   Chapter 3.5, Problem 38E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find y″ by implicit differentiation.38. x3 – y3 = 7

To determine

To find: The derivative y by implicit differentiation.

Explanation

Given:

The equation x3y3=7.

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x)  are both differentiable function, then dydx=dydududx.

(2) Quotient Rule: If f1(x) and f2(x) are both differentiable, then ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

Calculation:

Obtain the first derivative of the equation.

Consider the equation x3y3=7.

Differentiate the equation implicitly with respect to x,

ddx(x3y3)=ddx(7)ddx(x3)ddx(y3)=03x2ddx(y3)=0

Apply the chain rule (1) and simplify the terms,

3x2ddx(y3)=03x2[ddy(y3)dydx]=03x23y2dydx=0dydx=x2y2

Therefore, the derivative of the equation is dydx=x2y2.

Obtain the second derivative of the equation.

The first derivative is y(x)=x2y2

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