# How do you find the derivative of arcsin (x/2)?

Dec 1, 2016

$\frac{1}{\sqrt{4 - {x}^{2}}}$

#### Explanation:

Let $y = \arcsin \left(\frac{x}{2}\right)$. Rearranging this using the properties of inverse functions yields:

$\sin \left(y\right) = \frac{x}{2}$

Differentiate both sides with respect to $x$. The left hand side will need the chain rule.

The right hand side's derivative is $\frac{1}{2}$ since $\frac{x}{2} = \frac{1}{2} x$.

$\cos \left(y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

We can write $\cos \left(y\right)$ in terms of $\sin \left(y\right)$, which we know equals $\frac{x}{2}$. From ${\sin}^{2} \left(y\right) + {\cos}^{2} \left(y\right) = 1$, we see that $\cos \left(y\right) = \sqrt{1 - {\sin}^{2} \left(y\right)}$.

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

Since $\sin \left(y\right) = \frac{x}{2}$, we see that ${\sin}^{2} \left(y\right) = {x}^{2} / 4$.

$\sqrt{1 - {x}^{2} / 4} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

$\sqrt{\frac{4 - {x}^{2}}{4}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

Taking the $\sqrt{4}$ out of the denominator:

$\frac{1}{2} \sqrt{4 - {x}^{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

Multiplying both sides by $2$:

$\sqrt{4 - {x}^{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{4 - {x}^{2}}}$

Thus the derivative of $\arcsin \left(\frac{x}{2}\right)$ is $\frac{1}{\sqrt{4 - {x}^{2}}}$.