# What is the derivative of y=arcsin(3x )?

Oct 16, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\sqrt{1 - 9 {x}^{2}}}$

#### Explanation:

An alternative method that doesn't require knowing the derivative of $\arcsin \left(x\right)$:

$y = \arcsin \left(3 x\right)$

$\sin \left(y\right) = 3 x$

Differentiate both sides with respect to $x$. The derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$, so the derivative of $\sin \left(y\right)$ is $\cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$. The derivative of $3 x$ is $3$.

$\cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 3$

Solving for the derivative, $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\cos} \left(y\right)$

We know that $\sin \left(y\right) = 3 x$, so we can rewrite the function using all $x$ terms using the identity $\cos \left(y\right) = \sqrt{1 - {\sin}^{2} \left(y\right)}$, which comes from the Pythagorean Identity:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\sqrt{1 - {\sin}^{2} \left(y\right)}}$

Since $\sin \left(y\right) = 3 x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\sqrt{1 - {\left(3 x\right)}^{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\sqrt{1 - 9 {x}^{2}}}$