Prove that $\arcsin (x)$ differentiates to $\frac{1}{ \sqrt {1-x^2}}$

Question

I want to prove this. I have no idea where to start. How do I do it?

Answer 1

Let $y=\arcsin(x).$

Then $$ \sin(y)=x\tag{*}.$$

Now differentiate $(*)$ with respect to $x$, to give $$\cos(y) \cdot\frac{dy}{dx}=1,$$ from which we get $$\frac{dy}{dx}=\frac{1}{\color{green}{\cos(y)}}.$$

But if we use the identity $\sin^2(y)+\cos^2(y) \equiv 1$ and re-arrange it, we get $$\color{green}{\cos(y) \equiv \sqrt{1-\sin^2(y)}}$$

Subbing this into our $\frac{dy}{dx},$ we have $$\frac{dy}{dx}=\frac{1}{\sqrt{1-\sin^2(y)}}=\frac{1}{\sqrt{1-x^2}}.$$ $\square$