#### 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=3x+1x2+1 at
(1,2) is
y=−0.5x+2.5.

The equation of the normal line to the curve
y=3x+1x2+1 at
(1,2) is
y=2x.

#### Explanation

**Given:**

The equation of the curve
y=3x+1x2+1.

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). Quotient Rule: If
f1(t) and
f2(t) are both differentiable, then

ddt[f1(t)f2(t)]=f2(x)ddt[f1(t)]−f1(t)ddt[f2(t)][f2(t)]2

**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(3x+1x2+1)

Apply quotient rule (4), (2) and constant multiple rule (1),

dydx=(x2+1)ddx(3x+1)−(3x+1)ddx(x2+1)(x2+1)2=(x2+1)[3ddx(x)+ddx(1)]−(3x+1)[ddx(x2)+ddx(1)](x2+1)2=(x2+1)[3(1)+0]−(3x+1)[(2x2−1)+0](x2+1)2=3x2+3−6x2−2x(x2+1)2

Simplify further as,

3x2+3−6x2−2x(x2+1)2=−3x2+3−2x(x2+1)2

Therefore, the derivative of the function
y=3x+1x2+1 is
−3x2+3−2x(x2+1)2.

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

dydx|x=1=−3(1)2+3−2(1)(12+1)2=−3+3−2(2)2=−24=−12

Thus, the slope of the tangent line at
(1,2) is
m=−12 .

Substitute
(1,2) for
(x1,y1) and
−12 for *m* in the equation (1),

y−2=−12(x−1)y−2=−0.5x+0.5y=−0.5x+2.5

Therefore, the equation of tangent line to the curve
y=3x+1x2+1 at
(1,2) is
y=−0.5x+2.5.

Substitute
(1,2) for
(x1,y1) and
−12 for *m* in the equation (2),

y−2=−1−12(x−1)y−2=2(x−1)y=2x

Therefore, the equation of the normal line to the curve
y=3x+1x2+1 at
(1,2) is
y=2x.