#### To determine

**To find:** The derivative of the function h(t)=(t+1)23(2t2−1)3.

#### Answer

The derivative of h(t)=(t+1)23(2t2−1)3 is 23(t+1)−13(2t2−1)2(20t2+18t−1)_.

#### Explanation

**Given:**

The function is h(t)=(t+1)23(2t2−1)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)

**Product Rule:**

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

ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)] (2)

**Calculation:**

Obtain the derivative of h(t).

h′(t)=ddx[h(t)]=ddx[(t+1)23(2t2−1)3]

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

h′(t)=(t+1)23⋅ddt[(2t2−1)3]+(2t2−1)3⋅ddt[(t+1)23] (3)

Obtain the derivative ddt[(2t2−1)3] by using the power rule combined with the chain rule as shown equation (1).

ddt[(2t2−1)3]=3(2t2−1)3−1⋅ddt(2t2−1)=3(2t2−1)2(ddt(2t2)−ddt(1))=3(2t2−1)2(2(2t2−1)−(0))

Simplify further, the above derivative becomes

ddt[(2t2−1)3]=3(2t2−1)2(2(2t)−0)=3(2t2−1)2(4t)=12t(2t2−1)2

Thus, the derivative is ddt[(2t2−1)3]=12t(2t2−1)2 (4)

Obtain the derivative ddt[(t+1)23] by using the power rule combined with the chain rule as shown equation (1).

ddt[(t+1)23]=23(t+1)23−1⋅ddt(t+1)=23(t+1)2−33(ddt(t)+ddt(1))=23(t+1)−13((1t1−1)+(0))

Simplify further, the above derivative becomes

ddt[(t+1)23]=23(t+1)−13(t0+0)=23(t+1)−13(1)=23(t+1)−13

Thus, the derivative is ddt[(t+1)23]=23(t+1)−13 (5)

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

h′(t)=(t+1)2312t(2t2−1)2+(2t2−1)323(t+1)−13=(t+1)−13(2t2−1)2((t+1)12t+(2t2−1)23)=(t+1)−13(2t2−1)2(3(t+1)12t+2(2t2−1)3)=(t+1)−13(2t2−1)2(36t(t+1)+4t2−23)

Simplify further the above derivative becomes,

h′(t)=(t+1)−13(2t2−1)2(36t2+36t+4t2−23)=(t+1)−13(2t2−1)2(40t2+36t−23)=(t+1)−13(2t2−1)2(2(20t2+18t−1)3)=23(t+1)−13(2t2−1)2(20t2+18t−1)

Therefore, the derivative of h(t)=(t+1)23(2t2−1)3 is 23(t+1)−13(2t2−1)2(20t2+18t−1)_.