`F(t) = (3t - 1)^4 (2t + 1)^-3` Find the derivative of the function.

Textbook Question

Chapter 3, 3.4 - Problem 20 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

Posted on

`F(t)=(3t-1)^4(2t+1)^-3`

`F(t)=(3t-1)^4/(2t+1)^3`

Derivative can be found either by quotient rule or product rule.

Using quotient rule

`F'(t)=((2t+1)^3 d/(dt) (3t-1)^4 - (3t-1)^4 d/(dt) (2t+1)^3)/(2t+1)^6`

`F'(t) =((2t+1)^3 *(4(3t-1)^3*3 )- (3t-1)^4(3(2t+1)^2*2))/(2t+1)^6`

`F'(t)=((12(2t+1)^3(3t-1)^3) - (6(3t-1)^4(2t+1)^2))/(2t+1)^6`

`F'(t)=(6(3t-1)^3(2t+1)^2(2(2t+1)-(3t-1)))/(2t+1)^6`

`F'(t)=(6(3t-1)^3(2t+1)^2(t+3))/(2t+1)^6`

`F'(t)=(6(t+3)(3t-1)^3)/(2t+1)^4`

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gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

Using the quotient rule of derivatives:

`d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)`

`F'(t) = d/dt [(3t-1)^4 (2t+1)^(-3)] = 4(3t-1)^3 * 3* (2t+1)^(-3) `

`+ (3t-1)^4 * (-3) * (2t+1)^(-4) * 2`

`=(3t-1)^3 (2t-1)^-3 [12-6(3t-1)(2t+1)^-1]`

`=6(t+3)(3t-1)^3 (2t+1)^-4`

Hope this helps.

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