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 x3; (1, 0)

Answer 1

`y=3x-3`

Explanation:

We are given that

`y=ln x^3`

Point (1,0)

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

Differentiate w.r.t x

`(dy)/(dx)=(1)/(x^3)* 3x^2}=(3)/(x)`

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

Substitute x=1

`m=(dy)/(dx)=3`

Slope-point form:

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

`x_1=1,y_1=0`

By using this formula

The equation of tangent to the given graph

`y-0=3(x-1)=3x-3`

`y=3x-3`