#### To determine

**To find:** The differentiation of y=ccost+t2sint.

#### Answer

The differentiation of y=ccost+t2sint is t2cost+(2t−c)sint_.

#### Explanation

**Given:**

The function is y=ccost+t2sint.

**Formula used:**

**The Product Rule:**

If f1(t) and f2(t) are both differentiable functions, then the product rule is,

ddt[f1(t)f2(t)]=f1(t)⋅ddt[f2(t)]+f2(t)⋅ddt[f1(t)] (1)

**The Power Rule:**

If *n* is real number, then the power rule is ddt(tn)=ntn−1 (2)

**The Sum Rule:**

If f(x). and g(x) are both differentiations, then the summation rule is,

ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)] (3)

**The Constant Multiple Rule:**

If c is a constant and f(t) is differentiation function, then the constant multiple rule is,

ddt[cf(t)]=cddt[f(t)] (4)

**Calculation:**

Apply the sum rule as shown in equation (3).

ddt[ccost+t2sint]=ddt[ccost]+ddt[t2sint] (5)

Apply the constant multiple rule as shown in equation (4).

ddt[ccost]=cddt[cost]=−csint

Apply the product rule as shown in equation (1).

Substitute t2 for f1(t) and sint for f2(t) in equation (1).

ddt[t2sint]=t2ddt[sint]+sintddt[t2]

Apply the power rule as shown in equation (2).

ddt[t2sint]=t2(cost)+sint(2t2−1)=t2cost+2tsint

Substitute −csint for ddt[ccost] and t2cost+2tsint for ddt[t2sint] in equation (5).

ddt[ccost+t2sint]=−csint+t2cost+2tsint=t2cost+2tsint−csint=t2cost+(2t−c)sint

Therefore, the differentiation of y=ccost+t2sint is t2cost+(2t−c)sint_.