Question

If $y={\sin }^{-1}(x^2\sqrt{1-x}-\sqrt{x}\sqrt{1-x^4}),$ find $\frac{dy}{dx}.$

• A. $\frac{2x}{\sqrt{1-x^4}}-\frac{1}{2\sqrt{x}\sqrt{(1-x)}}$
• B. $\frac{2x}{\sqrt{1+x^4}}-\frac{1}{2\sqrt{x}\sqrt{(1-x)}}$
• C. $\frac{2x}{\sqrt{1-x^4}}-\frac{1}{2\sqrt{x}\sqrt{(1+x)}}$
• D. $\frac{2x}{\sqrt{1+x^4}}-\frac{1}{2\sqrt{x}\sqrt{(1+x)}}$

Answer 1

The correct option is A $\frac{2x}{\sqrt{1-x^4}}-\frac{1}{2\sqrt{x}\sqrt{(1-x)}}$

Let $y={\sin }^{-1}(x^2\sqrt{1-x}-\sqrt{x}\sqrt{1-x^4})$

put $x^2=\sin \theta$ and $\sqrt{x}=\sin \varphi$

$\Rightarrow y={\sin }^{-1}(\sin \theta .\cos \varphi -\sin \varphi \cos \theta )={\sin }^{-1}\sin (\theta -\varphi )=\theta -\varphi$

$\Rightarrow y={\sin }^{-1}{x}^2-{\sin }^{-1}\sqrt{x}$

$\therefore \frac{dy}{dx}=\frac{2x}{\sqrt{1-x^4}}-\frac{1}{2\sqrt{x}\sqrt{(1-x)}}$