differentiate (f(z)=\frac{5}{ln(5z)}). (f(z)=)

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 5$ and $v=\ln(5z)$.

Step2: Find $u^\prime$ and $v^\prime$

Since $u = 5$ is a constant, $u^\prime=0$. For $v=\ln(5z)$, using the chain - rule, if $y=\ln(u)$ and $u = 5z$, then $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dz}=5$. So $v^\prime=\frac{1}{5z}\times5=\frac{1}{z}$.

Step3: Apply quotient - rule

$f^\prime(z)=\frac{u^\prime v - uv^\prime}{v^{2}}=\frac{0\times\ln(5z)-5\times\frac{1}{z}}{(\ln(5z))^{2}}$.

Step4: Simplify the expression

$f^\prime(z)=-\frac{5}{z(\ln(5z))^{2}}$.

Answer:

$-\frac{5}{z(\ln(5z))^{2}}$