Question

Find the derivative. Simplify where possible.

41.\(f\left( x \right) = \tanh \sqrt x \)

Short answer

The derivative is \(f'\left( x \right) = \frac{{{{{\mathop{\rm sech}\nolimits} }^2}\sqrt x }}{{2\sqrt x }}\).

Step-by-step solution

Apply the derivative of composite functions

On applying the derivative to the given function\(f\left( x \right) = \tanh \sqrt x \), we get, as follows:

^.\(f'\left( x \right) = \frac{d}{{dx}}\tanh \sqrt x \frac{d}{{dx}}\left( {\sqrt x } \right)\)

Plug in the derivatives and simplify

The derivative of\(\tanh \sqrt x \)is \({\mathop{\rm sech}\nolimits} \sqrt x \) and that of \(\sqrt x \)is \(\frac{1}{{2\sqrt x }}\). On plugging this derivative and simplifying we get:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\tanh \sqrt x \frac{d}{{dx}}\left( {\sqrt x } \right)\\ &= {\sec ^2}\sqrt x \left( {\frac{1}{{2\sqrt x }}} \right)\\ &= \frac{{{{{\mathop{\rm sech}\nolimits} }^2}\sqrt x }}{{2\sqrt x }}\end{aligned}\)

So, \(f'\left( x \right) = \frac{{{{{\mathop{\rm sech}\nolimits} }^2}\sqrt x }}{{2\sqrt x }}\).