Question

Differentiate.

\(y = \left( {{z^2} + {e^z}} \right)\sqrt z \)

Short answer

The answer is \(y' = \frac{{5{z^2} + {e^z} + 2z{e^z}}}{{2\sqrt z }}\).

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 \(y = \left( {{z^2} + {e^z}} \right)\sqrt z \).

Differentiate the function with respect to \(z\) we get,

\(\begin{aligned}y' &= \frac{d}{{dz}}\left( {\left( {{z^2} + {e^z}} \right)\sqrt z } \right)\\ &= \left( {{z^2} + {e^z}} \right)\frac{d}{{dz}}\sqrt z + \sqrt z \frac{d}{{dz}}\left( {{z^2} + {e^z}} \right)\\ &= \left( {{z^2} + {e^z}} \right)\left( {\frac{1}{2}\frac{1}{{\sqrt z }}} \right) + \sqrt z \left( {2z + {e^z}} \right)\\ &= \frac{{\left( {{z^2} + {e^z}} \right)}}{{2\sqrt z }} + \sqrt z \left( {2z + {e^z}} \right)\\ &= \frac{{\left( {{z^2} + {e^z}} \right) + 2z\left( {2z + {e^z}} \right)}}{{2\sqrt z }}\\ &= \frac{{5{z^2} + 2z{e^z} + {e^z}}}{{2\sqrt z }}\end{aligned}\)

Hence \(y' = \frac{{5{z^2} + {e^z} + 2z{e^z}}}{{2\sqrt z }}\).