Proving Quotient Rule without Chain Rule Is there a proof of $(\frac{u}{v})' = \frac{vu'-v'u}{v^2}$ that doesn't utilize the Chain Rule?
 A: $\frac {d}{dx} \frac uv = \lim_\limits{a\to x} \frac {\frac {u(a)}{v(a)} - \frac{u(x)}{v(x)}}{a-x}\\
\lim_\limits{a\to x} \frac {u(a)v(x) - u(x)v(a)}{(v(a)v(x)(a-x)}\\
\lim_\limits{a\to x} \frac {u(a)v(x)-u(x)v(x) +u(x)v(x)- u(x)v(a)}{v(a)v(x)(a-x)}\\
\lim_\limits{a\to x} \frac {v(x)(u(a)-u(x)) -u(x)(v(a)- v(x))}{v(a)v(x)(a-x)}\\
\lim_\limits{a\to x} \frac {v(x)}{v(x)v(a)}\frac {u(a)-u(x)}{a-x} - \frac {u(x)}{v(x)v(a)}\frac {v(a)-v(x)}{a-x}\\
\frac {u'}{v} - \frac {uv'}{v^2}\\
\frac {u'v - uv'}{v^2}$
A: Let $w=\dfrac{u}{v}$. Then $u=vw$ and so $u'=v'w+vw'$. Therefore
$$
w'=\frac{u'-v'w}{v}= \frac{u'-v'\dfrac{u}{v}}{v}=\frac{u'v-uv'}{v^2}
$$
Only the product rule is needed.
A: Yes.  
$$\frac{1}{h}\left( \frac{f(x+h)}{g(x+h)} -\frac{f(x)}{g(x)}\right ) =
\frac{1}{h}\left( \frac{f(x+h)}{g(x+h)} -\frac{f(x)}{g(x+h)}+\frac{f(x)}{g(x+h)}-\frac{f(x)}{g(x)}\right ) $$
$$= \frac{f(x+h)-f(x)}{h}\cdot \frac{1}{g(x+h)}  +\frac{f(x)}{h}\cdot\frac{g(x)-g(x+h)}{g(x+h)g(x)}  $$
Let $h\to 0$ and this becomes
$$f'(x)\cdot\frac{1}{g(x)} +\frac{f(x)}{g(x)^2}\cdot (-g'(x))    $$
which reduces to the usual quotient rule.
A: Here is a proof that works for $u>0$ and $v>0$.
Let us consider identity:
$$\ln\left(\frac{u}{v}\right)=\ln(u)-\ln(v)$$
Let us differentiate both sides:
$$\dfrac{(\tfrac{u}{v})'}{(\tfrac{u}{v})}=\frac{u'}{u}-\frac{v'}{v}.$$
or
$$\left(\frac{u}{v}\right)'=\left(\frac{u}{v}\right)\left(\frac{u'v-uv'}{uv}\right)$$
whence the result.
