Help with the derivative of arcsinh

**craig** · June 19th 2009, 02:00 PM

Hi, I'm a little stuck on this problem.

Differentiate with respect to x, $\arcsin{(\frac{1}{x})}$ .

I know the derivative of $\arcsin$ is $\frac{1}{\sqrt{1-x^2}}$ , but I'm not too sure how to approach this problem.

Any pointers would be greatly appreciated

Thanks in advance

**Krizalid** · June 19th 2009, 02:10 PM

ya know what the chain rule is?

**Random Variable** · June 19th 2009, 02:22 PM

$y = \sin^{-1}\frac{1}{x}$

$\sin y = \frac{1}{x}$

$\cos y \ \frac{dy}{dx} = \text{-} \frac {1}{x^{2}}$

$\frac {dy}{dx} = \frac{\text{-1}}{x^{2} \cos y} = \frac{-1}{x^{2} \sqrt{x^2-1}}$

**craig** · June 19th 2009, 02:45 PM

Originally Posted by Random Variable

$y = \sin^{-1}\frac{1}{x}$

$\sin y = \frac{1}{x}$

$\cos y \ \frac{dy}{dx} = \text{-} \frac {1}{x^{2}}$

$\frac {dy}{dx} = \frac{\text{-1}}{x^{2} \cos y} = \frac{-1}{x^{2} \sqrt{x^2-1}}$

Thanks, I never though to do it though implicit differentiation.

**craig** · June 19th 2009, 02:45 PM

Originally Posted by Krizalid

ya know what the chain rule is?

Yep I know what the chain rule is, how would you use it in this situation?

**Chris L T521** · June 19th 2009, 02:54 PM

Originally Posted by craig

Yep I know what the chain rule is, how would you use it in this situation?

I think Kriz is referring to this: $\frac{\,d}{\,dx}\left[\sin^{-1}u\right]=\frac{1}{\sqrt{1-u^2}}\cdot\frac{\,du}{\,dx}$ . As you see here, chain rule is necessary.

**Random Variable** · June 19th 2009, 02:58 PM

Originally Posted by craig

Thanks, I never though to do it though implicit differentiation.

You can use implict differentiation to get the derivatives of all the basic inverse trigonometric functions.

For this problem it's not necessary if you already know the derivative of $\sin ^{-1} u$

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