differentiate $f(u)=\frac{9u}{7 + ln(2u)}$. $f(u)=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{g(u)}{h(u)}$, then $y'=\frac{g'(u)h(u)-g(u)h'(u)}{h(u)^2}$. Here, $g(u) = 9u$ and $h(u)=7+\ln(2u)$.

Step2: Find $g'(u)$

Differentiate $g(u)=9u$ with respect to $u$. Using the power - rule $\frac{d}{du}(au)=a$ (where $a = 9$), we get $g'(u)=9$.

Step3: Find $h'(u)$

Differentiate $h(u)=7+\ln(2u)$ with respect to $u$. The derivative of a constant is 0, so $\frac{d}{du}(7) = 0$. For $\ln(2u)$, using the chain - rule, if $y=\ln(2u)$ and let $t = 2u$, then $\frac{dy}{du}=\frac{dy}{dt}\cdot\frac{dt}{du}$. Since $\frac{d}{dt}(\ln t)=\frac{1}{t}$ and $\frac{dt}{du}=2$, we have $\frac{d}{du}(\ln(2u))=\frac{1}{2u}\cdot2=\frac{1}{u}$. So, $h'(u)=\frac{1}{u}$.

Step4: Substitute into quotient - rule

Substitute $g(u), g'(u), h(u), h'(u)$ into the quotient - rule formula: [ \begin{align*} f'(u)&=\frac{9(7 + \ln(2u))-9u\cdot\frac{1}{u}}{(7+\ln(2u))^2}\ &=\frac{63 + 9\ln(2u)-9}{(7+\ln(2u))^2}\ &=\frac{54 + 9\ln(2u)}{(7+\ln(2u))^2} \end{align*} ]

Answer:

$\frac{54 + 9\ln(2u)}{(7+\ln(2u))^2}$