Question

Find the derivative. Simplify where possible.

36.\(f\left( x \right) = {e^x}\cosh x\)

Short answer

The derivative is \(f'\left( x \right) = {e^x}\left( {\sinh x + \cosh x} \right)\).

Step-by-step solution

Apply the product rule

On applying the product rule to the given function\(f\left( x \right) = {e^x}\cosh x\), we get, as follows:

^.\(f'\left( x \right) = {e^x}\frac{d}{{dx}}\left( {\cosh x} \right) + \frac{d}{{dx}}\left( {{e^x}} \right)\cosh x\)

Plug in the derivatives and simplify

The derivative of \(\cosh x\) is \(\sinh x\) and \({e^x}\) is \({e^x}\). On plugging these derivatives and simplifying we get:

\(\begin{aligned}f'\left( x \right) &= {e^x}\frac{d}{{dx}}\left( {\cosh x} \right) + \frac{d}{{dx}}\left( {{e^x}} \right)\cosh x\\ &= {e^x}\sinh x + {e^x}\cosh x\\ &= {e^x}\left( {\sinh x + \cosh x} \right)\end{aligned}\)

Thus, the required answer is:

\(f'\left( x \right) = {e^x}\left( {\sinh x + \cosh x} \right)\)