Question

Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the given point. See Example 5.

y = ln x; (e, e)

Answer 1

`x-ey=e+e^2`

Step-by-step-explanation:

We are given that

`y=lnx`

Point (e,e)

We have to find the equation of tangent line to the given graph.

Differentiate w.r.t x

`(dy)/(dx)=(1)/(x)`

By using formula

`(d(lnx))/(dx)=(1)/(x)`

Substitute x=e

`m=(dy)/(dx)=(1)/(e)`

`m=(dy)/(dx)=(1)/(e)`

Slope-point form:

`y-y_1=m(x-x_1)`

`x_1=e,y_1=e`

By using this formula

The equation of tangent to the given graph

`y-e=(1)/(e)(x-e)`

`ey-e^2=x-e`

`x-ey=e+e^2`