In my textbook i have the problem:

f(x)=cos^-1(sin(x))

And the answer in the back of the textbook is -1

What i did was:

f'(x) = -1/(sqrt(1-sin(x)^2) * cos(x)

f'(x) = (-cos(x))/(sqrt(1-sin(x)^2))

But i can't see where i went wrong, any ideas?

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- December 30th 2012, 09:52 AMmariusgWhere am I going wrong?
In my textbook i have the problem:

f(x)=cos^-1(sin(x))

And the answer in the back of the textbook is -1

What i did was:

f'(x) = -1/(sqrt(1-sin(x)^2) * cos(x)

f'(x) = (-cos(x))/(sqrt(1-sin(x)^2))

But i can't see where i went wrong, any ideas? - December 30th 2012, 10:09 AMDevenoRe: Where am I going wrong?

(we can take the positive square root, because we only define an inverse for cosine on the interval [0,pi]).

note that:

f(x) = cos^{-1}(sin(x)) = cos^{-1}(cos(π/2 - x)) = π/2 - x.

it's pretty clear that for f(x) = π/2 - x, f'(x) = -1. - December 30th 2012, 12:05 PMSorobanRe: Where am I going wrong?
Hello, mariusg!

Quote:

Differentiate: .

And the answer in the back of the textbook is -1.

What i did was:

But i can't see where i went wrong . . . any ideas?

You did nothing wrong . . . except stop too soon.

You have: .

You know this, don't you?

. .

Got it?