# How do you differentiate (arcsinx)^2?

Nov 10, 2016

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

#### Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If $y = f \left(x\right)$ then $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

I was taught to remember that the differential can be treated like a fraction and that the "$\mathrm{dx}$'s" of a common variable will "cancel" (It is important to realise that $\frac{\mathrm{dy}}{\mathrm{dx}}$ isn't a fraction but an operator that acts on a function, there is no such thing as "$\mathrm{dx}$" or "$\mathrm{dy}$" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$ etc, or $\left(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\textcolor{red}{\cancel{\mathrm{dv}}}} \frac{\textcolor{red}{\cancel{\mathrm{dv}}}}{\textcolor{b l u e}{\cancel{\mathrm{du}}}} \frac{\textcolor{b l u e}{\cancel{\mathrm{du}}}}{\mathrm{dx}}\right)$

So with $y = {\left(\arcsin x\right)}^{2}$, Then:

$\left\{\begin{matrix}\text{Let "u=arcsinx & => & (du)/dx=1/sqrt(1-x^2) "(see below)" \\ "Then } y = {u}^{2} & \implies & \frac{\mathrm{dy}}{\mathrm{du}} = 2 u\end{matrix}\right.$

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$ we get:

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

NOTE - Derivative of $\arcsin \left(x\right)$
We can easily find $\frac{d}{\mathrm{dx}} \left(\arcsin x\right)$ using implicit differentiation:
Let $y = \arcsin x \iff \sin y = x$

$\therefore \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
Using ${\sin}^{2} y + {\cos}^{2} y \equiv 1 \implies {\cos}^{2} y = 1 - {x}^{2}$
$\therefore \cos y = \sqrt{1 - {x}^{2}}$
$\therefore \sqrt{1 - {x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{2}}}$