# What is the derivative of  y = Arcsin ((3x)/4)?

Jul 6, 2015

Use the Chain rule and the definition of the derivative of the arcsin of a function.

#### Explanation:

First, understand that $y = {\sin}^{-} 1 \left(3 \frac{x}{4}\right)$ is way of saying $y = f \left(g \left(x\right)\right)$

Second, let $f \left(m\right) = {\sin}^{-} 1 \left(m\right)$ (i.e., the "outside" part) and $m = g \left(x\right) = 3 \frac{x}{4}$ (i.e., the "inside" part).

Third, use the Chain Rule which states $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dm}} \cdot \frac{\mathrm{dm}}{\mathrm{dx}} = f ' \left(m\right) g ' \left(x\right)$

Fourth, find $\frac{\mathrm{dy}}{\mathrm{dm}}$

$\frac{\mathrm{dy}}{\mathrm{dm}} = f ' \left(m\right) = \frac{d}{\mathrm{dm}} \left({\sin}^{-} 1 \left(m\right)\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

Fifth, find $\frac{\mathrm{dm}}{\mathrm{dx}}$

$\frac{\mathrm{dm}}{\mathrm{dx}} = g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 \frac{x}{4}\right) = \frac{3}{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dm}} \frac{\mathrm{dm}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}} \cdot \frac{3}{4} = \frac{3}{4 \sqrt{1 - {x}^{2}}}$.