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http://planetmath.org/proofofproductrule

Here is a proof of the product rule. I understand how it all works, but I want to know the reasoning behind why whoever came up with this chose to add a $f(x+h)g(x) - f(x+h)g(x)$ in the numerator. I see it works out, but how can someone just think I should put this in to get a result that allows for a shortcut of taking the derivative of a product? Maybe it just takes a very very long time working with mathematics to be able to see such connections and that is all there is to it.

Dominick
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5 Answers5

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Look at the original expression $$f(x+h)g(x+h)-f(x)g(x).$$ Something very complicated is going on: we are trying to find how much the product changes when both factors change at once. Don't you think it might just possibly be easier to see what happens if we change one factor at a time? So we start with $$f(x)g(x)\tag1$$ and first change $f(x)$ to $f(x+h)$ $$f(x+h)g(x)\tag2$$ and after that change $g(x)$ to $g(x+h)$ $$f(x+h)g(x+h).\tag3$$ The increment from (1) to (2) is $$f(x+h)g(x)-f(x)g(x),$$ and the increment from (2) to (3) is $$f(x+h)g(x+h)-f(x+h)g(x),$$ and the total increment from (1) to (3) is $$[f(x+h)g(x+h)-f(x+h)g(x)]+[f(x+h)g(x)-f(x)g(x)].$$ The idea "change one variable at a time, rather than all at once" may be clever, but not inhumanly clever.

bof
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The reason that they put it in, as you probably already know, is so that it we can easily factor with it.

As for why someone thought of this, notice that if we did not add this in, there would be no connection between $f(x),g(x)$. When we add this connection it, it easily allows for factoring and simplification.

It's not that it takes very long to do it without it, it's that you cannot do it without it. You need that "connection", in order for the proof to simplify down and make sense.

K Split X
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  • So someone saw, "I need to relate these two parts $f(x+h)g(x+h)$ and $f(x)g(x)$ and I cannot progress further unless I find such a way. So I will add $f(x+h)g(x) - f(x+h)g(x)$." It seems like this would not be so intuitive to think that this is what should be added. I don't see it, but I am not very experienced with math. Is there any reasoning you can point out that will help me see how someone could come to the conclusion that this particular addition and subtraction will relate them? Why not $g(x+h)f(x) - g(x+h)f(x)$ or something? – Dominick Apr 23 '17 at 02:05
  • $g(x+h)f(x) - g(x+h)f(x)$ works just as well, actually – Ben Grossmann Apr 23 '17 at 02:07
  • That is interesting, however still the question remains is there any sort of reasoning about what prompted this that a "noobie" relative to such people that produce proofs could understand. It seems like the answer is just such people have a great deal of experience and they just see the connection. – Dominick Apr 23 '17 at 02:11
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    Yes you're right. The proofs of this nature that you do, the more "obvious" it becomes. You'll realise that all that step is doing is "adding zero". One of my favourite lecturers a few years ago said "All I ever do is add zero and multiply by one. That's all I ever do." – Harambe Apr 23 '17 at 02:14
  • @Dominick your right that someone saw that they needed to relate these parts. How they did it also makes sense. Since you cannot change the equation they decided to add a $0$. After you add the $0$, then we have the $f(x)g(x)$, and its negative correspondent, so we can then factor so we get the derivative form we are looking for. – K Split X Apr 23 '17 at 03:37
  • As for your question to why they added exactly $f(x+h)g(x)-f(x+h)g(x)$ is because the derivative has the form $\dfrac{f(x+h)-f(x)}{x-h}$, so when they add this, you neatly factor and can get the derivative forms – K Split X Apr 23 '17 at 03:41
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The logic of the product rule is to capture the intuition behind this image.

Visual Proof of Product Rule

A. Thomas Yerger
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This may not be what you're after, but there is a slightly different way of looking at a derivative which makes the proof of the product rule very natural. Let $f$ be a function which we want to differentiate at $c$. Say that $f$ is differentiable at $c$ if there is some continuous function $f_c$ such that $$f(x) = f(c) + (x - c) f_c(x)$$ and if such a function $f_c$ exists, say that the derivative of $f$ at $c$ is $f'(c) := f_c(c)$. Note that this agrees with the limit definition just by rearranging the equation, but this definition makes a lot of derivations much easier, and each step in the derivation is natural.

The rest of the answer is a proof that if $f$ and $g$ are differentiable at $c$, then their product $h(x) = f(x)g(x)$ has the derivative $h'(c) = f'(c) g(c) + f(c) g'(c)$ at $c$.

Now, suppose both $f$ and $g$ are differentiable at $c$. That means we have continuous functions $f_c$ and $g_c$ such that $$\begin{aligned} f(x) &= f(c) + (x - c) f_c(x) \\ g(x) &= g(c) + (x - c) g_c(x) \end{aligned}$$

Multiplying these gives $$f(x) g(x) = f(c) g(c) + (x - c)\left[ f_c(x) g(c) + f(c) g_c(x) + (x - c) f_c(x) g_c(x) \right] $$ So we've found that the function $h(x) = f(x) g(x)$ may be written $$h(x) = h(c) + (x - c) h_c(x)$$ where $h_c(x) = f_c(x) g(c) + f(c) g_c(x) + (x - c) f_c(x) g_c(x)$. So the derivative of $h$ at $c$ should be $$h'(c) := h_c(c) = f_c(c) g(c) + f(c) g_c(c) = f'(c) g(c) + f(c) g'(c)$$

Joppy
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If you find the trick of adding and subtracting $f(x+h) g(x) $ so surprising it is better to use a more direct approach. Since we have $$f'(x) =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ we have $$f(x+h) =f(x) +h\{f'(x)+\rho\} $$ where $\rho\to 0$ with $h$. Similarly $$g(x+h) =g(x) +h\{g'(x)+\rho'\} $$ where $\rho'\to 0$.

Now we can see that $$f(x+h) g(x+h) =f(x)g(x) + h\{f(x) g'(x) +g(x) f'(x) \} + h\{\rho'f(x)+\rho g(x)\} +h^2\{(f'(x)+\rho)(g'(x)+\rho')\}$$ and then $$\frac{f(x+h) g(x+h) - f(x) g(x)} {h} =f(x) g'(x) +g(x) f'(x) + \rho'g(x) +\rho f(x) +h\{(f'(x) +\rho) (g'(x) +\rho') \}$$ Now letting $h\to 0$ we get the desired result.

Paramanand Singh
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