#### To determine

**To find:** The equation of the tangent line to the curve at the point.

#### Answer

The equation of the tangent line to the curve y=2xsinx at (π2,π) is y=2x_.

#### Explanation

**Given:**

The equation of the curve is y=2xsinx and the point is (π2,π).

**Derivative rules:**

(1) Constant Multiple Rule: ddx[c⋅f(x)]=c⋅ddxf(x)

(2) Power Rule: ddx(xn)=nxn−1

(3) Product Rule: ddx(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(x))

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

**Calculation:**

The derivative of y is dydx, which is obtained as follows,

dydx=ddx(y) =ddx(2xsinx)

Apply the product rule (3),

dydx=2xddx[sinx]+sinxddx[2x]

Apply the constant multiple rule (1),

dydx=2xddx[sinx]+2sinxddx[x]

Apply the power rule (2) and simplify the expressions,

dydx=2x(cosx)+2sinx(1x1−1)=2xcosx+2sinx=2(xcosx+sinx)

Therefore, the derivative of the function y=2xsinx is 2(xcosx+sinx)_.

The slope of the tangent line at (π2,π) is,

m= dydx|x=π2=2(π2cosπ2+sinπ2) {Qcosπ2=0, sinπ2=1}=2(0+1)=2

Thus, the slope of the tangent line at (π2,π) is m=2_.

Substitute (π2,π) for (x1,y1), and 2 for *m* in equation (1),

(y−π)=2(x−π2)y−π=2x−πy=2x

Therefore, the equation of the tangent line to the curve y=2xsinx at (π2,π) is y=2x_.

#### To determine

**To sketch:** The given curve and the tangent line at the given point (π2,π).

#### Explanation

**Given:**

The curve is y=2xsinx and the tangent line at (π2,π) is y=2x.

**Graph:**

Use the online graphing calculator to draw the graph of the curve and the tangent line as shown below in Figure 1.

From Figure 1, it is observed that the equation of the tangent line touches the curve y=2xsinx at the point (π2,π).