ANSWER:
[tex]e^z\left(\frac{z^3+z^2+z-1}{z^2}\right)[/tex]
STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]f(z)=\left(\frac{z^2+1}{z}\right)e^z[/tex]
We calculate the derivative as follows:
[tex]\begin{gathered} \text{ We apply the following rule:} \\ \\ \left(f\cdot g\right)^{\prime}=f\:^{\prime}\cdot g+f\cdot g^{\prime} \\ \\ f=\frac{z^2+1}{z},\:g=e^z \\ \\ \frac{d}{dz}\left(\frac{z^2+1}{z}\right)e^z+\frac{d}{dz}\left(e^z\right)\frac{z^2+1}{z} \\ \\ \frac{d}{dz}\left(\frac{z^2+1}{z}\right) \\ \\ \text{We apply the following rule:} \\ \\ \left(\frac{f}{g}\right)^'=\frac{f\:'\cdot g-g'\cdot f}{g^2} \\ \\ \frac{d}{dz}\left(\frac{z^2+1}{z}\right)=\frac{\frac{d}{dz}\left(z^2+1\right)z-\frac{d}{dz}\left(z\right)\left(z^2+1\right)}{z^2}=\frac{2zz-1\cdot\left(z^2+1\right)}{z^2}=\frac{z^2-1}{z^2} \\ \\ \frac{d}{dz}\left(e^z\right)=e^z \\ \\ \frac{z^2-1}{z^2}e^z+e^z\frac{z^2+1}{z} \\ \\ \:e^z\left(\frac{z^2-1}{z^2}+\frac{z^2+1}{z}\right)=e^z\left(\frac{z^2-1+z^3+z}{z^2}\right)=e^z\left(\frac{z^3+z^2+z-1}{z^2}\right) \end{gathered}[/tex]
