Question

The derivative of $cose{c}^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is

• A. -4
• B. 4
• C. -1
• D. None of these

Answer 1

Answer: (A)

-4

Let $y=\frac{1}{2x^2-1}$

$\frac{d{\csc }^{-1}y}{d\sqrt{1-x^2}}=\frac{d{\csc }^{-1}y}{dx}\times \frac{dx}{d\sqrt{1-x^2}}$

$\frac{d{\csc }^{-1}y}{dx}=\frac{-1}{\left|y\right|\sqrt{y^2-1}}$

$\frac{d{\csc }^{-1}y}{dx}=\frac{-1}{\left|\frac{1}{2x^2-1}\right|\sqrt{{\left(\frac{1}{2x^2-1}\right)}^2-1}}\times (-1){(2x^2-1)}^{2}4x$

$\frac{d{\csc }^{-1}y}{dx}=\frac{2}{\sqrt{1-x^2}}$

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

$\frac{d{\csc }^{-1}y}{d\sqrt{1-x^2}}=-\frac{2}{x}$

When $x=\frac{1}{2}$

$\frac{d{\csc }^{-1}y}{d\sqrt{1-x^2}}=-4$