# How do you find the derivative of lnsin^-1(x)?

Nov 6, 2016

$\frac{d}{\mathrm{dx}} \left(\ln {\sin}^{-} 1 x\right) = \frac{1}{\left({\sin}^{-} 1 x\right) \sqrt{1 - {x}^{2}}}$

#### Explanation:

Use the derivative of natural logs rule:
$\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{u '}{u}$

So, find the derivative of ${\sin}^{-} 1 x$, and divide it by ${\sin}^{-} 1 x$.
The derivative of ${\sin}^{-} 1 x$ is $\frac{1}{\sqrt{1 - {x}^{2}}}$
$\frac{d}{\mathrm{dx}} \left(\ln {\sin}^{-} 1 x\right) = \frac{\frac{1}{\sqrt{1 - {x}^{2}}}}{{\sin}^{-} 1 x}$

After simplifying, we get
$\frac{d}{\mathrm{dx}} \left(\ln {\sin}^{-} 1 x\right) = \frac{1}{\left({\sin}^{-} 1 x\right) \sqrt{1 - {x}^{2}}}$

Nov 6, 2016

$\setminus \frac{1}{{\sin}^{-} 1 \left(x\right)} \setminus \cdot \setminus \frac{1}{\setminus} \sqrt{1 - {x}^{2}}$

#### Explanation:

This question is testing your knowledge of the chain rule.
The chain rule states that the derivative of a function$f \left(u\right)$ where $u$ is another function is$f ' \left(u\right) \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$, so the derivative of $\ln \left({\sin}^{-} 1 \left(x\right)\right)$ is $\setminus \frac{1}{{\sin}^{-} 1 \left(x\right)} \setminus \cdot \setminus \frac{\mathrm{dd}}{x} \left[{\sin}^{-} 1 \left(x\right)\right]$
And because $\setminus \frac{\mathrm{dd}}{x} \left[{\sin}^{-} 1 \left(x\right)\right] = \setminus \frac{1}{\setminus} \sqrt{1 - {x}^{2}}$,
$\setminus \frac{\mathrm{dd}}{x} \left[\ln \left({\sin}^{-} 1 \left(x\right)\right)\right] = \setminus \frac{1}{{\sin}^{-} 1 \left(x\right)} \setminus \cdot \setminus \frac{1}{\setminus} \sqrt{1 - {x}^{2}}$