ProofWiki:Jokes/Adult/Sex is Fun

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 $\ds \int e^x \rd x$.