# Derivative of inverse trig functions

1. Apr 12, 2014

### TheRedDevil18

1. The problem statement, all variables and given/known data

ln(sec^-1(3x^2 +1))

2. Relevant equations

3. The attempt at a solution

1/sec-1(3x2+1) * 1/(3x2+1)(sqrt(3x2+1)2-1) * 6x

Is this correct ?, do I just simplify from here ?

2. Apr 12, 2014

### Fredrik

Staff Emeritus
It looks wrong. There should be more trigonometric stuff in the result. How did you get that result? Do you know how to take the derivative of an inverse function: $$(f^{-1})'(x)=?$$

3. Apr 12, 2014

### Pranav-Arora

If you mean:
$$\frac{1}{\sec^{-1}(3x^2+1)}\frac{1}{(3x^2+1)\sqrt{(3x^2+1)^2-1}}(6x)$$
then yes, it is correct.

4. Apr 12, 2014

### Dick

It's a little hard to interpret exactly without more parantheses, but it looks ok to me. What kind of 'more trignonometric' stuff are you looking for?

5. Apr 12, 2014

### Fredrik

Staff Emeritus
I didn't try to work it all out, but I'm thinking that
$$\frac{d}{dx}\sec^{-1}(f(x))=(\sec^{-1})'(f(x))f'(x)=\frac{1}{\sec'(\sec^{-1}(f(x)))}f'(x)$$ and
$$\sec'(x)=\frac{d}{dx}\frac{1}{\cos x}=-\frac{1}{\cos^2x}(-\sin x).$$ So it looks like we get a big mess of "trigonometric stuff". But since you're both saying that he's right, I assume that I'm missing something.

6. Apr 12, 2014

### Dick

I'd do it the other way around. Since sec(arcsec(x))=x, sec'(arcsec(x))*arcsec'(x)=1. sec'=sec*tan and tan(arcsec(x))=sqrt(x^2-1). No trig left in the end.