Derivative for log

Question

I have the following problem: $$ \log \bigg( \frac{x+3}{4-x} \bigg) $$

I need to graph the following function so I will need a starting point, roots, zeros, stationary points, inflection points and local minimum and maximum and I need to know where the function grows and declines.

I calculated roots zeros $ x + 3= 0$, $x=-3$ and roots $ 4-x=0$,$x=-4 $. Now I sort of know how to graph the function from here but how do I get the stationary points do I have to find the derivative of $\log \left( \frac{x+3}{4-x} \right)$ or just $ \left( \frac{x+3}{4-x} \right)$.

I don't fully understand how to find the derivative of $\log$. Can i use the $\log(x)' = \frac{1}{x} $ rule here to get $ \frac{1}{\frac{x+3}{4-x}} $ and then find stationary points here ?

Answer 1

For the stationary points you need to find $\frac{d}{dx} \left(\log \bigg( \frac{x+3}{4-x} \bigg)\right)$. You need to use the chain rule here $\frac{d}{dx}\log(f(x))=\frac{1}{f(x)}\cdot f'(x)$ which would give: $$\frac{d}{dx} \left(\log \bigg( \frac{x+3}{4-x} \bigg)\right)= \frac{1}{\frac{x+3}{4-x}} \cdot \frac{d}{dx}\left(\frac{x+3}{4-x}\right)=\frac{4-x}{x+3}\left( \frac{(4-x)+(x+3)}{(4-x)^2}\right)=\frac{7}{(x+3)(4-x)}$$ Hence, the function has no stationary points. Also, $ \log \bigg( \frac{x+3}{4-x} \bigg)$ has zeros when $\frac{x+3}{4-x}=1$ because $\log(1)=0$. So for the zeroes you need to solve $x+3=4-x$ giving $x=\frac{1}{2}$. It will also have asymptotes where the derivative goes to infinity i.e. at $x=-3$ and $x=4$, the first going to $-\infty$ and the second to $+\infty$.

Answer 2

It may help to write

$$f(x):=\frac{x+3}{4-x}=-1+\frac{7}{4-x}\Longrightarrow f'(x)=\frac{7}{(4-x)^2}\Longrightarrow$$

$$\left(\log f(x)\right)'=\frac{4-x}{x+3}\frac{7}{(4-x)^2}=\frac{7}{(x+3)(4-x)}$$

Since, by the chain rule, we have

$$(\log f(x))'=\frac{f'(x)}{f(x)}$$