find the derivative of\n\n$f(z)=\frac{sqrt{z}}{(e)^z}$.\n\n$f(z)=$

Explanation:

Step1: Rewrite the function

Rewrite $\sqrt{z}=z^{\frac{1}{2}}$, so $f(z)=\frac{z^{\frac{1}{2}}}{e^{z}}$.

Step2: Apply quotient - rule

The quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = z^{\frac{1}{2}}$, $u^\prime=\frac{1}{2}z^{-\frac{1}{2}}$, $v = e^{z}$, $v^\prime=e^{z}$.

Step3: Substitute values

$f^\prime(z)=\frac{\frac{1}{2}z^{-\frac{1}{2}}\cdot e^{z}-z^{\frac{1}{2}}\cdot e^{z}}{(e^{z})^{2}}$.

Step4: Simplify the expression

Factor out $e^{z}$ from the numerator: $f^\prime(z)=\frac{e^{z}(\frac{1}{2}z^{-\frac{1}{2}}-z^{\frac{1}{2}})}{e^{2z}}$. Then cancel out $e^{z}$ in the numerator and denominator, getting $f^\prime(z)=\frac{\frac{1}{2}z^{-\frac{1}{2}}-z^{\frac{1}{2}}}{e^{z}}=\frac{\frac{1}{2\sqrt{z}}-\sqrt{z}}{e^{z}}=\frac{1 - 2z}{2\sqrt{z}e^{z}}$.

Answer:

$\frac{1 - 2z}{2\sqrt{z}e^{z}}$