Derivative of inverse trig functions

[TheRedDevil18]

Homework Statement

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

Homework Equations

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]

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: $$(f^{-1})'(x)=?$$

 

[Saitama]

Homework Statement

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

Homework Equations

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]

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]

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]

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.

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.