Chapter 4, Problem 80RE

To determine

To find: The $\frac{d^{2}y}{d{x}^2}$ at the point $(2,\sqrt{3)}$ .

Answer to Problem 80RE

The value of $\frac{d^{2}y}{d{x}^2}$ at the point $(2,\sqrt{3)}$ is $\frac{1}{3\sqrt{3}}$ .

Explanation of Solution

Given information: The given equation is $x^2-y^2=1$ .

Calculation: The given equation is $x^2-y^2=1$ .

Firstly find the first derivative:

  $\begin{array}{l}2x-2y\frac{dy}{dx}=0\\ -2y\frac{dy}{dx}=-2x\\ \frac{dy}{dx}=\frac{-2x}{-2y}=\frac{x}{y}\end{array}$

Then find the second derivative:

  $\begin{array}{l}\frac{dy}{dx}=\frac{x}{y}\\ \frac{d^{2}y}{d{x}^2}=\frac{y\frac{d}{dx}(x)-x\frac{d}{dx}(y)}{y^2}\\ \frac{d^{2}y}{d{x}^2}=\frac{y(1)-x\frac{dy}{dx}}{y^2}\\ \mathrm{Put}\frac{dy}{dx}=\frac{x}{y}\\ \frac{d^{2}y}{d{x}^2}=\frac{y-x\frac{x}{y}}{y^2}=\frac{\frac{y^2-x^2}{y}}{y^2}\\ \frac{d^{2}y}{d{x}^2}=\frac{y^2-x^2}{y\times y^2}=\frac{y^2-x^2}{y^3}\\ \mathrm{Put}x=2\mathrm{and}y=\sqrt{3}\\ \frac{d^{2}y}{d{x}^2}=\frac{2^2-{\sqrt{3}}^2}{{\sqrt{3}}^3}=\frac{4-3}{3\sqrt{3}}=\frac{1}{3\sqrt{3}}\end{array}$

Thus, the value of $\frac{d^{2}y}{d{x}^2}$ at the point $(2,\sqrt{3)}$ is $\frac{1}{3\sqrt{3}}$ .