Thread: Finding f'(x)

I don't even know where to start with this:

d/dx f(sin(x))=1/(cos(x))

find a formula for f'(x)

This is a review question and I've tried expanding f'(sin(x)) but nothing

I just need a hint to how I get started on this problem

Oh I know the answer to f'(x) is 1/(1-x^2) but I got it from trial and error is there a mathematical way

d/dx f(sin(x)) = f'(sin(x))cos(x) = 1/cos(x), from where f'(sin(x)) = 1/cos^2(x) = 1/(1 - sin^2(x)). I think this implies that f'(x) = 1/(1-x^2), but, strictly speaking, only for |x| <= 1.

dy/dx = 1/cos x = sec x
dy = sec x dx
y = ∫ sec x dx = ln|sec x + tan x| + C

Originally Posted by ppark

I don't even know where to start with this:

d/dx f(sin(x))=1/(cos(x))

Let u= sin(x). Then du/dx= cos(x). By the chain rule, so that equation becomes which is the same as . Replacing "u" with "x", that is .

find a formula for f'(x)

This is a review question and I've tried expanding f'(sin(x)) but nothing


I just need a hint to how I get started on this problem

Oh I know the answer to f'(x) is 1/(1-x^2) but I got it from trial and error is there a mathematical way

I feel I should point out that "trial and error" is a perfectly valid "mathematical way" of solving a problem!

Originally Posted by johnny

dy/dx = 1/cos x = sec x
dy = sec x dx
y = ∫ sec x dx = ln|sec x + tan x| + C

I think you misunderstood the question.