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 1E

  (a)

To determine

To calculate: The first derivative of the function xy+2x+3x2=4 by implicit differentiation.

Expert Solution

Answer to Problem 1E

The first derivative of the function is y'=y6x2x .

Explanation of Solution

Given information:

The function xy+2x+3x2=4 .

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 xy+2x+3x2=4 .

Differentiate both sides with respect to x ,

  ddx(xy+2x+3x2)=ddx(4)ddx(xy)+ddx(2x)+ddx(3x2)=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(xy+2x+3x2)=ddx(4)ddx(xy)+ddx(2x)+ddx(3x2)=0y+xy'+2+6x=0

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

  y+xy'+2+6x=0xy'=y6x2y'=y6x2x

Thus, the first derivative of the function is y'=y6x2x .

  (b)

To determine

To calculate: The first derivative of the function xy+2x+3x2=4 by explicit differentiation.

Expert Solution

Answer to Problem 1E

The first derivative of the function is y'=3x24x2 .

Explanation of Solution

Given information:

The function xy+2x+3x2=4 .

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 .

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 xy+2x+3x2=4 .

Isolate the value of y on left hand side on the above equation.

Subtract the terms 2x+3x2 on both sides of the equation.

  xy=2x3x2+4y=2x3x2+4x

Differentiate both sides with respect to x ,

  ddx(y)=ddx(2x3x2+4x)y'=ddx(2x3x2+4x)

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

  y'=x(2x3x2+4)'(2x3x2+4)(x)'x2y'=x(26x)(2x3x2+4)1x2y'=2x6x2+2x+3x24x2y'=3x24x2

Thus, the first derivative of the function is y'=3x24x2 .

  (c)

To determine

To verify: The solutions obtained in parts (a) and (b) are consistent.

Expert Solution

Explanation of Solution

Given information:

The derivative of the function xy+2x+3x2=4 by explicit differentiation is y'=3x24x2 and by implicit differentiation is y'=y6x2x .

Consider the derivative of the function xy+2x+3x2=4 that is by explicit differentiation it is y'=3x24x2 and by implicit differentiation it is y'=y6x2x .

In order to verify that derivative obtained implicitly and explicitly is same substitute the value of y from the function in the value of derivative found by implicit differentiation.

Consider the function xy+2x+3x2=4 .

Isolate the value of y on left hand side on the above equation.

Subtract the terms 2x+3x2 on both sides of the equation.

  xy=2x3x2+4y=2x3x2+4x

The first derivative of the function by implicit differentiationis y'=y6x2x .

Now,

  y'=(2x3x2+4x)6x2x=2x+3x246x22xx2=3x24x2

Since, the derivative found implicitly is equal to derivative found explicitly so both the derivatives are equal.

Hence, the solutions obtained in parts (a) and (b) are consistent.

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