Question

Differentiate.

\(y = {e^p}\left( {p + p\sqrt p } \right)\)

Short answer

The answer is \(y' = {e^p}\left( {1 + \frac{3}{2}\sqrt p + p + p\sqrt p } \right)\).

Step-by-step solution

Derivative of multiplication of two function

If a function is multiplication of two function the we can differentiate the function by the following rule:

Let the function be \(h\left( x \right) = f\left( x \right)g\left( x \right)\) then the derivativewill be

\(\begin{aligned}h'\left( x \right) &= \frac{d}{{dx}}h\left( x \right)\\ &= \frac{d}{{dx}}f\left( x \right)g\left( x \right)\\ &= f\left( x \right)\frac{d}{{dx}}g\left( x \right) + g\left( x \right)\frac{d}{{dx}}f\left( x \right)\end{aligned}\)

Finding the derivative of the given function

The function is given by \(y = {e^p}\left( {p + p\sqrt p } \right)\).

Differentiate the function with respect to \(p\) we get

\(\begin{aligned}y' &= \frac{d}{{dp}}\left( {{e^p}\left( {p + p\sqrt p } \right)} \right)\\ &= {e^p}\frac{d}{{dp}}\left( {p + p\sqrt p } \right) + \left( {p + p\sqrt p } \right)\frac{d}{{dp}}{e^p}\\ &= {e^p}\left( {1 + \sqrt p + p \times \frac{1}{{2\sqrt p }}} \right) + \left( {p + p\sqrt p } \right){e^p}\\ &= {e^p}\left( {1 + \sqrt p + \frac{p}{{2\sqrt p }} + p + p\sqrt p } \right)\\ &= {e^p}\left( {1 + \frac{{2p + p}}{{2\sqrt p }} + p + p\sqrt p } \right)\end{aligned}\)

Further solving we get

\(\begin{aligned}y' &= {e^p}\left( {1 + \frac{{3p}}{{2\sqrt p }} + p + p\sqrt p } \right)\\ &= {e^p}\left( {1 + \frac{{3\sqrt p }}{2} + p + p\sqrt p } \right)\end{aligned}\)

Hence \(y' = {e^p}\left( {1 + \frac{3}{2}\sqrt p + p + p\sqrt p } \right)\).