Question

find the indicated derivative and simplify.
y for y = \frac{\log_{9}x}{3 + x}
y = square

Explanation:

Step1: Recall quotient - rule and log - derivative

The quotient rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Also, the derivative of $\log_{a}x=\frac{1}{x\ln a}$. Here, $u = \log_{9}x=\frac{\ln x}{\ln 9}$ and $v=3 + x$.

Step2: Find $u'$ and $v'$

The derivative of $u=\frac{\ln x}{\ln 9}$ is $u'=\frac{1}{x\ln 9}$ (since the derivative of $\ln x$ is $\frac{1}{x}$ and $\frac{1}{\ln 9}$ is a constant). The derivative of $v = 3 + x$ is $v'=1$.

Step3: Apply the quotient - rule

$y'=\frac{u'v - uv'}{v^{2}}=\frac{\frac{1}{x\ln 9}(3 + x)-\log_{9}x\times1}{(3 + x)^{2}}$.
Simplify the numerator: $\frac{3 + x}{x\ln 9}-\log_{9}x=\frac{3 + x}{x\ln 9}-\frac{\ln x}{\ln 9}$. Get a common denominator $x\ln 9$: $\frac{3 + x-x\ln x}{x\ln 9}$.
So, $y'=\frac{3 + x - x\ln x}{x\ln 9(3 + x)^{2}}$.

Answer:

$\frac{3 + x - x\ln x}{x\ln 9(3 + x)^{2}}$