Derivative of Hyperbolic Cosine

In this tutorial we shall prove the derivative of the hyperbolic cosine function.

Let the function be of the form $y = f\left( x \right) = \cosh x$

By the definition of the hyperbolic function, the hyperbolic cosine function is defined as
$\cosh x = \frac{{{e^x} + {e^{ – x}}}}{2}$

Now taking this function for differentiation, we have
$\cosh x = \frac{{{e^x} + {e^{ – x}}}}{2}$

Differentiating both sides with respect to the variable $x$, we have
$\begin{gathered} \frac{d}{{dx}}\cosh x = \frac{d}{{dx}}\left( {\frac{{{e^x} + {e^{ – x}}}}{2}} \right) \\ \Rightarrow \frac{d}{{dx}}\left( {\cosh x} \right) = \frac{1}{2}\frac{d}{{dx}}\left( {{e^x} + {e^{ – x}}} \right) \\ \Rightarrow \frac{d}{{dx}}\left( {\cosh x} \right) = \frac{1}{2}\left[ {\frac{d}{{dx}}\left( {{e^x}} \right) + \frac{d}{{dx}}\left( {{e^{ – x}}} \right)} \right] \\ \end{gathered}$

Using the formula of exponential differentiation $\frac{d}{{dx}}{e^x} = {e^x}$, we have
$\begin{gathered} \Rightarrow \frac{d}{{dx}}\left( {\cosh x} \right) = \frac{1}{2}\left[ {{e^x} – {e^{ – x}}} \right] \\ \Rightarrow \frac{d}{{dx}}\left( {\cosh x} \right) = \frac{{{e^x} – {e^{ – x}}}}{2} \\ \end{gathered}$

By definition, $\sinh x = \frac{{{e^x} – {e^{ – x}}}}{2}$, we get
$\Rightarrow \frac{d}{{dx}}\left( {\cosh x} \right) = \sin hx$

Example: Find the derivative of $y = f\left( x \right) = \cosh {x^3}$

We have the given function as
$y = \cosh {x^3}$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\cosh {x^3}$

Using the rule, $\frac{d}{{dx}}\left( {\cosh x} \right) = \sin hx$, we get
$\begin{gathered} \frac{{dy}}{{dx}} = \sinh {x^3}\frac{d}{{dx}}{x^3} \\ \Rightarrow \frac{{dy}}{{dx}} = \sinh {x^3}\left( {3{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = 3{x^2}\sinh {x^3} \\ \end{gathered}$