#### 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=3x+6cosx at (π3,π+3) is y=(3−33)x+3+π3_.

#### Explanation

**Given:**

The equation of the curve is, y=3x+6cosx and the point is (π3,π+3).

**Derivative rules:**

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

(2) Power Rule: ddx(xn)=nxn−1

(3) Sum Rule: ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(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(3x2−x3)

Apply the sum rule (3)

ddx[3x+6cosx]=ddx[3x]+ddx[6cosx]

Apply the constant multiple rule(1).

ddx[3x+6cosx]=3ddx[x]+6ddx[cosx]

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

ddx[3x+6cosx]=3(1x1−1)+6(−sinx)=3−6sinx

Therefore, the derivative of the function y=3x+6cosx is 3−6sinx_.

The slope of the tangent line at (π3,π+3) is,

m= dydx|x=π3 =3−6sinπ3 (Qsinπ3=32)=3−6(32)=3−33

Thus, the slope of the tangent line at (π3,π+3) is m=3−33_.

Substitute (π3,π+3) for (x1,y1) and 3−33 for *m* in equation (1),

y−(π+3)=(3−33)(x−π3)y−π−3=(3−33)x−(3−33)π3y−π−3=(3−33)x−π+3πy−3=(3−33)x+3π

Add 3 on both sides and simplify further,

y−3+3=(3−33)x+3π+3y=(3−33)x+3π+3

Therefore, the equation of the tangent line to the curve y=3x+6cosx is y=(3−33)x+3+3π.

#### To determine

**To sketch:** The given curve and the tangent line at the given point (π3,π+3).

#### Explanation

**Given:**

The curve is y=3x+6cosx and the tangent line at (π3,π+3) is y=(3−33)x+3+3π.

**Graph:**

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

From Figure 1, it is observed that the equation of the tangent line touches the curve y=3x+6cosx at the point (π3,π+3)**.**