#### To determine

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

#### Answer

The equation of tangent line to the curve
y=x+x at
(1,2) is
y=1.5x+0.5.

The equation of the normal line to the curve
y=x+x at
(1,2) is
y−2=−23(x−1).

#### Explanation

**Given:**

The equation of the curve
y=x+x.

The curve passing through the point
(1,2).

**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))

(4). Derivative of natural exponential function:
ddx(ex)=ex

**Formula used:**

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

The equation of normal line at
(x1,y1) is,
y−y1=−1m(x−x1)
. (2)

Here, *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(x+x)

Apply sum rule (3) and constant multiple rule (1),

dydx=ddx(x)+ddx(x)

Apply rule (4) and the power rule (2),

dydx=ddx(x)+ddx(x12)=1+12x12−1=1+12x−12=1+12x

Therefore, the derivative of the function
y=x4+2ex is
1+12x.

The slope of the tangent line at
(1,2) is,

m=dydx|x=1=1+121=1+12=32

Thus, the slope of the tangent line at
(1,2) is
m=32 .

Substitute
(1,2) for
(x1,y1) and 1.5 for *m* in the equation (1),

y−2=32(x−1)y−2=1.5x−1.5y=1.5x+0.5

Therefore, the equation of tangent line to the curve
y=x+x at
(1,2) is
y=1.5x+0.5.

Substitute
(1,2) for
(x1,y1) and 1.5 for *m* in the equation (2),

y−2=−1(32)(x−1)y−2=−23(x−1)

Therefore, the equation of the normal line to the curve
y=x+x at
(1,2) is
y−2=−23(x−1).