Question

#54, I keep getting a different answer please take your time

Answer 1

`e^z\left((z^3+z^2+z-1)/(z^2)\right)`

Explanation:

We have the following function:

`f(z)=\left((z^2+1)/(z)\right)e^z`

We calculate the derivative as follows:

`\begin{gathered} \text{ We apply the following rule:} \\ \\ \left(f\cdot g\right)^(\prime)=f\:^(\prime)\cdot g+f\cdot g^(\prime) \\ \\ f=(z^2+1)/(z),\:g=e^z \\ \\ (d)/(dz)\left((z^2+1)/(z)\right)e^z+(d)/(dz)\left(e^z\right)(z^2+1)/(z) \\ \\ (d)/(dz)\left((z^2+1)/(z)\right) \\ \\ \text{We apply the following rule:} \\ \\ \left((f)/(g)\right)^'=(f\:'\cdot g-g'\cdot f)/(g^2) \\ \\ (d)/(dz)\left((z^2+1)/(z)\right)=((d)/(dz)\left(z^2+1\right)z-(d)/(dz)\left(z\right)\left(z^2+1\right))/(z^2)=(2zz-1\cdot\left(z^2+1\right))/(z^2)=(z^2-1)/(z^2) \\ \\ (d)/(dz)\left(e^z\right)=e^z \\ \\ (z^2-1)/(z^2)e^z+e^z(z^2+1)/(z) \\ \\ \:e^z\left((z^2-1)/(z^2)+(z^2+1)/(z)\right)=e^z\left((z^2-1+z^3+z)/(z^2)\right)=e^z\left((z^3+z^2+z-1)/(z^2)\right) \end{gathered}`