#### To determine

**To evaluate:**

1. Equation of the tangent line of * y* at the points (1, 0) and (e,1e)

2. Plot * y* and its tangent line at (1, 0) and (e,1e)

#### Answer

1. The equation of tangent line of *y* at (1, 0) is: y=x−1

2. The equation of tangent line of *y* at (e,1e) is: y=1e

3. Illustrate tangent by drawing graph of *y* and its tangent.

#### Explanation

**Formula used:**

1. Quotient rule: uv'=u'v-uv'v2

2. Equation of line with slope *m* and passing point (x1,y1) is: y−y1=m(x−x1)

**Calculation:**

For calculating tangent lines we have to find slope of the tangent lines. We know that the slope of the tangent line is nothing but the derivative of the function at the tangent point. So, first we have to calculate y'.

Consider,

y=lnxx

Differentiate *y* with respect to x, we get

y'=ddx[lnx]x−lnxddx[x]x2 (By product rule)=1x2[1x×x−lnx×1] =1x2[1−lnx] =1−lnxx2

Thus, y'=1−lnxx2 ……(1)

**(i) Equation of tangent line of ***y* at (1, 0):

Put x=1 in equation (1), we get

y'(1)=1−ln11 =1-0 =1

Therefore, slope of the tangent line at the point (1, 0) is y'(1)=1. So, by formula 2, equation of line with slope y'(1)=1 and passing point (1, 0) is: (y−0)=1(x−1) i.e.

y=x−1

**Conclusion:** Thus, the equation of the tangent of *y* at (1, 0) is: y=x−1

**(ii) Equation of tangent line of ***y* at (e,1e):

Put x=e in equation (1), we get

y'(e)=1−lnee2 =1−1e2 =0

Therefore, slope of the tangent line at the point (e,1e) is y'(e)=0. So, by formula 2, equation of line with slope y'(e)=0 and passing point (e,1e) is: (y−1e)=0(x−e) i.e.

y=1e

**Conclusion:** Thus, the equation of the tangent of *y* at (e,1e) is: y=1e

From this graph also we can see the tangent lines at the points (1, 0) and (e,1e).