# Thread: relative extrema

1. ## relative extrema

Can someone help me with the first derivative of this one please.

$f(x)=7x+ln(-4x)$

2. Take the derivative of $f(x)=7x+\ln(-4x)$ using the chain rule for ln(-4x),

$\frac{d}{dx}\left(7x+\ln(-4x)\right)=7+\frac{1}{x}$.

Set the derivative equal to zero and solve for x,

$7+\frac{1}{x}=0\rightarrow x=-\frac{1}{7}$.

This is the x-coordinate for your sole relative maximum. Plug this back into the original function,

$\ln\left(\frac{4}{7}\right)-1=f_{max}(x)$

So your relative maximum occurs at $(x,f(x))=\left(-\frac{1}{7},\ln\left(\frac{4}{7}\right)-1\right)$.

Pretty sure that's what you were asking for.