What is the best way to prove general differentiation product rule w/o going back to limit interpretation?

Question

I am trying to put some flow in how concepts and rules are established in basic differentiation: which comes first and which comes second. Particularly, my question is what the best way is to prove product rule in general w/o going to the very definition of differentiation. Here is my approach.

The 4 basic foundations needed are below.

1. Definition of differentiation: $f' (x)=\lim\limits_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$
2. Chain rule: $f'(x) = \frac {d f} {dx} = \frac {df_1} {df_2} \frac {df_2} {df_3} \frac {df_3} {dx}$
3. Derivative of $ e^x: \frac{d}{dx} e^x = e^x$
4. Derivative of $\ln x: \frac{d}{dx}\ln x = \frac{1}{x}$

Then we prove product rule : $(uv)'=u'v+uv'$ in the following steps.

Let $y=u\cdot v$

$ln(y) = \ln(u)+\ln(v) $ - property of logarithm

$\frac{1}{y}y'=\frac{1} {u}u'+\frac{1} {v}u'$ - differentiation on both sides with chain rule, $\frac{d}{dy} ln (y) =\frac{1}{y} $

$y'=y\frac{1} {u}u'+y\frac{1} {v}u'$ - multiply both sides by $y$

$(uv)'=u'v+uv'$ - replace $y$ with $uv$

The above process can be easily applied to $(uvw)'=u'vw+uv'w+uvw^\prime$ and so on...

Any one has suggestions for different and better approaches?

Answer 1

One good way of doing differential rules is to replace each variable with "variable + differential". Then, at the end, any differential multiplied by a differential can be eliminated (you can think of it as being an infinitely small decimal multiplied by another infinitely small decimal - the value is then infinitely-infinitely small).

So, below, we start off with $z$ and then solve for $dz$. Step 4 eliminates $z$ itself by subtracting the first equation.

$$z = x\cdot y\\ z + dz = (x + dx)(y + dy) \\ z + dz = xy + x\,dy + y\,dx + dx\,dy \\ dz = x\,dy + y\,dx + dx\,dy \\ dz \simeq x\,dy + y\,dx $$

Answer 2

Making johnnyb's approach more rigorous. My answer goes to definitions (against the request to avoid this), but it is just to show why johnnyb's approach actually works.

By the definition of a derivative $$f(x+h)=f(x)+f'(x)h+o(|h|),\ g(x+h)=g(x)+g'(x)h+o(|h|)$$ $$f(x+h)g(x+h)=f(x)g(x)+(f(x)g(x))'h+o(|h|)$$ but also from the first line $$f(x+h)g(x+h)=(f(x)+f'(x)h+o(|h|))(g(x)+g'(x)h+o(|h|))=$$ $$=f(x)g(x)+[f'(x)g(x)+f(x)g'(x)]h+o(|h|)$$ and this holds as $g'(x)f'(x)h^2+g(x)o(|h|)+g'(x)o(|h|)h+o(|h|)(f(x)+f'(x)h+o(|h|))\in o(|h|)$ because $$\lim|\frac{g'(x)f'(x)h^2+g(x)o(|h|)+g'(x)o(|h|)h+o(|h|)(f(x)+f'(x)h+o(|h|))}{h}|=0$$ So from the uniqueness of derivative we have $$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$$

There is also a definition of derivative that simplifies this kind of derivations (and complicates other kinds), but i can't recall it now.

Answer 3

The best way is to go back to the limit definition, as this only involves elementary means.

(When using alternatives approaches, make sure that you don't invoke the product rule indirectly - unknowingly - as this would make a circular argument. So you should check the definitions of the exponential and the logarithms, and how their derivatives are established, as well as how the chain rule is obtained.)

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This said, you ought to know the process named as the logarithmic derivative,

$$\frac{(tuvw)'}{tuvw}=(\log(tuvw))'=(\log(t)+\log(u)+\log(v)+\log(w))'=\frac{t'}t+\frac{u'}u+\frac{v'}v+\frac{w'}w$$

and

$$(tuvw)'=tuvw\left(\frac{t'}t+\frac{u'}u+\frac{v'}v+\frac{w'}w\right).$$