Derivative of Hyperbolic Sine

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

Let the function of the form

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

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

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


Now take 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}


Differentiation 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}

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