# What is the derivative of y=xarcsin(x^2)?

Aug 29, 2015

${y}^{'} = \frac{2 {x}^{2}}{\sqrt{1 - {x}^{4}}} + \arcsin \left({x}^{2}\right)$

#### Explanation:

To make the calculations more interesting, I'll assume that you don't know what the derivative of $\arcsin \left(x\right)$ is.

You can differentiate this function by using implicit differentiation. Start by isolating $\arcsin \left({x}^{2}\right)$ on one side

$\frac{y}{x} = \arcsin \left({x}^{2}\right) \text{ } \textcolor{b l u e}{\left(1\right)}$

This is equivalent to

$\sin \left(\frac{y}{x}\right) = {x}^{2} \text{ } \textcolor{b l u e}{\left(2\right)}$

Differentiate both sides with respect to $x$

$\frac{d}{\mathrm{dx}} \left(\sin \left(\frac{y}{x}\right)\right) = \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

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

Rearrange this to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ isolated on one side

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{\cos \left(\frac{y}{x}\right) \cdot \frac{1}{x}} + \frac{y}{x} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\cos \left(\frac{y}{x}\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\cos \left(\frac{y}{x}\right)}}}}$

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

Use the trigonometric identity

$\textcolor{b l u e}{{\sin}^{2} x + {\cos}^{2} x = 1}$

To write $\cos \left(\frac{y}{x}\right)$ as a function of $\sin \left(\frac{y}{x}\right)$

${\cos}^{2} x = 1 - {\sin}^{2} x$

$\sqrt{{\cos}^{2} x} = \sqrt{1 - {\sin}^{2} x}$

$\cos x = \sqrt{1 - {\sin}^{2} x}$

Use this identity and equations $\textcolor{b l u e}{\left(1\right)}$ and $\textcolor{b l u e}{\left(2\right)}$ to write

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{2 {x}^{2}}{\sqrt{1 - {x}^{4}}} + \arcsin \left({x}^{2}\right)}$