Finding the derivative of $y=\sin(\cos^{-1}(2x))$ Question
Find the derivative of $y=\sin(\cos^{-1}(2x))$.
My Working
So I let $u=\cos^{-1}(2x)$, so
\begin{align}
\frac{du}{dx}&=-\frac{1}{\sqrt{1-(2x)^2}}\times2\\
&=-\frac{2}{\sqrt{1-4x^2}}
\end{align}
Also,
$$\frac{dy}{du}=\cos u$$
I feel like I should use the chain rule somewhere, but I am not sure where. Could anyone please help? Thanks!


This differentiation can also be accomplished in a couple of other ways by constructing the right triangle implied by the argument of the sine function. If we call $ \ \theta \ = \ \cos^{-1}(2x) \ \ , \ $ then $ \ \cos \theta \ = \ 2x \ = \ \frac{2x}{1} \ \ . \ $ With the hypotenuse of the right triangle taken as $ \ 1 \ \ , \ $ the "leg" opposite to $ \ \theta \ $ has length $ \  \sqrt{1 - (2x)^2} \ = \ \sqrt{1 - 4x^2} \ \ . $
We can now write our function as $ \ y \ = \ \sin \theta \ = \ \sin(\cos^{-1}(2x)) \ = \ \frac{\sqrt{1 - 4x^2}}{1} \ = \ (1 - 4x^2)^{1/2} \ \ . \ $  The derivative is then $$   y' \ \ = \ \ \frac{1}{2} · (1 - 4x^2)^{-1/2} · (1 - 4x^2)' \ \ = \ \ \frac{1}{2} · (- \ 8x) · (1 - 4x^2)^{-1/2} \ \ = \ \ \frac{- \ 4x}{\sqrt{1 - 4x^2}} \ \ .  $$
If we instead begin by re-arranging the expression as $ \ \sin^{-1} y \ = \ \cos^{-1}(2x) \ \ $ (an equation of two angles), implicit differentiation of the left side produces
$$ \frac{1}{\sqrt{1 - y^2}} \ · \ y' \ \ = \ \ -\frac{1}{\sqrt{1 - (2x)^2}} · (2x)' \ \ \Rightarrow \ \ y' \ \ = \ \ -\frac{2 · \sqrt{1 - y^2}}{\sqrt{1 - 4x^2}} \ \ . $$
We then "clear" $ \ y \ $ from the expression for the derivative by applying our earlier result from the constructed right triangle:
$$ y' \ \ = \ \ -\frac{2 · \sqrt{1 - [ \ (1 - 4x^2)^{1/2} \ ]^2}}{\sqrt{1 - 4x^2}} \ \ = \ \ -\frac{2 · \sqrt{1 - (1 - 4x^2)}}{\sqrt{1 - 4x^2}} $$ $$ = \ \ -\frac{2 · \sqrt{ 4x^2 }}{\sqrt{1 - 4x^2}} \ \ = \ \  -\frac{4x}{\sqrt{1 - 4x^2}} \ \ . $$
