The slope of the normal to the curve $y=x^3-4 \sin x$ at x = 0 is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{4}$
$y=x^3-4 \sin x$
$⇒\frac{dy}{dx}=3x^2-4\cos x$
and,
$\left.\frac{dy}{dx}\right|_{x=0}=3(0)^2-4\cos (0)$
$=-4$
∴ Slope of the normal, $m=\frac{1}{4}$ $[m×\frac{dy}{dx}=-1]$