The derivative of the function x 3 + y 3 = 1 by implicit differentiation.

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 3E
To determine

To calculate: The derivative of the function x3+y3=1 by implicit differentiation.

Expert Solution

The derivative of the function is x2y2 .

Explanation of Solution

Given information:

The function x3+y3=1 .

Formula used:

Thechain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .

Power rule for differentiation is ddxxn=nxn1 .

Calculation:

Consider the function x3+y3=1 .

Differentiate both sides with respect to x ,

ddx(x3+y3)=ddx(1)ddx(x3)+ddx(y3)=0

Recall that power rule for differentiation is ddxxn=nxn1 and chain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .

Apply it. Also observe that y is a function of x,

ddx(x3+y3)=ddx1ddx(x3)+ddx(y3)=03x2+3y2dydx=03y2dydx=3x2

Divide both sides by 3y2 and simplify,

3y2dydx=3x2dydx=3x23y2dydx=x2y2

Thus, the derivative of the function is x2y2 .

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