions
f(x) = sin(arccos(4x))
What is f'(x)?

f(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] 
Thanks. The next question I have is pretty similar.
f(x) =cos(arcsin(2x))
Find f'(x) 
It is so similar that you should be able to do it the same way. Draw the triangle and you have sqrt(14x^2)
f(x) = (14x^2)^.5
f'(x) = .5 [(14x^2)^.5] (8x)
= 4x/sqrt(14x^2)