What is the slope of the line tangent to the graph of #y=ln(x/2)# at x=4?

1 Answer
Feb 9, 2016

#y=1/4x-1+ln(2)#

Explanation:

First, find the point that the tangent line will intercept by plugging in #x=4#.

#y=ln(4/2)=ln(2)#

The point of tangency is #(4,ln(2))#.

Differentiating #y# will be simpler if we make use of the following logarithm rule:

#ln(a/b)=ln(a)-ln(b)#

Thus, we can redefine the function as

#y=ln(x)-ln(2)#

When we differentiate this, recall that #ln(2)# is a constant and can be ignored. Thus, the derivative of #y# is equivalent to just the derivative of #ln(x)#, which is #1"/"x#.

#y'=1/x#

The slope of the tangent line is equal to the value of the derivative when #x=4#, which is

#y'=1/4#

We know that tangent line has a slope #1"/"4# and passes through the point #(4,ln(2))#. These can be related as a line in point-slope form:

#y-ln(2)=1/4(x-4)#

Which can be rewritten as

#y=1/4x-1+ln(2)#

Graphed are the function and its tangent line:

graph{(y-ln(x/2))(y-ln(2)-(x-4)/4)=0 [-2.48, 13.32, -4.53, 3.37]}