# How do you differentiate y = arcsin(x/2)?

Jun 18, 2015

Apply the chain rule to the derivative of $\arcsin$.

#### Explanation:

You may want a more full treatment of Differentiating Inverse Sine

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

Applying the chain rule, we get:

$\frac{d}{\mathrm{dx}} \left(\arcsin u\right) = \frac{1}{\sqrt{1 - {u}^{2}}} \frac{\mathrm{du}}{\mathrm{dx}}$

In this question $u = \frac{x}{2}$, so $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2}$.

We get

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

We're done with calculus, but this can be 'cleaned up' algebraically:

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

= 1/(2 sqrt (1-x^2/4)

= 1/(2 sqrt ((4-x^2)/4)

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

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

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