Question

Slope of normal to the curve $y=x^2-\frac{1}{x^2}$ at $(-1,0)$​ is ​

• A. $\frac{1}{4}$
• B. $-\frac{1}{4}$
• C. $4$
• D. $-4$

Answer 1

The correct option is A

$\frac{1}{4}$

$\frac{dy}{dx}=2x+\frac{2}{x^3}\Rightarrow \frac{dy}{d{x}_{(-1,0)}}=-4\therefore -\frac{dx}{dy}=\frac{1}{4}$