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 2E

  (a)

To determine

To calculate: The first derivative of the function cosx+y=5 by implicit differentiation.

Expert Solution

Answer to Problem 2E

The first derivative of the function is y'=2sinxy .

Explanation of Solution

Given information:

The function cosx+y=5 .

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 .

Calculation:

Consider the function cosx+y=5 .

Differentiate both sides with respect to x ,

  ddx(cosx+y)=ddx(5)ddx(cosx)+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(cosx+y)=ddx(5)ddx(cosx)+ddx(y)=0sinx+12yy'=0

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

  sinx+12yy'=012yy'=sinxy'=2sinxy

Thus, the first derivative of the function is y'=2sinxy .

  (b)

To determine

To calculate: The first derivative of the function cosx+y=5 by explicit differentiation.

Expert Solution

Answer to Problem 2E

The first derivative of the function is y'=2sinx(5cosx) .

Explanation of Solution

Given information:

The function cosx+y=5 .

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 cosx+y=5 .

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

Subtract the terms cosx on both sides of the equation.

  y=5cosx

Square both the sides of the above expression,

  y=(5cosx)2=25+cos2x10cosx

Differentiate both sides with respect to x ,

  ddx(y)=ddx(25+cos2x10cosx)y'=ddx(25)+ddx(cos2x)ddx(10cosx)

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,

  y'=02cosxsinx+10sinxy'=2sinx(5cosx)

Thus, the first derivative of the function is y'=2sinx(5cosx) .

  (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 cosx+y=5 by explicit differentiation is y'=2sinx(5cosx) and by implicit differentiation is y'=2sinxy .

Consider the derivative of the function cosx+y=5 that is by explicit differentiation it is y'=2sinx(5cosx) and by implicit differentiation it is y'=2sinxy .

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 cosx+y=5 .

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

Subtract the terms cosx on both sides of the equation.

  y=5cosx

The first derivative of the function by implicit differentiation is y'=2sinxy .

Now,

  y'=2sinxyy'=2sinx(5cosx)

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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