Answer:
[tex]y=3x-3[/tex]
Step-by-step explanation:
We are given that
[tex]y=ln x^3[/tex]
Point (1,0)
We have to find the equation of tangent line to the given graph.
Differentiate w.r.t x
[tex]\frac{dy}{dx}=\frac{1}{x^3}\times 3x^2}=\frac{3}{x}[/tex]
[tex]\frac{d(lnx)}{dx}=\frac{1}{x}[/tex]
Substitute x=1
[tex]m=\frac{dy}{dx}=3[/tex]
Slope-point form:
[tex]y-y_1=m(x-x_1)[/tex]
[tex]x_1=1,y_1=0[/tex]
By using this formula
The equation of tangent to the given graph
[tex]y-0=3(x-1)=3x-3[/tex]
[tex]y=3x-3[/tex]