Mdthbs

So I tried to derive the product rule without adding $f(x)g(x+Delta x)-f(x)g(x+Delta x)$ as we used to. Instead I started deriving it directly and ran into a strange conclusion that $(uv)'=u'v$. The derivation looks like this:

begin{align}
(uv)' & = lim_{Delta xto0} frac{u(x+Delta x)v(x+Delta x)-u(x)v(x)}{Delta x} \
& = lim_{Delta xto0} frac{u(x+Delta x)v(x+Delta x)}{Delta x}-lim_{Delta xto0} frac{u(x)v(x)}{Delta x} \
& = lim_{Delta xto0} frac{u(x+Delta x)}{Delta x}lim_{Delta xto0} v(x+Delta x)-lim_{Delta xto0} frac{u(x)}{Delta x}lim_{Delta xto0} v(x) \
& = lim_{Delta xto0} frac{u(x+Delta x)}{Delta x}v(x)-lim_{Delta xto0} frac{u(x)}{Delta x}v(x) \
& = v(x)(lim_{Delta xto0} frac{u(x+Delta x)}{Delta x}-lim_{Delta xto0} frac{u(x)}{Delta x}) \
& = v(x)lim_{Delta xto0} frac{u(x+Delta x)-u(x)}{Delta x} \
& = u'v
end{align}

Apparently there is a mistake somewhere, but I can't figure out where exactly. Any ideas?

I find your notation very cumbersome. Now, taking into account that differentiability implies continuity, we can write

$$frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}=frac{f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)}{x-x_0}=$$

$$=f(x)frac{g(x)-g(x_0)}{x-x_0}+frac{f(x)-f(x_0)}{x-x_0}g(x)xrightarrow[xto x_0]{}f(x_0)g'(x_0)+f'(x_0)g(x_0)$$

Complete details. The above is the easiest proof I know of the product rule for derivative. The trick in the first step is also used in general limits.

This may not be exactly what you are looking for, but one easy way to arrive at the product rule is to use logarithms. So, if $y=uv$ then

begin{align} log{y}&=log{uv}&\
&=log{u} +log{v}end{align}

Hence,

$frac{y'}{y}=frac{u'}{u}+frac{v'}{v}$

Now, multiply by $y$ and you get $y'=u'v+v'u$ I mention this approach only as an alternative since other posters have already given satisfactory explanations using the difference quotient definition. With regard to where you went wrong in your original posting, it is in the second line. You have writen a limit of the form $infty -infty$ which is indeterminate. Remember, $x$ is fixed, so $u(x), v(x)$ are also fixed, therefore, the factor $frac{u(x+Delta x)}{Delta x}$ of the first term in line 4 and $frac{u(x)}{Delta x}$ of the second term of line 4 have infinite limits as $Delta x rightarrow 0$. But really, you already had this error present from line 2.

Taking the idea that a $u'$ has to do with the change in $u$ over the change in $x$, write $Delta u = u(x+Delta x) - u(x).$
Likewise write $Delta v = v(x+Delta x) - v(x).$

That is, $u(x+Delta x) = u(x) + Delta u$ and $v(x+Delta x) = v(x) + Delta v.$
Now plug this into your first formula and plod along without knowing where we're going until we get there.

begin{align}
(uv)'
&= lim_{Delta xto0} frac{u(x + Delta x)v(x + Delta x) - u(x)v(x)}{Delta x} \
&= lim_{Delta xto0} frac{(u(x) + Delta u)(v(x) + Delta v) - u(x)v(x)}{Delta x} \
&= lim_{Delta xto0} frac{(u(x)v(x) + (Delta u)v(x) + u(x)(Delta v)
+ (Delta u)(Delta v)) - u(x)v(x)}{Delta x} \
&= lim_{Delta xto0} frac{(Delta u)v(x) + u(x)(Delta v)
+ (Delta u)(Delta v)}{Delta x} \
end{align}

Now observe that
begin{align}
lim_{Delta xto0} frac{Delta u}{Delta x}
&= lim_{Delta xto0} frac{u(x+Delta x) - u(x)}{Delta x} = u'(x), \
lim_{Delta xto0} frac{Delta v}{Delta x}
&= lim_{Delta xto0} frac{v(x+Delta x) - v(x)}{Delta x} = u'(x), \
lim_{Delta xto0} frac{(Delta u)(Delta v)}{Delta x}
&= left(lim_{Delta xto0} frac{Delta u}{Delta x}right)
left(lim_{Delta xto0} Delta vright) = 0
end{align}
since $lim_{Delta xto0} Delta v = 0.$
Putting all of this together,

begin{align}
(uv)'
&= v(x)lim_{Delta xto0} frac{Delta u}{Delta x}
+ u(x)lim_{Delta xto0} frac{Delta v}{Delta x}
+ lim_{Delta xto0} frac{(Delta u)(Delta v)}{Delta x} \
&= v(x) u'(x) + u(x) v'(x) + 0 \
&= u' v + u v'.
end{align}

No tricks with adding and subtracting a mysterious term or other great inspiration, just distribution of multiplication over addition and beating on the monster until it's dead.

Personally I like the more inspired solutions, but sometimes when you're stuck for inspiration you can just push your way through.

Let $h>0$, then
$$frac{u(x+h)v(x+h) - u(x)v(x)}{h} = frac{u(x+h)}{h}v(x+h) -frac{u(x)v(x)}{h}$$
$$=frac{u(x+h)}{h}v(x+h) +frac{u(x)[v(x+h) - v(x) - v(x+h)]}{h}$$
$$= frac{u(x+h) + u(x)}{h}v(x+h) - frac{u(x)[v(x) +v(x+h)] }{h}$$
$$= frac{u(x+h) + u(x)}{h}v(x+h) - frac{u(x)[v(x) -v(x+h) + 2v(x+h)]}{h}$$
$$= frac{u(x+h) - u(x)}{h}v(x+h) + frac{v(x+h) - v(x)}{h}u(x)$$
taking the limit gives
$$(uv)' = u'v + v'u$$

Remember you may only use the additive and multiplicative rules for limits when you can justify that those limits do have finite convergence.   So we just need to arrange the limit definition of $[uv]'$ into that for $ucdot v'+u'cdot v$ (or vice versa).   This is fairly straight forward, as long as you recognise that: $u(x)=limlimits_{hto 0}u(x{+}h)$.

$$begin{align}[uv'+u'v](x)&=u(x)left(lim_{hto 0}dfrac{v(x{+}h)-v(x)}{h}right)+left(lim_{hto 0}dfrac{u(x{+}h)-u(x)}{h}right)v(x)
\[1ex]&= left(lim_{hto 0}u(x{+}h)right)left(lim_{hto 0}dfrac{v(x{+}h)-v(x)}{h}right)+left(lim_{hto 0}dfrac{u(x{+}h)-u(x)}{h}right)v(x)
\[1ex]&=lim_{hto 0}dfrac{u(x{+}h)left(v(x{+}h)-v(x)right)+left(u(x{+}h)-u(x)right)v(x)}{h}\[1ex]&=lim_{hto x}dfrac{u(x{+}h),v(x{+}h)-u(x{+}h),v(x)+u(x{+}h),v(x)-u(x),v(x)}{h}\[1ex]&=lim_{hto 0}dfrac{u(x{+}h),v(x{+}h)-u(x),v(x)}{h}\[3ex][uv'+u'v](x)&= [uv]'(x)end{align}$$