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]}