Proofs of the product and quotient rules
Sum Rule: Suppose F (x) = f (x) + g(x), where f and g are differentiable functions. Then
F 0 (x) = f 0 (x) + g 0 (x).
Proof.
F 0 (x)
=
=
=
=
=
=
F (x + h) − F (x)
h
[f (x + h) + g(x + h)] − [f (x) + g(x)]
lim
h→0
h
f (x + h) − f (x) + g(x + h) − g(x)
lim
h→0
h
f (x + h) − f (x) g(x + h) − g(x)
lim
+
h→0
h
h
f (x + h) − f (x)
g(x + h) − g(x)
lim
+ lim
h→0
h→0
h
h
0
0
f (x) + g (x)
lim
h→0
Product Rule: Suppose F (x) = f (x)g(x), where f and g are differentiable functions. Then
F 0 (x) = f 0 (x)g(x) + f (x)g 0 (x).
Proof.
F 0 (x)
=
=
=
=
=
=
=
F (x + h) − F (x)
h
f (x + h)g(x + h) − f (x)g(x)
lim
h→0
h
f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x)
lim
h→0
h
[f (x + h) − f (x)]g(x + h) + f (x)[g(x + h) − g(x)]
lim
h→0
h
f (x + h) − f (x)
g(x + h) − g(x)
lim
g(x + h) + f (x)
h→0
h
h
f (x + h) − f (x)
g(x + h) − g(x)
lim
lim g(x + h) + lim f (x)
lim
h→0
h→0
h→0
h→0
h
h
f 0 (x)g(x) + f (x)g 0 (x)
lim
h→0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(1) Write the definition of the derivative of F .
(2) Rewrite F in terms of f and g.
(3) Add and subtract the same quantity, f (x)g(x + h). We could also have used f (x + h)g(x).
(4) Group like terms. Factor g(x + h) from the first pair of summands. Factor f (x) from the second pair.
(5) Rearrange the sum into two separate quotients.
(6) Since we assume f and g are differentiable (this implies g is continuous also), all of the limits in this step
exist, so take the limit of each term individually and then multiply and add.
(7) Take the limits, noting the derivatives of f and g, respectively. Also note that since g is continuous,
lim g(x+h) = g lim (x + h) = g(x).
h→0
h→0
Quotient Rule: Suppose F (x) =
f (x)
, where f and g are differentiable functions. Then
g(x)
F 0 (x) =
f 0 (x)g(x) − f (x)g 0 (x)
.
[g(x)]2
Proof.
F 0 (x)
=
=
=
=
=
=
=
=
lim
h→0
lim
F (x + h) − F (x)
h
f (x)
f (x+h)
g(x+h) − g(x)
h
f (x + h)g(x) − f (x)g(x + h)
lim
h→0
h[g(x)g(x + h)]
f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h)
lim
h→0
h[g(x)g(x + h)]
[f (x + h) − f (x)]g(x) − f (x)[g(x + h) − g(x)]
lim
h→0
h[g(x)g(x + h)]
g(x+h)−g(x)
f (x+h)−f (x)
g(x)
−
f
(x)
h
h
lim
h→0
[g(x)g(x + h)]
(x)
limh→0 f (x+h)−f
(limh→0 g(x)) − (limh→0 f (x)) limh→0
h
(8)
(9)
h→0
(limh→0 g(x)) (limh→0 g(x + h))
f 0 (x)g(x) − f (x)g 0 (x)
[g(x)]2
(10)
(11)
(12)
(13)
g(x+h)−g(x)
h
(14)
(15)
(8) Write the definition of the derivative of F .
(9) Rewrite F in terms of f and g.
(10) Rewrite over the common denominator g(x)g(x + h).
(11) Add and subtract the same quantity, f (x)g(x).
(12) Group like terms. Factor g(x) from the first pair of summands. Factor −f (x) from the second pair.
(13) Rearrange the sum, incorporating h into the denominators of quotients in the numerator.
(14) Since we assume f and g are differentiable, all of the limits in this step exist, so take the limit of each
term individually and then multiply and add.
(15) Take the limits, noting the derivatives of f and g, respectively. Also note that since g is continuous,
lim g(x+h) = g lim (x + h) = g(x).
h→0
h→0