the derivative of f(x) = log6(∛((10 - x)/(-2x - 5))) using the properties\nde your answer below:

Explanation:

Step1: Use logarithm property

First, rewrite $y = f(x)=\log_6\left(\sqrt[3]{\frac{10 - x}{-2x - 5}}\right)$ as $y=\frac{1}{3}\log_6\left(\frac{10 - x}{-2x - 5}\right)$ using the property $\log_a(b^c)=c\log_a(b)$. Then, by the property $\log_a\left(\frac{m}{n}\right)=\log_a(m)-\log_a(n)$, we have $y = \frac{1}{3}\left[\log_6(10 - x)-\log_6(-2x - 5)\right]$.

Step2: Apply chain - rule for derivative

The derivative of $\log_a(u)$ with respect to $x$ is $\frac{1}{u\ln(a)}\cdot\frac{du}{dx}$. For $y_1=\frac{1}{3}\log_6(10 - x)$, let $u = 10 - x$, then $\frac{dy_1}{dx}=\frac{1}{3}\cdot\frac{1}{(10 - x)\ln(6)}\cdot(-1)=-\frac{1}{3(10 - x)\ln(6)}$. For $y_2=\frac{1}{3}\log_6(-2x - 5)$, let $v=-2x - 5$, then $\frac{dy_2}{dx}=\frac{1}{3}\cdot\frac{1}{(-2x - 5)\ln(6)}\cdot(-2)=\frac{2}{3(-2x - 5)\ln(6)}$.

Step3: Find the derivative of $y$

Since $y=y_1 - y_2$, then $y^\prime=\frac{dy}{dx}=\frac{2}{3(-2x - 5)\ln(6)}-\frac{1}{3(10 - x)\ln(6)}$.

Answer:

$\frac{2}{3(-2x - 5)\ln(6)}-\frac{1}{3(10 - x)\ln(6)}$