ProofWiki:Jokes/Adult/Sex is Fun

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Joke

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

Then:

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

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