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 8E
To determine

To calculate: The derivative of the function y5+x2y3=1+yex2 by implicit differentiation.

Expert Solution

Answer to Problem 8E

The derivative of the function is 2xy(ex2y2)5y4+3x2y2ex2 .

Explanation of Solution

Given information:

The function y5+x2y3=1+yex2 .

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 y5+x2y3=1+yex2 .

Differentiate both sides with respect to x ,

  ddx(y5+x2y3)=ddx(1+yex2)ddx(y5)+ddx(x2y3)=ddx(1)+ddx(yex2)

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 terms 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(y5+x2y3)=ddx(1+yex2)ddx(y5)+ddx(x2y3)=ddx(1)+ddx(yex2)5y4y'+2xy3+3x2y2y'=0+ex2y'+2xyex2

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

  5y4y'+2xy3+3x2y2y'=0+ex2y'+2xyex2(5y4+3x2y2ex2)y'=2xy3+2xyex2y'=2xy3+2xyex25y4+3x2y2ex2y'=2xy(ex2y2)5y4+3x2y2ex2

Thus, the derivative of the function is 2xy(ex2y2)5y4+3x2y2ex2 .

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