# Thread: Help with the derivative of arcsinh

1. ## Help with the derivative of arcsinh

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

2. ya know what the chain rule is?

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

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

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

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

4. 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.

5. 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?

6. 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.

7. 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 $\sin ^{-1} u$