# How do I find the derivative of  y=xsin^(-1)x + sqrt(1-x^2)?

Aug 22, 2015

${y}^{'} = \arcsin x$

#### Explanation:

You can differentiate this function by using the product rule and the chain rule, provided that you know that

d/dx(arcsinx) = 1/(sqrt(1-x^2)

So, your function can be written like this

$y = x \cdot \arcsin x + \sqrt{1 - {x}^{2}}$

You will use the product rule to differentiate $x \cdot \arcsin x$, and the chain rule to differentiate $\sqrt{u}$, with $u = 1 - {x}^{2}$. This will get you

$\frac{d}{\mathrm{dx}} \left(x \cdot \arcsin x\right) = \left[\frac{d}{\mathrm{dx}} \left(x\right)\right] \cdot \arcsin x + x \cdot \frac{d}{\mathrm{dx}} \left(\arcsin x\right)$

d/dx(x * arcsinx) = 1 * arcsinx + x * 1/(sqrt(1-x^2)

and

$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{d}{\mathrm{du}} \sqrt{u} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

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

$\frac{d}{\mathrm{dx}} \left(\sqrt{1 - {x}^{2}}\right) = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \frac{1}{\sqrt{1 - {x}^{2}}} \cdot \left(- \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x\right)$

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

The derivative of $y$ will thus be

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

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

${y}^{'} = \arcsin x + \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{x}{\sqrt{1 - {x}^{2}}}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{x}{\sqrt{1 - {x}^{2}}}}}}$

${y}^{'} = \textcolor{g r e e n}{\arcsin x}$