Derivative of $y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]$
Put $\sqrt{x}=\sin\alpha,\;\sqrt{a}=\sin\beta$
$$
y=\sin^{-1}\bigg[\sqrt{x-ax}-\sqrt{a-ax}\bigg]=\sin^{-1}\bigg[\sqrt{x(1-a)}-\sqrt{a(1-x)}\bigg]\\
=\sin^{-1}\bigg[\sin\alpha|\cos\beta|-|\cos\alpha|\sin\beta\bigg]
$$
My reference gives the solution $\dfrac{1}{2\sqrt{x}\sqrt{1-x}}$, but is it a complete solution ?
How do I proceed further with my attempt ?
Best Answer
Assume $0 < x < 1$ and $0 < a < 1$. Then, $0< \alpha < \frac\pi2$ and $0< \beta< \frac\pi2$. Continue with what you have,
$$ y=\sin^{-1}\bigg(\sin\alpha\cos\beta-\cos\alpha\sin\beta\bigg)=\sin^{-1}[\sin(\alpha-\beta)]=\alpha-\beta $$
Since $\alpha = \sin^{-1}\sqrt x$ and $(\sin^{-1}t)' = \frac1{\sqrt{1- t^2}}$, the derivative is,
$$\frac{dy}{dx} = \frac{d\alpha}{dx}=\frac12 \frac 1{\sqrt x} \frac1{\sqrt{1- x}}=\frac1{2\sqrt{x(1- x)}} $$