Question: If f(3)=7, f'(3)= -4, g(3)= -9, and g'(3)=6, what is (f/g)'(3).
This is use of the quotient rule, which is,
let y = u/v, where u and v are both functions of x, then
dy/dx = {v.du/dx - u.dv/dx} / v^2
So, if y = f/g, then
y' = {g.f' - f.g') / g^2, and
y'(3) = (f/g)'(3) = {g(3).f'(3) - f(3).g'(3)} / (g(3))^2
(f/g)'(3) = {(-9).(-4) - (7).(6)} / (-9)^2
(f/g)'(3) = { 36 - 42} / 81
(f/g)'(3) = -6/81 = -2/27
Answer: (f/g)'(3) =2/27