Where am I going wrong?

• Dec 30th 2012, 10:52 AM
mariusg
Where 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?
• Dec 30th 2012, 11:09 AM
Deveno
Re: Where am I going wrong?
$\frac{1}{\sqrt{1 - \sin^2(x)}} = \frac{1}{\sqrt{\cos^2(x)}} = \frac{1}{\cos(x)}$

(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.
• Dec 30th 2012, 01:05 PM
Soroban
Re: Where am I going wrong?
Hello, mariusg!

Quote:

Differentiate: . $f(x)\:=\:\cos^{-1}(\sin x)$

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

What i did was:

$f'(x) \:=\: \frac{-1}{\sqrt{1-\sin^2x}}\cdot\cos x \:=\: \frac{-\cos x}{\sqrt{1-\sin^2x}}$

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

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

You have: . $f'(x) \;=\;\frac{-\cos x}{\sqrt{1-\sin^2x}}$

You know this, don't you?
. . $\sin^2\!x + \cos^2\!x \:=\:1 \quad\Rightarrow\quad \cos^2\!x \:=\:1 - \sin^2\!x \quad\Rightarrow\quad \cos x \:=\:\sqrt{1-\sin^2\!x}$

Got it?