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