stuck in finding $\frac{d(\sin^{-1}(x))}{dx}$

Question

what is wrong with the following reasoning? $$\sin(\sin^{-1}(x))=x$$ This is of the form $f(g(x))=h(x)$. Then, by implicit differentiation: $$(f(g(x)))'=f'(g(x))\cdot g'(x)=h'(x) $$ Or in this case: $$\cos(\sin^{-1}(x))\frac{d(\sin^{-1}(x))}{dx} = 1$$ Then $$\frac{d(\sin^{-1}(x))}{dx} = \frac{1}{\cos(\sin^{-1}(x))}$$ However I do not think this is correct. I wonder if I made a conceptual mistake in the implicit differentiation. Perhaps there are some extra steps needed to get to the result $$\frac{d(\sin^{-1}(x))}{dx} = \frac{1}{\sqrt{1-x^2}}$$ If anyone could help me out that would be awesome! Thanks in advance.

Answer 1

Note we are making the usual assumptions on $x$ here.

Note that since $\sin(\arcsin x)=x$ and $\cos^2t+\sin^2t=1$ we have:$$\cos(\arcsin x)^2+\sin(\arcsin x)^2=1\\\cos(\arcsin x)^2=1-\sin(\arcsin x)^2\\\cos(\arcsin x)^2=1-x^2\\\cos(\arcsin x)=\sqrt{1-x^2}$$ Can you see now?

Answer 2

Let $y=\arcsin x=\sin^{-1}x$. Then $x=\sin y$, and so diffrentiating both sides of this equality w.r.t. $x$ and using chain rule we have $1=\cos y\frac{dy}{dx}$. Solving for $\frac {dy}{dx}$ and using the fact that $\cos y=\sqrt{1-x^2}$, (from the right triangle), we get $\frac {dy}{dx}=\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$.