## Calculus (3rd Edition)

$(\frac{f}{g})'(1) = -1$
We know that: $(\frac{f}{g})' = \frac{gf' - fg'}{g^2}$ By substituting the values we are given, we get: $(\frac{f}{g})'(1) = \frac{2*2 - 2*4}{2^2}$ Which can then simplify to: $(\frac{f}{g})'(1) = -1$