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 32E
To determine

To calculate: The second derivative of the function x+y=1 by implicit differentiation.

Expert Solution

Answer to Problem 32E

The second derivative of the function is y''=12xx .

Explanation of Solution

Given information:

The function x+y=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 .

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 x+y=1 .

Differentiate both sides with respect to x ,

  ddx(x+y)=ddx(1)ddx(x)+ddx(y)=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(x+y)=ddx(1)ddx(x)+ddx(y)=012x+12yy'=01x+1yy'=0

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

  1x+1yy'=01yy'=1xy'=yx

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

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(yx)y''=(x)ddx(y)(y)ddx(x)(x)2y''=x(12yy')+y(12x)xy''=xy'2y+y2xx

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

  y''=x2y(yx)+y2xx=12++y2xx=x+y2xx

According to the question x+y=1 , substitute the value in the above expression,

  y''=x+y2xx=12xx

Thus, the second derivative of the function is y''=12xx .

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