#### To determine

**To find:** The equation of the tangent line to the curve
y=2xx+1 at the point
(1,1).

#### Answer

The equation of tangent line to the curve
y=2xx+1 at
(1,1) is
y=x2+12_.

#### Explanation

**Given:**

The curve
y=2xx+1.

The point is
(1,1).

**Derivative rules:**

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

(2) Power Rule:
ddx(xn)=nxn−1

(3) 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

(4) Difference 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(2xx+1)

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

dydx=(x+1)2ddx(x)−2xddx(x+1)(x+1)2=(x+1)2(1)−2x(1)(x+1)2=2x+2−2x(x+1)2=2(x+1)2

Therefore, the derivative of *y* is
2(x+1)2.

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

m=dydx|x=1 =2(1+1)2=222=12

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

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

y−1=12(x−1)y−1=x2−12y=x2−12+1y=x2+12

Therefore, the equation of tangent line to the curve
y=2xx+1 at
(1,1) is
y=x2+12_.