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 89E

a.

To determine

Explain the proof of given relation.

Expert Solution

Explanation of Solution

Given:

The given relationis ddx(sinnxcosnx)=nsinn1xcos(n+1)x , where n is positive integer.

Calculation:

  ddx(sinnxcosnx)=nsinn1xcos(n+1)x

  dydt=ddt(2t3+3t2+1)

From L.H.S

  ddx(sinnxcosnx)

Applyproductrule (fg)'=gf'+fg' .

  =cosnxddx(sinnx)+sinnxddx(cosnx)

Use derivative rule ddx(xn)=nxn1 , ddx(sinx)=cosx and ddx(coskx)=ksinkx .

  =cosnxddx(sinnx)+sinnxddx(cosnx)=cosnx(nsinn1xcosx)+sinnx(nsinnx)=nsinn1x(cosxcosnxsinnxsinx)=nsinn1xcos(n+1)x=R.H.S

Hence ddx(sinnxcosnx)=nsinn1xcos(n+1)x proved

b.

To determine

Find the derivative formula for the given equation and compare with part (a).

Expert Solution

Answer to Problem 89E

The formula is ddx(cosnxcosnx)=ncosn1xsin(n+1)x .

Explanation of Solution

Given:

The given equationis y=cosnxcosnx .

Calculation:

  y=cosnxcosnx

  dydx=ddx(cosnxcosnx)

Apply product rule (fg)'=gf'+fg' .

  dydx=cosnxddx(cosnx)+cosnxddx(cosnx)

Use derivative rule ddx(xn)=nxn1 , ddx(cosx)=sinx and ddx(coskx)=ksinkx .

  dydx=cosnx(ncosn1xsinx)+cosnx(nsinnx)dydx=ncosn1x(sinxcosnx+sinnxcosx)dydx=ncosn1xsin(n+1)x

Hence the formula is ddx(cosnxcosnx)=ncosn1xsin(n+1)x .

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