need little help with this: First i must decide the inverse to:
f(x)=1/cos(x)
so i think the inverse will be : x = arccos(1/y) right ?
then i need help with derivate it
Not necessarily. With the inverse trig functions, a very common procedure to find the derivative of the inverse trig functions is to use implicit differentiation. This is how, for example,
$\displaystyle \dfrac{d}{dx}\,\sin^{-1}(x)=\dfrac{1}{\sqrt{1-x^{2}}}$ is done.
You have $\displaystyle y=\sin^{-1}(x),$ implying that $\displaystyle \sin(y)=x.$
Implicitly differentiating yields
$\displaystyle \cos(y)\,y'=1\quad\Rightarrow\quad y'=\dfrac{1}{\cos(y)}.$
Now, if you draw the correct triangle, you will see that
$\displaystyle \cos(y)=\sqrt{1-x^{2}}.$ (Or you can just use the Pythagorean theorem.)
Does that help?
Also, just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
Here is how you could also allow for an inner function like 1/y.
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
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