How to find instantaneous rate of change for #f(x) = ln(x)# at x=5?

Question

Emilia Aguilar · Accepted Answer

#1/5#

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-lnx)=0 [-0.365, 17.415, -3.44, 5.45]}/(y-ln(5)-1/5(x-5))

Scarlett Allison · Answer

undefined To find the instantaneous rate of change for $ f(x) = \ln(x) $ at $ x = 5 $, we can use the derivative of the natural logarithm function. The derivative of $ \ln(x) $ with respect to $ x $ is $ \frac{1}{x} $. Therefore, the instantaneous rate of change of $ f(x) $ at $ x = 5 $ is equal to $ \frac{1}{5} $, or 0.2.