#### To determine

**To find:** The derivative of the function F(t)=t2t3+1.

#### Answer

The derivative of F(t)=t2t3+1 is F′(t)=2t(t3+1)12−3t42(t3+1)−12(t3+1)_.

#### Explanation

**Given:**

The function is F(t)=t2t3+1.

**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(t2t3+1)

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

F′(t)=t3+1ddt[t2]−t2ddt[t3+1][t3+1]2 (3)

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

ddt[t2]=2t2−1⋅dtdt=2t⋅(1)=2t

Thus, the derivative is ddt[t2]=2t.

Obtain the derivative ddt[t3+1] using the power rule combined with the chain rule as shown equation (1),

ddt[t3+1]=ddt[(t3+1)12]=12(t3+1)12−1(ddt(t3)+ddt(1))=12(t3+1)1−22((3t2)+(0))=3t22(t3+1)−12

Thus, the derivative is ddt[t3+1]=3t22(t3+1)−12.

Substitute 2t for ddt[t2] and 3t22(t3+1)−12 for ddt[t3+1] in equation (3),

F′(t)=t3+1(2t)−t2(3t22(t3+1)−12)[t3+1]2=2t(t3+1)12−3t42(t3+1)−12[(t3+1)12]2=2t(t3+1)12−3t42(t3+1)−12(t3+1)

Therefore, The derivative of F(t)=t2t3+1 is F′(t)=2t(t3+1)12−3t42(t3+1)−12(t3+1)_.