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 20E
To determine

Find the derivative of the given function.

Expert Solution

Answer to Problem 20E

The derivative of the given function is h'(t)=12t2(t41)2(t3+1)3[2t4+t1] .

Explanation of Solution

Given:

The given function is h(t)=(t41)3(t3+1)4 .

Calculation:

  h(t)=(t41)3(t3+1)4

Apply product rule.

  (fg)=f'g+fg'

  ddth(t)=ddt(t41)3(t3+1)4+(t41)3ddt(t3+1)4

Apply chain rule.

  df(a)dx=dfdadadx

Let f=a3,a=t41 and f=a4,a=t3+1

  h'(t)=[dda(a3)ddt(t41)](t3+1)4+(t41)3[dda(a4)ddt(t3+1)]

Apply sum/difference rule.

  f±g=f'±g'

  h'(t)=[dda(a3){ddt(t4)ddt(1)}](t3+1)4+(t41)3[dda(a4){ddt(t3)+ddt(1)}] Use derivative rule.

  ddx(xn)=nxn1 .

  h'(t)=[3a2{4t30}](t3+1)4+(t41)3[4a3{3t2+0}]h'(t)=[12a2t3](t3+1)4+(t41)3[12a3t2]

Substitute the value of a=t41 and a=t3+1 .

  h'(t)=[12(t41)2t3](t3+1)4+(t41)3[12(t3+1)3t2]h'(t)=12t3(t41)2(t3+1)4+12t2(t41)3(t3+1)3h'(t)=12t2(t41)2(t3+1)3[t(t3+1)+(t41)]h'(t)=12t2(t41)2(t3+1)3[t4+t+t41]h'(t)=12t2(t41)2(t3+1)3[2t4+t1]

Hence the derivativeof the given function is h'(t)=12t2(t41)2(t3+1)3[2t4+t1] .

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