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.
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)