ProofWiki:Jokes/Sex is Fun

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Joke

Let $e^{x / n} = \dfrac \d {\d x} f \paren u$.

Then:

\(\displaystyle \sqrt [n] {e^x} =\) \(=\) \(\displaystyle \dfrac \d {\d x} f \paren u\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\sqrt [n] {e^x} }\) \(=\) \(\displaystyle \paren {\dfrac \d {\d x} f \paren u}^n\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle e^x\) \(=\) \(\displaystyle \paren {\dfrac \d {\d x} f \paren u}^n\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int e^x\) \(=\) \(\displaystyle \int \paren {\dfrac \d {\d x} f \paren u}^n\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int e^x\) \(=\) \(\displaystyle f \paren u^n\) $\quad$ $\quad$

Purists are entitled of course to quibble that the left hand side should really read $\displaystyle \int e^x \rd x$.