Question

If $y={\sin }^{-1}[x\sqrt{x-1}-\sqrt{x}\sqrt{1-x^2}]$, find $\frac{dy}{dx}$.

Answer 1

Consider the given expression.

$y={\sin }^{-1}[x\left(\sqrt{x-1}\right)-\sqrt{x}\sqrt{1-x^2}]$

Let $y={\sin }^{-1}\left(g\left(x\right)\right)$, where $g\left(x\right)=x\sqrt{x-1}-\sqrt{x}\sqrt{1-x^2}$. Then,

$\frac{dy}{dx}=\frac{dy}{dg\left(x\right)}\times \frac{dg\left(x\right)}{dx}$

Now,

$\frac{dy}{dg\left(x\right)}=\frac{1}{\sqrt{1-{\left[g\left(x\right)\right]}^2}}$

As,

${\left\{g\left(x\right)\right\}}^2={(x\sqrt{x-1}-\sqrt{x}\sqrt{1-x^2})}^2$

${\left\{g\left(x\right)\right\}}^2=x^2(x-1)+x(1-x^2)-2x\sqrt{x}\sqrt{(x-1)(1-x^2)}$

${\left\{g\left(x\right)\right\}}^2=x^3-x^2+x-x^3-2x\sqrt{x^2-x^4-x+x^3}$

${\left\{g\left(x\right)\right\}}^2=x-x^2-2x\sqrt{x^2-x^4-x+x^3}$

And,

$\frac{dg\left(x\right)}{dx}=\sqrt{x-1}+x\frac{1}{2\sqrt{x-1}}-\sqrt{x}\frac{-2x}{2\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}}\sqrt{1-x^2}$

$\frac{dg\left(x\right)}{dx}=\sqrt{x-1}+\frac{x}{2\sqrt{x-1}}+\frac{2x\sqrt{x}}{2\sqrt{1-x^2}}-\frac{\sqrt{1-x^2}}{2\sqrt{x}}$

Therefore,

$\frac{dy}{dx}=\frac{\sqrt{x-1}+\frac{x}{2\sqrt{x-1}}+\frac{2x\sqrt{x}}{2\sqrt{1-x^2}}-\frac{\sqrt{1-x^2}}{2\sqrt{x}}}{\sqrt{1-x+x^2+2x\sqrt{x^2-x^4-x+x^3}}}$