Nancy Richmond

## Answered question

2022-03-26

Derivative for log
I have the following problem:
$\mathrm{log}\left(\frac{x+3}{4-x}\right)$
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 $\mathrm{log}\left(\frac{x+3}{4-x}\right)$ 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 ${\mathrm{log}\left(x\right)}^{\prime }=\frac{1}{x}$ rule here to get $\frac{1}{\frac{x+3}{4-x}}$ and then find stationary points here ?

### Answer & Explanation

Korbin Rivera

Beginner2022-03-27Added 11 answers

For the stationary points you need to find $\frac{d}{dx}\left(\mathrm{log}\left(\frac{x+3}{4-x}\right)\right)$. You need to use the chain rule here $\frac{d}{dx}\mathrm{log}\left(f\left(x\right)\right)=\frac{1}{f\left(x\right)}\cdot {f}^{\prime }\left(x\right)$ which would give:
$\frac{d}{dx}\left(\mathrm{log}\left(\frac{x+3}{4-x}\right)\right)=\frac{1}{\frac{x+3}{4-x}}\cdot \frac{d}{dx}\left(\frac{x+3}{4-x}\right)=\frac{4-x}{x+3}\left(\frac{\left(4-x\right)+\left(x+3\right)}{{\left(4-x\right)}^{2}}\right)=\frac{7}{\left(x+3\right)\left(4-x\right)}$
Hence, the function has no stationary points. Also, $\mathrm{log}\left(\frac{x+3}{4-x}\right)$ has zeros when $\frac{x+3}{4-x}=1$ because $\mathrm{log}\left(1\right)=0$. 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 $-\mathrm{\infty }$ and the second to $+\mathrm{\infty }$

Laylah Hebert

Beginner2022-03-28Added 15 answers

It may help to write
$f\left(x\right)\phantom{\rule{0.222em}{0ex}}=\frac{x+3}{4-x}=-1+\frac{7}{4-x}\to {f}^{\prime }\left(x\right)=\frac{7}{{\left(4-x\right)}^{2}}\to$
${\left(\mathrm{log}f\left(x\right)\right)}^{\prime }=\frac{4-x}{x+3}\frac{7}{{\left(4-x\right)}^{2}}=\frac{7}{\left(x+3\right)\left(4-x\right)}$
Since, by the chain rule, we have
${\left(\mathrm{log}f\left(x\right)\right)}^{\prime }=\frac{{f}^{\prime }\left(x\right)}{f\left(x\right)}$

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