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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 ?

Sooraj S
  • 7,573

3 Answers3

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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,

$$\frac{dy}{dx} = \frac{d\alpha}{dx}=\frac12 \frac 1{\sqrt x} \frac1{\sqrt{1- x}}=\frac1{2\sqrt{x(1- x)}} $$

Quanto
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Using $$ (\arcsin u)'=\frac1{\sqrt{1-u^2}}u' $$ one has \begin{eqnarray} \frac{dy}{dx}&=&\frac{1}{\sqrt{1-(\sqrt{x-ax}-\sqrt{a-ax})^2}}(\sqrt{x-ax}-\sqrt{a-ax})'\\ &=&\frac1{\sqrt{1-(x-ax)-(a-ax)+2\sqrt{a(1-a)x(1-x)}}}\bigg[\frac{\sqrt{1-a}}{2\sqrt x}+\frac{\sqrt a}{2\sqrt{1-x}}\bigg]\\ &=&\frac{1}{\sqrt{(1-a)(1-x)+ax+2\sqrt{a(1-a)x(1-x)}}}\cdot\frac{\sqrt{(1-a)(1-x)}+\sqrt{ax}}{2\sqrt{x(1-x)}}\\ &=&\frac{1}{2\sqrt{x(1-x)}}, \end{eqnarray} where $$ 1-(x-ax)-(a-ax)=(1-a)(1-x)+ax $$ is used.

xpaul
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You can also use the $\arcsin$ addition formula to get

\begin{eqnarray} f(x) &=& \arcsin\left(\sqrt{x-ax}-\sqrt{a-ax}\right)=\\ &=&\arcsin\left(\sqrt x \sqrt{1-a}-\sqrt a \sqrt{1-x}\right)=\\ &=&\arcsin\sqrt x - \arcsin \sqrt a. \end{eqnarray}

And now, easily, $$f'(x) = \left(\arcsin \sqrt x\right)'.$$

dfnu
  • 7,528