If g'(3) = 4 and h'(3) = -1 , find f '(3) for f(x) = 5g(x) + 3h(x) + 2
Initial equation
$\displaystyle f(x) ~~=~~ 5~g(x) ~~+~~ 3~h(x)$
Take the derivative
$\displaystyle \frac{d}{dx}[ ~f(x)~] ~~=~~ \frac{d}{dx}[~5~g(x) ~~+ ~~3~h(x)~]$
Split the derivative on the right side
$\displaystyle \frac{d}{dx}[ ~f(x)~] ~~=~~ \frac{d}{dx}[~5~g(x)~] ~~+ ~~\frac{d}{dx}[~3~h(x)~]$
Pull coefficients out
$\displaystyle \frac{d}{dx}[ ~f(x)~] ~~=~~ 5~\frac{d}{dx}[~g(x)~] ~~+ ~~3~\frac{d}{dx}[~h(x)~]$
Switch to Newtonian notation (use a slash for prime instead of d/dx)
$\displaystyle f\prime (x) ~~=~~ 5~g\prime(x) ~~+~~ 3~h\prime (x)$
Now plug in x=3
$\displaystyle f\prime (3) ~~=~~ 5~g\prime(3) ~~+~~ 3~h\prime (3)$
And substitute your answers with the data given to you
$\displaystyle f\prime (3) ~~=~~ 5(4) ~~+~~ 3(-1)$
And simplify
$\displaystyle f\prime (3) ~~=~~ 17$