What is the general formula for the derivative of the natural logarithm?

1 Answer
Lovecraft
Nov 21, 2015

For #ln(f(x)) = y# we have #dy/dx = f^'(x)/f(x)#

In case of the simple case of #ln(kx) = y# we can simplify that to #dy/dx = 1/x#

Explanation:

Well, in general terms

#y = ln(kx)# if #k != 0# we have

#dy/dx = 1/x#

Since #ln(kx) = ln(k) + ln(x)# and the constant will get dropped. Now, since the log is often used for implicit differentiation in functions with lots of multiplications, divisions and exponents, it might be better to say

If we have #y = ln(f(x))#

#dy/dx = f^'(x)/f(x)#

This is often used because we can do this:

#y = f(x)#

#ln(y) = ln(f(x))#

#y^'/y = f^'(x)/f(x)#

#y^' = f^'(x)#

But if we have lots of multiplications, division or exponents, the log properties will help us deal with them in a more expeditious manner.

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