# Thread: derivative of trig function

1. ## derivative of trig function

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

2. $\dfrac{1}{\cos(x)}=\sec(x)$

what is the derivative of $\sec(x)$?

3. i think it will be tan(x)cos(x) ?

but if the question is to find out the inverses derivate, don´t that mean that i must first get the inverse and then derivate the inverse, not 1/cos(x) ?

4. Originally Posted by paulaa
i think it will be tan(x)cos(x) ?

but if the question is to find out the inverses derivate, don´t that mean that i must first get the inverse and then derivate the inverse, not 1/cos(x) ?
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,

$\dfrac{d}{dx}\,\sin^{-1}(x)=\dfrac{1}{\sqrt{1-x^{2}}}$ is done.

You have $y=\sin^{-1}(x),$ implying that $\sin(y)=x.$

Implicitly differentiating yields

$\cos(y)\,y'=1\quad\Rightarrow\quad y'=\dfrac{1}{\cos(y)}.$

Now, if you draw the correct triangle, you will see that

$\cos(y)=\sqrt{1-x^{2}}.$ (Or you can just use the Pythagorean theorem.)

Does that help?

5. Also, just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case theta), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

Here is how you could also allow for an inner function like 1/y.

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