BuyFind

Single Variable Calculus: Concepts...

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

Single Variable Calculus: Concepts...

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

Solutions

Chapter 3.5, Problem 6E
To determine

To calculate: The derivative of the function 2x3+x2yxy3=2 by implicit differentiation.

Expert Solution

Answer to Problem 6E

The derivative of the function is 6x22xy+2y3x23y2 .

Explanation of Solution

Given information:

The function 2x3+x2yxy3=2 .

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 .

Product rule for differentiation is ddx(fg)=f'(x)g(x)+f(x)g'(x) where f and g are functions of x .

Calculation:

Consider the function 2x3+x2yxy3=2 .

Differentiate both sides with respect to x ,

  ddx(2x3+x2yxy3)=ddx(2)ddx(2x3)+ddx(x2y)ddx(xy3)=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) .

Also for the second and third term of the above expression, apply the product rule for differentiation.

Recall that product rule for differentiation is ddx(fg)=f'(x)g(x)+f(x)g'(x) where f and g are functions of x .

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

  ddx(2x3+x2yxy3)=ddx(2)ddx(2x3)+ddx(x2y)ddx(xy3)=06x2+x2y'+2xy2y33y2y'=0x2y'3y2y'=6x22xy+2y3

Isolate the value of y' on left hand side and simplify,

  x2y'3y2y'=6x22xy+2y3(x23y2)y'=6x22xy+2y3y'=6x22xy+2y3x23y2

Thus, the derivative of the function is 6x22xy+2y3x23y2 .

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