Question:
Write the slope of the normal to the curve $y=\frac{1}{x}$ at the point $\left(3, \frac{1}{3}\right)$.
Solution:
Given that $y=\frac{1}{x}$
On differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x}=-\frac{1}{x^{2}}$
Now, slope of the tangent at $\left(3, \frac{1}{3}\right)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{9}$
Slope of normal $=9$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.