Proof
The derivative h (x) is given by the limit formula:
h(x + ∆x) − h(x)
h (x) = lim ,
∆x→0 ∆x
provided the limit exists. We now express h using the product of f and g,
h(x + ∆x) − h(x)
h (x) = lim
∆x→0 ∆x
f (x + ∆x)g(x + ∆x) − f (x)g(x)
= lim .
∆x→0 ∆x
To make further progress, we need to relate our limit formula to the limit formulas for the
derivatives f (x) and g (x), namely
f (x + ∆x) − f (x) g(x + ∆x) − g(x)
f (x) = lim and g (x) = lim .
∆x→0 ∆x ∆x→0 ∆x
To relate these formulas to the limit for h (x), we use the “trick” of adding and subtracting
the term f (x)g(x + ∆x) in the numerator, and then simplifying:
f (x + ∆x)g(x + ∆x) − f (x)g(x)
h (x) = lim
∆x→0 ∆x
f (x + ∆x)g(x + ∆x) + (f (x)g(x + ∆x) − f (x)g(x + ∆x)) − f (x)g(x)
= lim
∆x→0 ∆x
f (x + ∆x) − f (x) g(x + ∆x) − g(x)
= lim g(x + ∆x) + f (x)
∆x→0 ∆x ∆x
f (x + ∆x) − f (x) g(x + ∆x) − g(x)
= lim lim g(x + ∆x) + f (x) lim
∆x→0 ∆x ∆x→0 ∆x→0 ∆x
= f (x)g(x) + f (x)g (x).
The last two steps are justified by assuming that f (x) and g (x) exist.