Derivative of Hyperbolic Sine

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

Let the function be of the form

y = f\left( x \right) = \sinh x

By the definition of the hyperbolic function, the hyperbolic sine function is defined as

\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}

Now taking this function for differentiation, we have

\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}

Differentiating both sides with respect to the variable x, we have

\begin{gathered} \frac{d}{{dx}}\sinh x = \frac{d}{{dx}}\left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right) \\ \Rightarrow \frac{d}{{dx}}\left( {\sinh x} \right) = \frac{1}{2}\frac{d}{{dx}}\left( {{e^x} - {e^{ - x}}} \right) \\ \Rightarrow \frac{d}{{dx}}\left( {\sinh 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( {\sinh x} \right) = \frac{1}{2}\left[ {{e^x} + {e^{ - x}}} \right] \\ \Rightarrow \frac{d}{{dx}}\left( {\sinh x} \right) = \frac{{{e^x} + {e^{ - x}}}}{2} \\ \end{gathered}

By definition, \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}, we get

 \Rightarrow \frac{d}{{dx}}\left( {\sinh x} \right) = \cos hx

Example: Find the derivative of

y = f\left( x \right) = \sinh \sqrt {x + 1}

We have the given function as

y = \sinh \sqrt {x + 1}

Differentiating with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}\sinh \sqrt {x + 1}

Using the rule, \frac{d}{{dx}}\left( {\sinh x} \right) = \cos hx, we get

\begin{gathered} \frac{{dy}}{{dx}} = \cos h\sqrt {x + 1} \frac{d}{{dx}}\sqrt {x + 1} \\ \Rightarrow \frac{{dy}}{{dx}} = \cos h\sqrt {x + 1} \frac{1}{{2\sqrt {x - 1} }}\frac{d}{{dx}}\left( {x - 1} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\cos h\sqrt {x + 1} }}{{2\sqrt {x - 1} }} \\ \end{gathered}