Product rule

The product rule is a rule of differentiation which states that for product of differentiable function's $u, v$:

$\frac{d}{dx}(u \cdot v)=u\frac{dv}{dx}+\frac{du}{dx}v$

In prime notation:

$(uv)'=uv'+u'v$

In the case of three terms multiplied together, the rule becomes

$\frac{d}{dx}(u \cdot v \cdot w)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}$

It is one of the most common differentiation rules used for functions of combination, and is also very simple to apply. For instance, consider the function $f(x)=x\sin(x)$. The derivative is easily found:

$f'(x)=x\frac{d}{dx}(\sin(x))+\sin(x)\frac{d}{dx}(x)$

$f'(x)=x\cos(x)+\sin(x)$

Proof

Both of the following limits exist, as $f(x), g(x)$ are differentiable

$f'(x) = \lim_{h \rightarrow 0} \dfrac{f(x+h) -f(x)}{h}$

$g'(x) = \lim_{h \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

Then from first principles

$(f(x)g(x))' =\lim_{h \rightarrow 0} \dfrac{f(x+h)g(x+h) - f(x)g(x)}{h}$

Adding and subtracting a $f(x+h)g(x)$ changes nothing, but allows for algebraic manipulation

$(f(x)g(x))' =\lim_{h \rightarrow 0} \dfrac{f(x+h)g(x+h) -f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h}$

Now, factoring, separating the fractions and distributing the limits with the algebra of limits yields

$(f(x)g(x))' =\lim_{h \rightarrow 0} f(x+h)\dfrac{g(x+h) -g(x)}{h} +\lim_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} g(x)$

We can then use turn both of these into a product of limits and it is easy to see that

$(f(x)g(x))' = f(x)g'(x) + f'(x)g(x)$

This completes the proof.