Logarithmic differentiation

Using Faà di Bruno's formula, the n-th order logarithmic derivative is,

$${\displaystyle {\frac {d^{n}}{dx^{n}}}\ln f(x)=\sum _{m_{1}+2m_{2}+\cdots +nm_{n}=n}{\frac {n!}{m_{1}!\,m_{2}!\,\cdots \,m_{n}!}}\cdot {\frac {(-1)^{m_{1}+\cdots +m_{n}-1}(m_{1}+\cdots +m_{n}-1)!}{f(x)^{m_{1}+\cdots +m_{n}}}}\cdot \prod _{j=1}^{n}\left({\frac {f^{(j)}(x)}{j!}}\right)^{m_{j}}.}$$

Using this, the first four derivatives are,

$${\displaystyle {\begin{aligned}{\frac {d^{2}}{dx^{2}}}\ln f(x)&={\frac {f''(x)}{f(x)}}-\left({\frac {f'(x)}{f(x)}}\right)^{2}\\[1ex]{\frac {d^{3}}{dx^{3}}}\ln f(x)&={\frac {f^{(3)}(x)}{f(x)}}-3{\frac {f'(x)f''(x)}{f(x)^{2}}}+2\left({\frac {f'(x)}{f(x)}}\right)^{3}\\[1ex]{\frac {d^{4}}{dx^{4}}}\ln f(x)&={\frac {f^{(4)}(x)}{f(x)}}-4{\frac {f'(x)f^{(3)}(x)}{f(x)^{2}}}-3\left({\frac {f''(x)}{f(x)}}\right)^{2}+12{\frac {f'(x)^{2}f''(x)}{f(x)^{3}}}-6\left({\frac {f'(x)}{f(x)}}\right)^{4}\end{aligned}}}$$

A natural logarithm is applied to a product of two functions

$${\displaystyle f(x)=g(x)h(x)}$$

to transform the product into a sum

$${\displaystyle \ln(f(x))=\ln(g(x)h(x))=\ln(g(x))+\ln(h(x)).}$$

Differentiating by applying the chain and the sum rules yields

$${\displaystyle {\frac {f'(x)}{f(x)}}={\frac {g'(x)}{g(x)}}+{\frac {h'(x)}{h(x)}},}$$

and, after rearranging, yields[5]

$${\displaystyle f'(x)=f(x)\times \left\{{\frac {g'(x)}{g(x)}}+{\frac {h'(x)}{h(x)}}\right\}=g(x)h(x)\times \left\{{\frac {g'(x)}{g(x)}}+{\frac {h'(x)}{h(x)}}\right\}=g'(x)h(x)+g(x)h'(x),}$$

which is the product rule for derivatives.

A natural logarithm is applied to a quotient of two functions

$${\displaystyle f(x)={\frac {g(x)}{h(x)}}}$$

to transform the division into a subtraction

$${\displaystyle \ln(f(x))=\ln \left({\frac {g(x)}{h(x)}}\right)=\ln(g(x))-\ln(h(x))}$$

Differentiating by applying the chain and the sum rules yields

$${\displaystyle {\frac {f'(x)}{f(x)}}={\frac {g'(x)}{g(x)}}-{\frac {h'(x)}{h(x)}},}$$

and, after rearranging, yields

$${\displaystyle f'(x)=f(x)\times \left\{{\frac {g'(x)}{g(x)}}-{\frac {h'(x)}{h(x)}}\right\}={\frac {g(x)}{h(x)}}\times \left\{{\frac {g'(x)}{g(x)}}-{\frac {h'(x)}{h(x)}}\right\}={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}},}$$

which is the quotient rule for derivatives.

For a function of the form

$${\displaystyle f(x)=g(x)^{h(x)}}$$

the natural logarithm transforms the exponentiation into a product

$${\displaystyle \ln(f(x))=\ln \left(g(x)^{h(x)}\right)=h(x)\ln(g(x))}$$

Differentiating by applying the chain and the product rules yields

$${\displaystyle {\frac {f'(x)}{f(x)}}=h'(x)\ln(g(x))+h(x){\frac {g'(x)}{g(x)}},}$$

and, after rearranging, yields

$${\displaystyle f'(x)=f(x)\times \left\{h'(x)\ln(g(x))+h(x){\frac {g'(x)}{g(x)}}\right\}=g(x)^{h(x)}\times \left\{h'(x)\ln(g(x))+h(x){\frac {g'(x)}{g(x)}}\right\}.}$$

The same result can be obtained by rewriting f in terms of exp and applying the chain rule.

Using capital pi notation, let

$${\displaystyle f(x)=\prod _{i}(f_{i}(x))^{\alpha _{i}(x)}}$$

be a finite product of functions with functional exponents.

The application of natural logarithms results in (with capital sigma notation)

$${\displaystyle \ln(f(x))=\sum _{i}\alpha _{i}(x)\cdot \ln(f_{i}(x)),}$$

and after differentiation,

$${\displaystyle {\frac {f'(x)}{f(x)}}=\sum _{i}\left[\alpha _{i}'(x)\cdot \ln(f_{i}(x))+\alpha _{i}(x)\cdot {\frac {f_{i}'(x)}{f_{i}(x)}}\right].}$$

Rearrange to get the derivative of the original function,

$${\displaystyle f'(x)=\overbrace {\prod _{i}(f_{i}(x))^{\alpha _{i}(x)}} ^{f(x)}\times \overbrace {\sum _{i}\left\{\alpha _{i}'(x)\cdot \ln(f_{i}(x))+\alpha _{i}(x)\cdot {\frac {f_{i}'(x)}{f_{i}(x)}}\right\}} ^{[\ln(f(x))]'}.}$$