# What is the derivative of this function y=sin^-1(x/3)?

Apr 6, 2017

$\frac{1}{\sqrt{9 - {x}^{2}}}$

#### Explanation:

I am going to assume that the ${\sin}^{-} 1 \left(x\right)$ mean $\arcsin \left(x\right)$ instead of $\csc \left(x\right)$.

Looking at the differentiation of trigonometric functions, the derivative of $\arcsin \left(x\right)$ is $\frac{1}{\sqrt{1 - {x}^{2}}}$.

Proof:
If $y = \arcsin \left(x\right)$, then $x = \sin \left(y\right)$. Differentiating both sides with respect to $x$, we get $1 = \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$. Then, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right)$, which is equal to $\frac{1}{\sqrt{1 - {\sin}^{2} \left(y\right)}}$. But we said that $y = \arcsin \left(x\right)$, so $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$.

In order to differentiate $\arcsin \left(\frac{x}{3}\right)$, we use the chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, with $y = \arcsin \left(\frac{x}{3}\right)$ and $u = \frac{x}{3}$.

Then, $\frac{d}{\mathrm{dx}} \left(\arcsin \left(\frac{x}{3}\right)\right) = \frac{d}{d \left(\frac{x}{3}\right)} \left(\arcsin \left(\frac{x}{3}\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{x}{3}\right) = \frac{1}{\sqrt{1 - {\left(\frac{x}{3}\right)}^{2}}} \cdot \frac{1}{3} = \frac{1}{3 \sqrt{1 - \frac{1}{9} {x}^{2}}} = \frac{1}{\sqrt{{3}^{2} \left(1 - \frac{1}{9} {x}^{2}\right)}} = \frac{1}{\sqrt{9 - {x}^{2}}}$.