# Derivative of inverse trig functions

## Homework Statement

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

## 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 ?

Fredrik
Staff Emeritus
Gold Member
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: $$(f^{-1})'(x)=?$$

## Homework Statement

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

## 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 ?

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. Dick
Homework Helper
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: $$(f^{-1})'(x)=?$$

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?

Fredrik
Staff Emeritus
Gold Member
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?
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.

Dick
$$\frac{d}{dx}\sec^{-1}(f(x))=(\sec^{-1})'(f(x))f'(x)=\frac{1}{\sec'(\sec^{-1}(f(x)))}f'(x)$$ 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.