Logarithm derivatives

01A logarithm's sensitivity#

A natural logarithm changes quickly near zero and more slowly at large positive inputs. In likelihood calculations, that means an already tiny probability can have a large effect on log probability when it changes.

For $f(x)=\ln x$, the derivative is $1/x$ on $x>0$. One way to see the relationship is to write $y=\ln x$, so $e^y=x$. Differentiating gives $e^y y'=1$, hence $y'=1/x$.

At $x=2$, the slope is $1/2$. At $x=0.1$, it is $10$. These are slopes of the logarithm, not its output values.

02Divide the inner derivative by the inner value#

For $f(x)=\ln g(x)$, apply the chain rule where $g(x)>0$:

For $f(x)=\ln(2x+3)$, the inner derivative is $2$. The derivative is $2/(2x+3)$ and the real domain is $x>-3/2$. At $x=1$, the derivative is $2/5$.

For $h(x)=-\ln x$, the minus sign scales the derivative to $-1/x$. At $x=1/4$, this is $-4$. Increasing a positive probability decreases its negative log loss.

03Carry the domain into the answer#

The real logarithm is undefined at zero and at negative inputs. A denominator formula alone does not extend its domain. For example, $1/x$ exists at $x=-1$, but it is not a derivative of the real function $\ln x$ there.

Taking the logarithm of a product can simplify a later derivative, but any rewritten expression still has to respect its original domain. Check the inner value before differentiating a logarithmic expression.