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 34E
To determine

To calculate: The second derivative of the function x4+y4=a4 by implicit differentiation.

Expert Solution

Answer to Problem 34E

The second derivative of the function is y''=3x2a4y7 .

Explanation of Solution

Given information:

The function x4+y4=a4 .

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 .

Quotient rule for differentiation is ddx(fg)=g(x)f'(x)f(x)g'(x)[g(x)]2 where f and g are functions of x .

Calculation:

Consider the function x4+y4=a4 .

Differentiate both sides with respect to x ,

  ddx(x4+y4)=ddx(a4)ddx(x4)+ddx(y4)=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(x4+y4)=ddx(a4)ddx(x4)+ddx(y4)=04x3+4y3y'=0x3+y3y'=0

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

  x3+y3y'=0y3y'=x3y'=x3y3

Therefore, the value of first derivative of the function is y'=x3y3 .

Now, again differentiate the above expression with respect to x . Recall that quotient rule for differentiation is ddx(fg)=g(x)f'(x)f(x)g'(x)[g(x)]2 where f and g are functions of x .

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

  ddx(y')=ddx(x3y3)y''=(y3)ddx(x3)(x3)ddx(y3)(y3)2y''=y3(3x2)+x3(3y2y')y6y''=3x2y3+3x3y2y'y6

Now, substitute the value y'=x3y3 in the above expression.

  y''=3x2y3+3x3y2(x3y3)y6=3x2y3+3x3(x3y)y6=3x2y43x6y7=3x2(y4+x4)y7

According to the question x4+y4=a4 , substitute the value in the above expression,

  y''=3x2(y4+x4)y7=3x2a4y7

Thus, the second derivative of the function is y''=3x2a4y7 .

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