# Finding f'(x)

• April 11th 2011, 10:07 AM
ppark
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
• April 11th 2011, 10:43 AM
emakarov
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.
• April 11th 2011, 08:19 PM
johnny
dy/dx = 1/cos x = sec x
dy = sec x dx
y = ∫ sec x dx = ln|sec x + tan x| + C
• April 12th 2011, 09:35 AM
HallsofIvy
Quote:

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, $\frac{df}{dx}= \frac{df}{du}\frac{du}{dx}= cos(x)\frac{df}{du}$ so that equation becomes $cos(x)\frac{df}{dx}= \frac{1}{cos(x)}$ which is the same as $\frac{df}{du}= \frac{1}{cos^2(x)}= \frac{1}{1- sin^2(x)}= \frac{1}{1- u^2}$. Replacing "u" with "x", that is $\frac{df}{dx}= \frac{1}{1- x^2}$.

Quote:

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!
• April 12th 2011, 09:38 AM
HallsofIvy
Quote:

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.