   Chapter 3.4, Problem 20E

Chapter
Section
Textbook Problem

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

To determine

To find:  The derivative of the function F(t)=(3t1)4(2t+1)3.

Explanation

Given:

The function is F(t)=(3t1)4(2t+1)3.

Result used:

The Power Rule combined with the Chain Rule:

If n is any real number and g(x) is differentiable function, then

ddx[g(x)]n=n[g(x)]n1g(x) (1)

Quotient Rule:

If f(t). and g(t) are both differentiable function, then

ddt[f(t)g(t)]=g(t)ddt[f(t)]f(t)ddt[g(t)][g(t)]2 (2)

Calculation:

Obtain the derivative of F(t).

F(t)=ddt(F(t))=ddt[(3t1)4(2t+1)3]=ddt[(3t1)4(2t+1)3]

Apply the quotient rule as shown in equation (2),

F(t)=(2t+1)3ddt[(3t1)4](3t1)4ddt[(2t+1)3][(2t+1)3]2 (3)

Obtain the derivative ddt[(3t1)4] by using the power rule combined with the chain rule as shown equation (1).

ddt[(3t1)4]=4(3t1)41ddt(3t1)=4(3t1)3(ddt(3t)ddt(1))=4(3t1)3(3(1t11)(0))=4(3t1)3(3(t0)0)

Simplify further, the above derivative becomes

ddt[(3t1)4]=4(3t1)3(3(1))=4(3t1)3(3)=12(3t1)3

Thus, the derivative is ddt[(3t1)4]=12(3t1)3 (4)

Obtain the derivative ddt[(2t+1)3] by using the power rule combined with the chain rule as shown equation (1)

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