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 3.4, Problem 92E

a.

To determine

Explain the proof of the given relation using chain rule.

Expert Solution

Explanation of Solution

Given:

The given relationis ddx|x|=x|x| , where |x|=x2 .

Calculation:

  ddx|x|=x|x|

From L.H.S

  ddx|x|

  |x|=x2

  ddx|x|=ddxx2

Apply chain rule.

  df(a)dx=dfdadadx

Let f=a,a=x2

  ddx|x|=dda(a)ddx(x2)

Use derivative rule.

  ddx(x)=12xandddx(xn)=nxn1 .

  ddx|x|=12a2xddx|x|=xa

Substitute the value of |x|=x2 .

  ddx|x|=xx2ddx|x|=x|x|=R.H.S

Hence ddx|x|=x|x| proved.

b.

To determine

Find the derivative of the given function and sketch the graph of function and its derivative.

Expert Solution

Answer to Problem 92E

The derivative of the function is f'(x)=sinxcosx|sinx| .

Explanation of Solution

Given:

The given function is f(x)=|sinx| , where |x|=x2 .

Calculation:

  f(x)=|sinx|

  ddx[f(x)]=ddx|sinx|

where |sinx|=sin2x .

Apply chain rule.

  df(a)dx=dfdadadx

Let f=a,a=sin2x

  f'(x)=dda(a)ddx(sin2x)

Use derivative rule.

  ddx(x)=12x,ddx(xn)=nxn1andddx(sinx)=cosx .

  f'(x)=12a2sinxcosxf'(x)=sinxcosxa

Substitute the value of |sinx|=sin2x .

  f'(x)=sinxcosxsin2xf'(x)=sinxcosx|sinx|

The graph of the function and its derivative is given below.

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.4, Problem 92E , additional homework tip  1

Green line shows the graph of the function and blue line shows the graph of its derivative.

Hence the derivative of the function is f'(x)=sinxcosx|sinx| .

c.

To determine

Find the derivative of the given function and sketch the graph of function and its derivative.

Expert Solution

Answer to Problem 92E

The derivative of the function is g'(x)=xcos|x||x| .

Explanation of Solution

Given:

The given function is g(x)=sin|x| , where |x|=x2 .

Calculation:

  g(x)=sin|x|

  ddx[g(x)]=ddx(sin|x|)

where |x|=x2 .

Apply chain rule.

  df(a)dx=dfdadadx

Let f=sina,a=x2

  g'(x)=dda(sina)ddx(x2)

Use derivative rule.

  ddx(x)=12x,ddx(xn)=nxn1andddx(sinx)=cosx .

  g'(x)=cosa2x2x2g'(x)=cosaxx2

Substitute the value of |x|=x2 .

  g'(x)=cosx2xx2g'(x)=xcos|x||x|

The graph of the function and its derivative is given below.

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.4, Problem 92E , additional homework tip  2

Green line shows the graph of the function and blue line shows the graph of its derivative.

Hence the derivative of the function is g'(x)=xcos|x||x| .

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