derivative of trig function

**paulaa** · February 10th 2011, 10:40 AM

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

**harish21** · February 10th 2011, 10:48 AM

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

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

**paulaa** · February 10th 2011, 11:02 AM

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) ?

**Ackbeet** · February 10th 2011, 11:12 AM

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?

**tom@ballooncalculus** · February 10th 2011, 12:18 PM

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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