If f(x) = (5x+4)^-1 then, f'(x) = ?

[UMstudent]

I'm having a little trouble with this mostly just finding f'(x) for any equation. I can find f'(at some point), but not at x so any help on how to do this would be great, thanks.


It would also be helpful to find g(x) = 6e^(x*cos(x)), then g'(x) = ?

 

[galactus]

You could use the product rule or quotient rule

Quotient rule: \(\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}\)

\(\displaystyle \L\\\frac{1}{5x+4}\)

\(\displaystyle \L\\\frac{(5x+4)(0)-(1)(5)}{(5x+4)^{2}}\)

 

[UMstudent]

so would the quotient rule also work for the g(x) = 6e^(x*cos(x)), then g'(x) = ? problem?