Can someone help me with the first derivative of this one please.
$\displaystyle f(x)=7x+ln(-4x)$
Take the derivative of $\displaystyle f(x)=7x+\ln(-4x)$ using the chain rule for ln(-4x),
$\displaystyle \frac{d}{dx}\left(7x+\ln(-4x)\right)=7+\frac{1}{x}$.
Set the derivative equal to zero and solve for x,
$\displaystyle 7+\frac{1}{x}=0\rightarrow x=-\frac{1}{7}$.
This is the x-coordinate for your sole relative maximum. Plug this back into the original function,
$\displaystyle \ln\left(\frac{4}{7}\right)-1=f_{max}(x)$
So your relative maximum occurs at $\displaystyle (x,f(x))=\left(-\frac{1}{7},\ln\left(\frac{4}{7}\right)-1\right)$.
Pretty sure that's what you were asking for.