   Chapter 3.6, Problem 36E ### 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 = (ln x)/x at the points (1, 0) and (e, 1/e). Illustrate by graphing the curve and its tangent lines.

To determine

To find: The equation of the tangent lines to the curve at the points.

Explanation

Given:

The equation is y=lnxx.

The points are (1,0) and (e,1e).

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:

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

dydx=ddx(y)=ddx(lnxx)

Apply the quotient rule ddx(fg)=gddx(f)fddx(g)(g)2 ,

dydx=xddx(lnx)lnxddx(x)(x)2=x(1x)lnx(1x11)(x)2=1lnxx2

Therefore, the derivative of y is dydx=1lnxx2.

The slope of the tangent line at (1,0) is computed as follows,

m1=dydx|x=1=1ln1(1)2=10(1)2  (Qln1=0)=1

Thus, the slope of the tangent line at (1,0) is m1=1

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