temujinzujb

2023-03-26

What is the derivative of $y=\mathrm{arcsin}\left(\frac{3x}{4}\right)$?

### Answer & Explanation

awesomeamber802x

Use the Chain rule and the definition of the derivative of the arcsin of a function.
First, understand that $y={\mathrm{sin}}^{-1}\left(3\frac{x}{4}\right)$ is way of saying $y=f\left(g\left(x\right)\right)$
Second, let $f\left(m\right)={\mathrm{sin}}^{-1}\left(m\right)$ (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 $\frac{dy}{dx}=\frac{dy}{dm}\cdot \frac{dm}{dx}=f\prime \left(m\right)g\prime \left(x\right)$
Fourth, find $\frac{dy}{dm}$
$\frac{dy}{dm}=f\prime \left(m\right)=\frac{d}{dm}\left({\mathrm{sin}}^{-1}\left(m\right)\right)=\frac{1}{\sqrt{1-{x}^{2}}}$
Fifth, find $\frac{dm}{dx}$
$\frac{dm}{dx}=g\prime \left(x\right)=\frac{d}{dx}\left(3\frac{x}{4}\right)=\frac{3}{4}$
Hence, multiply your results:
$\frac{dy}{dx}=\frac{dy}{dm}\frac{dm}{dx}=\frac{1}{\sqrt{1-{x}^{2}}}\cdot \frac{3}{4}=\frac{3}{4\sqrt{1-{x}^{2}}}$.

Do you have a similar question?

Recalculate according to your conditions!