#### To determine

**To find:** The derivative of the function y=xsin1x.

#### Answer

The derivative of y=xsin1x is y′=sin1x−1xcos(1x).

#### Explanation

**Given:**

The function is y=xsin1x.

**Result used:**

**The Chain Rule:**

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

F′(x)=f′(g(x))⋅g′(x) (1)

**Derivative Rule:**

(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 y.

y′=ddx(y)=ddx(xsin1x)

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

y′=xddx(sin1x)+sin1xddx(x)=xddx(sinx−1)+sin1x(1)

=xddx(sinx−1)+sin1x (2)

Obtain the derivative ddx(sinx−1) using the chain rule as shown in equation (1)

Let g(x)=x−1 and f(u)=sinu where u=g(x).

ddx(sinx−1)=f′(g(x))⋅g′(x) (3)

The derivative f′(g(x)) is computed as follows,

f′(g(x))=f′(u)=ddu(f(u))=ddu(sinu)=cosu

Substitute u=x−1 in above equation,

f′(g(x))=cos1x

Thus, the derivative of f′(g(x))=cos1x.

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

g′(x)=ddx(1x)=ddx(x−1)=−(1x−1−1)=−x−2

Thus, the derivative of g(x) is g′(x)=−1x2.

Substitute cos1x for f′(g(x)) and −1x2 for g′(x) in equation (3)

f′(g(x))⋅g′(x)=cos1x(−1x2)=−cos1xx2

Thus, the derivative of ddx(sinx−1)=−cos1xx2.

Substitute ddx(sinx−1)=−cos1xx2 in equation (2)

y′=x(−cos1xx2)+sin1x=sin1x−1xcos1x

Therefore, the derivative of y=xsin1x is y′=sin1x−1xcos(1x).