#### To determine

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

#### Answer

The derivative of F(t)=(3t−1)4(2t+1)−3 is F′(x)=6(3t−1)3(t+3)(2t+1)4_.

#### Explanation

**Given:**

The function is F(t)=(3t−1)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)]n−1g′(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[(3t−1)4(2t+1)−3]=ddt[(3t−1)4(2t+1)3]

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

F′(t)=(2t+1)3⋅ddt[(3t−1)4]−(3t−1)4⋅ddt[(2t+1)3][(2t+1)3]2 (3)

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

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

Simplify further, the above derivative becomes

ddt[(3t−1)4]=4(3t−1)3(3(1))=4(3t−1)3(3)=12(3t−1)3

Thus, the derivative is ddt[(3t−1)4]=12(3t−1)3 (4)

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

ddt[(2t+1)3]=3(2t+1)3−1ddt(2t+1)=3(2t+1)2(ddt(2t)+ddt(1))=3(2t+1)2(2(1t1−1)+(0))=3(2t+1)2(2(1t0)+0)

Simplify further, the above derivative becomes

ddt[(2t+1)3]=3(2t+1)2(2(1))=3(2t+1)2(2)=6(2t+1)2

Thus, the derivative is ddt[(2t+1)3]=6(2t+1)2 (5)

Substitute equations (3) and (4) in equation (3)

F′(t)=(2t+1)3(12(3t−1)3)−(3t−1)4(6(2t+1)2)[(2t+1)3]2=12(2t+1)3((3t−1)3)−6(3t−1)4((2t+1)2)[(2t+1)3]2=(3t−1)3(2t+1)2(12(2t+1)−6(3t−1))(2t+1)6=(3t−1)3(24t+12−18t+6)(2t+1)−2(2t+1)6

Simplify further, the above derivative becomes

F′(t)=(3t−1)3(6t+18)(2t+1)6−2=6(3t−1)3(t+3)(2t+1)4

Therefore, the derivative of F(t)=(3t−1)4(2t+1)−3 is F′(x)=6(3t−1)3(t+3)(2t+1)4_.