#### 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=1+x3 at (2,3) is y=2x−1_.

#### Explanation

**Given:**

The function is y=1+x3.

The point is (2,3).

**Result used:**

**Chain Rule:**

If *h* is differentiable at *x* and *g* is differentiable at h(x), then the composite function F=g∘h defined by F(x)=g(h(x)) is differentiable at *x* and F′ is given by the product,

F′(x)=g′(h(x))⋅h′(x) (1)

**Formula used:**

The equation of the tangent line at (x1,y1) is y−y1=m(x−x1) (2)

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(1+x3)=ddx(1+x3)12

Let h(x)=1+x3 and g(u)=u12 where u=h(x).

Apply the chain rule as shown in equation (1),

dydx=g′(h(x))⋅h′(x) (3)

The derivative g′(h(x)) is computed as follows,

g′(h(x))=g′(u)=ddu(g(u))=dduu12=12u12−1

Simplify further and obtain the derivative,

g′(h(x))=12u1−22=12u−12=12u12=12u

Substitute u=1+x3 in the above equation,

g′(h(x))=121+x3

Thus, the derivative is g′(h(x))=121+x3.

The derivative of h(x) is computed as follows,

h′(x)=ddx(1+x3)=ddx(1)+ddx(x3)=(0)+(3x3−1)=3x2

Thus, the derivative is h′(x)=3x2.

Substitute 121+x3 for g′(h(x)) and 3x2 for h′(x) in equation (3),

dydx=121+x3(3x2)=3x221+x3=3x221+x3

Therefore, the derivative of y=1+x3 is dydx=3x221+x3_.

The slope of the tangent line at (2,3) is computed as follows,

m=dydx|x=2=3(2)221+(2)3 =3⋅429 =3⋅23=2

Thus the slope of the tangent line at (2,3) is m=2.

Substitute (2,3) for (x1,y1) and m=2 in equation (2),

(y−3)=2(x−2)y−3=2x−4y=2x−4+3y=2x−1

Therefore, the equation of the tangent line to the curve y=1+x3 at (2,3) is y=2x−1_.