#### To determine

**To find:** The derivative of f(t)=tsinπt.

#### Answer

The derivative of f(t)=tsinπt is tπcosπt+sinπt_.

#### Explanation

**Given:**

The function is f(t)=tsinπt.

**Derivative Rules:**

**The Chain Rule:**

If *g* is differentiable at *t* and *f* is differentiable at g(t), then the composite function F=f∘g defined by F(t)=f(g(t)) is differentiable at *x* and F′ is given by the product

F′(t)=f′(g(t))⋅g′(t) (1)

**(1) Power Rule:** ddx(xn)=nxn−1

**(2) Product Rule:** ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)]

**Calculation:**

Obtain the derivative of f(t).

f′(t)=ddt(f(t))=ddt(tsinπt)

Apply the product rule (2) and the power rule (1),

f′(t)=tddt(sinπt)+sinπtddt(t)=tddt(sinπt)+sinπt(1t1−1)=tddt(sinπt)+sinπt(1)

f′(t)=tddt(sinπt)+sinπt (2)

Obtain the derivative ddt(sinπt) by using the chain rule as shown in equation (1).

Let g(t)=πt and h(u)=sinu where u=g(t)

Then, ddt(sinπt)=h′(g(t))⋅g′(t) (3)

The derivative h′(g(t)) is computed as follows,

h′(g(t))=f′(u)=ddu(f(u))=ddu(sinu)=cosu

Substitute u=πt in the above equation,

h′(g(t))=cosπt.

The derivative of g(t) is computed as follows,

g′(t)=ddx(g(t))=ddx(πt)=π

Thus, the derivative of g(t) is g′(t)=π.

Substitute cosπt for h′(g(t)) and π for g′(t) in equation (3),

ddt(sinπt)=(cosπt)⋅(π)=πcosπt

Thus, the derivative is ddt(sinπt)=πcosπt.

Substitute ddt(sinπt)=πcosπt in equation (2),

f′(t)=t(πcosπt)+sinπt=tπcosπt+sinπt

Therefore, the derivative of f(t)=tsinπt is f′(t)=tπcosπt+sinπt_.