Question

If $y={\sin }^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, then $dy/dx$ equals -

• A. $\frac{2}{1+x^2}$
• B. $\frac{1}{\sqrt{1-x^2}}$
• C. $\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$

In this we would be using the formula

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

In this formula if we put $y=\sqrt{x}$ then we get,

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

We would get our equation.

So, we can write our given equation as:

$y=\arcsin \left(x\right)-\arcsin \left(\sqrt{x}\right)$

And we know,

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

So, differentiating the given equation y we get,

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

Option [C]