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}