#### To determine

**To find:** The point on which the curve is the tangent line perpendicular to the given line.

#### Answer

The point on which the curve is the tangent line perpendicular to the given line is (4, 3).

#### Explanation

**Given:**

The equation of the curve y=1+2x.

The equation of the line 6x+2y=1 .

**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)

**Derivative rules:**

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

(2) Power Rule: ddx(xn)=nxn−1

**Calculation:**

The derivative of y is dydx, which is obtained as follows,

dydx=ddx(y) =ddx(1+2x)

Let h(x)=1+2x and g(u)=u where u=h(x).

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

dydx=g′(h(x))⋅h′(x) (2)

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

g′(h(x))=g′(u)=ddu(g(u))=ddu(u)=ddu(u)12

Apply the power rule (2),

g′(h(x))=12(u)12−1=12(u)1−22=12(u)−12=12u

Substitute u=1+2x in the above equation,

g′(h(x))=121+2x

Thus, the derivative g′(h(x)) is g′(h(x))=121+2x.

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

h′(x)=ddx(1+2x)=ddx(1)+ddx(2x)

Apply the constant multiple rule (1),

h′(x)=ddx(1)+2ddx(x)=0+2(1x1−1)=2

Thus, the derivative h′(x) is h′(x)=2.

Substitute 121+2x for g′(h(x)) and 2 for h′(x) in equation (2),

dydx=121+2x⋅2=221+2x=11+2x

Therefore, the derivative of the curve *y* is dydx=11+2x .

Obtain the slope of the line 6x+2y=1 as follows,

Rewrite the line equation as slope intercept form.

6x+2y=12y=−6x+1y=−62x+12y=−3x+12

Therefore, the slope of the line is m1=−3.

Obtain the point on the curve is if slope of tangent line is perpendicular to slope of the line 6x+2y=1.

Note that, if two lines are perpendicular with slopes are m1 and m2, then the product of their slopes is m1m2=−1.

The required slope (m2) of the tangent line is perpendicular to the given line 6x+2y=1.

That is, m2=−1m1.

m2=−1(−3)=13

Since slope of the tangent line is dydx=11+2x.

11+2x=133=1+2x

Take square on both sides,

9=1+2x9−1=2x8=2xx=4

Substitute x=4 in the curve y=11+2x as follows,

y=1+2(4)=1+8=9=3

Thus, the required point is (4,3).

Therefore, the point on which the curve is the tangent line perpendicular to the given line is (4, 3).