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?
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?
$\displaystyle \frac{1}{\sqrt{1 - \sin^2(x)}} = \frac{1}{\sqrt{\cos^2(x)}} = \frac{1}{\cos(x)}$
(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.
Hello, mariusg!
Differentiate: .$\displaystyle f(x)\:=\:\cos^{-1}(\sin x)$
And the answer in the back of the textbook is -1.
What i did was:
$\displaystyle 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: .$\displaystyle f'(x) \;=\;\frac{-\cos x}{\sqrt{1-\sin^2x}} $
You know this, don't you?
. . $\displaystyle \sin^2\!x + \cos^2\!x \:=\:1 \quad\Rightarrow\quad \cos^2\!x \:=\:1 - \sin^2\!x \quad\Rightarrow\quad \cos x \:=\:\sqrt{1-\sin^2\!x}$
Got it?