The product rule for derivatives: proof and examples
Why the derivative of a product is not the product of the derivatives: an intuition with rectangles, the proof, examples, the quotient rule and an application to electrical power.
The derivative of a sum is the sum of the derivatives. It is tempting to think the same holds for a product, that 
. It does not, and one example is enough to see it: with 
the product is 
, whose derivative is 
, while 
. The correct rule is Leibniz’s:
![\[ (f \cdot g)'(x) = f'(x)\,g(x) + f(x)\,g'(x). \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-c495acc42273d50b6bea1ef06d11c572_l3.png "Rendered by QuickLaTeX.com")
An idea with rectangles
Think of 
as the area of a rectangle with sides 
and 
. When 
increases a little, side grows by 
and side by 
. The new area is the old one plus two strips, 
and 
, plus a small corner square 
. The square is the product of two small quantities, so it is negligible compared with the strips: divide by 
, take the limit, and what remains is exactly 
.
The proof
Assume and are differentiable at . The difference quotient of the product is
![\[ \frac{f(x+h)g(x+h) - f(x)g(x)}{h}. \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-7e67eb8cf9b7ce597ef51a7c025bd0ff_l3.png "Rendered by QuickLaTeX.com")
The trick is to add and subtract the same term 
in the numerator:
![\[ \frac{f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)}{h} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-cc84331d07327c5b23da6b1e4e94f7e9_l3.png "Rendered by QuickLaTeX.com")
and grouping the terms:
![\[ = g(x+h)\,\frac{f(x+h) - f(x)}{h} + f(x)\,\frac{g(x+h) - g(x)}{h}. \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-1e27b8faddef5276f63f105274b3f6d7_l3.png "Rendered by QuickLaTeX.com")
Now let 
:
and 
by definition of derivative;
because a differentiable function is also continuous. This is the step people often forget to justify.
By the rules for limits of sums and products we get 
. ∎
Why does differentiable imply continuous? For 
we can write 
. The first factor tends to 
, the second to 0, so the difference tends to 0.
Examples
.
.
, consistent with 
.
Consequences
Three or more factors. Applying the rule twice: 
. Each term differentiates one factor and leaves the others alone.
Powers. With 
repeated 
times you get back 
: terms, all equal to 
.
Quotients. Writing 
and knowing that 
, you obtain the quotient rule 
.
Integration by parts. Integrating the product rule gives 
, the integration by parts method.
An example from circuit theory
The instantaneous power absorbed by a component is 
. If voltage and current both change with time, the rate at which the power changes is
![\[ \frac{dp}{dt} = \frac{dv}{dt}\,i + v\,\frac{di}{dt}. \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-0821dccfc5de737f5e99780c62f53f31_l3.png "Rendered by QuickLaTeX.com")
In a capacitor, where 
, the stored energy 
has derivative 
, which is exactly the power: a small consistency check that goes through the product rule.
