Question

find $\frac{dy}{dx}$ when $y = \frac{log(x)}{6+log(x)}$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = \log(x)$ and $v=6+\log(x)$. The derivative of $\log(x)$ with respect to $x$ is $\frac{1}{x\ln(10)}$ (assuming base - 10 logarithm; if it's the natural logarithm $\ln(x)$, the derivative is $\frac{1}{x}$). So, $u'=\frac{1}{x\ln(10)}$ and $v'=\frac{1}{x\ln(10)}$.

Step2: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{\frac{1}{x\ln(10)}(6 + \log(x))-\log(x)\frac{1}{x\ln(10)}}{(6+\log(x))^{2}}$.

Step3: Simplify the numerator

The numerator $\frac{1}{x\ln(10)}(6 + \log(x))-\log(x)\frac{1}{x\ln(10)}=\frac{6+\log(x)-\log(x)}{x\ln(10)}=\frac{6}{x\ln(10)}$.
So, $\frac{dy}{dx}=\frac{6}{x\ln(10)(6 + \log(x))^{2}}$.

Answer:

$\frac{6}{x\ln(10)(6+\log(x))^{2}}$