Chapter 3.7, Problem 1E

To determine

To find $y^{\prime }$ , where

  $8x^2+y^2=10$

Answer to Problem 1E

  $y^{\prime }=-\frac{8x}{y}$

Explanation of Solution

Given:

  $8x^2+y^2=10$

Concept Used:

• Chain rule of differentiation is: $\frac{d}{dx}f\left[g\left(x\right)\right]=\frac{d}{dg}f\left(g\right)\frac{dg}{dx}$ , where u is a differentiable function at x and f is a differentiable function at $u\left(x\right)$
• Power rule of differentiation formula

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

To calculate $y^{\prime }$ , where

  $8x^2+y^2=10$

Use implicit differentiation. For that differentiate both side of the equation with respect to x as

  $\frac{d}{dx}\left[8x^2+y^2\right]=\frac{d}{dx}10$

Now solve it further using chain rule of differentiation, as

  $\begin{array}{l}\frac{d}{dx}\left[8x^2+y^2\right]=\frac{d}{dx}10\\ 8\frac{d}{dx}{x}^2+\frac{d}{dx}{y}^2=0\\ 8\left(2x\right)+2y\frac{d}{dx}y=0,\text{ using chain rule}\\ 16x+2y\frac{dy}{dx}=0\\ 2y\frac{dy}{dx}=-16x\\ \frac{dy}{dx}=-\frac{16x}{2y}\\ \Rightarrow y^{\prime }=-\frac{8x}{y}\end{array}$

Hence,

  $y^{\prime }=-\frac{8x}{y}$