- #1

- 408

- 1

## Homework Statement

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

## Homework Equations

## The Attempt at a Solution

1/sec

^{-1}(3x

^{2}+1) * 1/(3x

^{2}+1)(sqrt(3x

^{2}+1)

^{2}-1) * 6x

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

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- Thread starter TheRedDevil18
- Start date

- #1

- 408

- 1

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

1/sec

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

- #2

Fredrik

Staff Emeritus

Science Advisor

Gold Member

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- #3

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## Homework Statement

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

## Homework Equations

## The Attempt at a Solution

1/sec^{-1}(3x^{2}+1) * 1/(3x^{2}+1)(sqrt(3x^{2}+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.

- #4

Dick

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Homework Helper

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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

Fredrik

Staff Emeritus

Science Advisor

Gold Member

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I didn't try to work it all out, but I'm thinking thatIt'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?

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

Dick

Science Advisor

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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.

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