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