Logarithmic differentiation

Applications

Products

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 $${\displaystyle \frac{f^{\prime }(x)}{f(x)}=\frac{g^{\prime }(x)}{g(x)}+\frac{h^{\prime }(x)}{h(x)},}$$ and, after rearranging, yields $${\displaystyle f^{\prime }(x)=f(x)\times \left\{\frac{g^{\prime }(x)}{g(x)}+\frac{h^{\prime }(x)}{h(x)}\right\}=g(x)h(x)\times \left\{\frac{g^{\prime }(x)}{g(x)}+\frac{h^{\prime }(x)}{h(x)}\right\}=g^{\prime }(x)h(x)+g(x)h^{\prime }(x),}$$ which is the product rule for derivatives.

Quotients

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 $${\displaystyle \frac{f^{\prime }(x)}{f(x)}=\frac{g^{\prime }(x)}{g(x)}-\frac{h^{\prime }(x)}{h(x)},}$$ and, after rearranging, yields $${\displaystyle f^{\prime }(x)=f(x)\times \left\{\frac{g^{\prime }(x)}{g(x)}-\frac{h^{\prime }(x)}{h(x)}\right\}=\frac{g(x)}{h(x)}\times \left\{\frac{g^{\prime }(x)}{g(x)}-\frac{h^{\prime }(x)}{h(x)}\right\}=\frac{g^{\prime }(x)h(x)-g(x)h^{\prime }(x)}{h(x{)}^2},}$$

which is the quotient rule for derivatives.

Functional exponents

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 $${\displaystyle \frac{f^{\prime }(x)}{f(x)}=h^{\prime }(x)\ln (g(x))+h(x)\frac{g^{\prime }(x)}{g(x)},}$$ and, after rearranging, yields $${\displaystyle f^{\prime }(x)=f(x)\times \left\{h^{\prime }(x)\ln (g(x))+h(x)\frac{g^{\prime }(x)}{g(x)}\right\}=g(x{)}^{h(x)}\times \left\{h^{\prime }(x)\ln (g(x))+h(x)\frac{g^{\prime }(x)}{g(x)}\right\}.}$$ The same result can be obtained by rewriting f in terms of exp and applying the chain rule.

General case

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, $${\displaystyle \frac{f^{\prime }(x)}{f(x)}=\sum _i\left[{\alpha }_i^{\prime }(x)\cdot \ln (f_i(x))+{\alpha }_i(x)\cdot \frac{f_i^{\prime }(x)}{f_i(x)}\right].}$$ Rearrange to get the derivative of the original function, $${\displaystyle f^{\prime }(x)=\stackrel{f(x)}{\overbrace{\prod _i(f_i(x){)}^{{\alpha }_i(x)}}}\times \stackrel{[\ln (f(x)){]}^{\prime }}{\overbrace{\sum _i\left\{{\alpha }_i^{\prime }(x)\cdot \ln (f_i(x))+{\alpha }_i(x)\cdot \frac{f_i^{\prime }(x)}{f_i(x)}\right\}}}.}$$