Logarithmic differentiation

Using capital pi notation,

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

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,

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

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

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

Using this, the first four derivatives are,

${\displaystyle {\frac {d^{2}}{dx^{2}}}\ln f(x)={\frac {f''(x)}{f(x)}}-\left({\frac {f'(x)}{f(x)}}\right)^{2}}$

${\displaystyle {\frac {d^{3}}{dx^{3}}}\ln f(x)={\frac {f'''(x)}{f(x)}}-3{\frac {f'(x)f''(x)}{f(x)^{2}}}+2\left({\frac {f'(x)}{f(x)}}\right)^{3}}$

${\displaystyle {\frac {d^{4}}{dx^{4}}}\ln f(x)={\frac {f''''(x)}{f(x)}}-4{\frac {f'(x)f'''(x)}{f(x)^{2}}}-3\left({\frac {f''(x)}{f(x)}}\right)^{2}+12{\frac {f'(x)^{2}f''(x)}{f(x)^{3}}}-6\left({\frac {f'(x)}{f(x)}}\right)^{4}}$

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[7]

${\displaystyle f'(x)=f(x)\times {\Bigg \{}{\frac {g'(x)}{g(x)}}+{\frac {h'(x)}{h(x)}}{\Bigg \}}=g(x)h(x)\times {\Bigg \{}{\frac {g'(x)}{g(x)}}+{\frac {h'(x)}{h(x)}}{\Bigg \}}.}$

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

${\displaystyle \ln(f(x))=\ln {\Bigg (}{\frac {g(x)}{h(x)}}{\Bigg )}=\ln(g(x))-\ln(h(x))\,\!}$

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

${\displaystyle f'(x)=f(x)\times {\Bigg \{}{\frac {g'(x)}{g(x)}}-{\frac {h'(x)}{h(x)}}{\Bigg \}}={\frac {g(x)}{h(x)}}\times {\Bigg \{}{\frac {g'(x)}{g(x)}}-{\frac {h'(x)}{h(x)}}{\Bigg \}}.}$

After multiplying out and using the common denominator formula the result is the same as after applying the quotient rule directly to ${\displaystyle f(x)}$.

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

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

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