   Chapter 3.2, Problem 54E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find equations of the tangent lines to the curve y = x − 1 x + 1 that are parallel to the line x – 2y = 2.

To determine

To find: The equation of tangent lines to the curve that are parallel to given line.

Explanation

Given:

Curve y=x1x+1 and line x2y=2.

Derivative rules:

(1) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

(2) Power rule: ddx(xn)=nxn1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(4) Difference rule: ddx(fg)=ddx(f)ddx(g)

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

Calculation:

Obtain the slope of the tangent line to the curve.

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

dydx=ddx(x1x+1)

Apply the quotient rule (1) and simplify the terms,

dydx=(x+1)ddx(x1)(x1)ddx(x+1)[x+1]2

Apply the derivative rule (3),(4) and (2),

dydx=(x+1)[ddx(x)ddx(1)](x1)[ddx(x)+ddx(1)][x+1]2=(x+1)(x1)[1+0][x+1]2=x+1x+1[x+1]2=2(x+1)2

Thus, the derivative of curve y is dydx=2(x+1)2

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