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

Answer:

$x-ey=e+e^2$

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

$\frac{dy}{dx}=\frac{1}{x}$

By using formula

$\frac{d(lnx)}{dx}=\frac{1}{x}$

Substitute x=e  

$m=\frac{dy}{dx}=\frac{1}{e}$

$m=\frac{dy}{dx}=\frac{1}{e}$

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  

$y-e=\frac{1}{e}(x-e)$

$ey-e^2=x-e$

$x-ey=e+e^2$