Derivative of Hyperbolic Cosine Function

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $u$ be a differentiable real function of $x$.

Then:

$\map {\dfrac \d {\d x} } {\cosh u} = \sinh u \dfrac {\d u} {\d x}$

where $\cosh$ is the hyperbolic cosine and $\sinh$ is the hyperbolic sine.


Proof

\(\displaystyle \map {\frac \d {\d x} } {\cosh u}\) \(=\) \(\displaystyle \map {\frac \d {\d u} } {\cosh u} \frac {\d u} {\d x}\) Chain Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \sinh u \frac {\d u} {\d x}\) Derivative of Hyperbolic Cosine

$\blacksquare$


Also see


Sources