Derivative of the logarithm

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \log(z) = \dfrac{1}{z},$$ where $\log$ denotes the logarithm.

Proof

By the definition, $$\log(z) = \displaystyle\int_1^z \dfrac{1}{z} \mathrm{d}z.$$ Using the fundamental theorem of calculus, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \log(z) = \dfrac{1}{z},$$ as was to be shown.

References

• 1964: {{ #if: |{{{2}}}|Milton Abramowitz}}{{#if: Irene A. Stegun|{{#if: |, {{ #if: |{{{2}}}|Irene A. Stegun}}{{#if: |, [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]{{#if: |, [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]}}| and [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]}}| and {{ #if: |{{{2}}}|Irene A. Stegun}}}}|}}: [[Book:Milton Abramowitz/Handbook of mathematical functions{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Handbook of mathematical functions{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}}|{{#if: | ({{{ed}}} ed.)}}}}]]{{#if: | (translated by [[Mathematician:{{{translated}}}|{{ #if: |{{{2}}}|{{{translated}}}}}]])}}{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: findme | ... (previous)|}}{{#if: nth derivative of logarithm | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}: $4.1.46$