What is the equation of the tangent line to y = ln x at the point where x = 2 ?

Question

I got
y = x/4 - 1 + 2 ln 2

I'm not sure this is right

Answer 1

Slope = derivative of ln x at x = 2
Slope = m = 1/x = 1/2

When x = 2, y = ln 2
y - y1 = m (x - x1)
y - ln 2 = 1/2 (x - 2)
y = x/2 -1 + ln 2

Answer 2

Your function is extra acceptable written as (a million/2)e^x the point of tangency is (Ln(3), 3/2) y' = (a million/2)e^x certain, that is an same function So the slope of the tangent line at (Ln(3), 3/2) is y = 3x/2 + 3/2(a million - Ln(3))

Answer 3

y = ln x

dy/dx = 1/x. and at x= 2, m= ½

The equation of tangent is y = mx +c

Therefore, y = (1/2) x +c.

Since it passes through (2, ln2)

ln2 = 0.5(2) +c, or c= ln2-1.

Therefore the equation is y = (1/2) x + (ln2-1).


The first answer is correct.

Answer 4

The Equation is just y = ln 2, the answer being, put it in a calculator

Answer 5

tangent line can be given by
dy/dx = d (ln x) / dx
dy/dx = 1/x
when x = 2
dy/dx = 1/2


dy/dx = (Y - y) / (X - x)
@ (x ,y) = (2 , ln2)
dy/dx = (Y - ln2 ) / (X - 2)
2 (Y-ln2) = X -2
2y - 2 ln2 = x -2
y = x/2 - 1 + 2 ln2