The instantaneous rate of change of the function #f# at #x=a# is expressible through #f'(a)#, since this is the slope (rate of change) of the tangent line at that point.
So, for this question, we must know that #d/dxln(x)=1/x#; that is, the derivative of #ln(x)# is #1/x#. This is a very well known fact and can also be shown through another very well known derivative: #d/dxe^x=e^x#.
Anyway, we see that #f(x)=ln(x)#, so the derivative is #f'(x)=1/x#.
The instantaneous rate of change at #x=5# is #f'(5)#, and #f'(5)=1/5#.
This means that when the point #(5,ln(5))#, which lies on the graph of #ln(x)#, is included in a line with a slope of #1/5#, the line will be tangent at the point #(5,ln(5))# and the line's slope of #1/5# represents how fast the function #ln(x)# is changing at #x=5#.
That line is #y-ln(5)=1/5(x-5)#, or:
graph{(y-ln(5)-1/5(x-5))(y-lnx)=0 [-0.365, 17.415, -3.44, 5.45]}
