Derivation for quotient rule help I'm trying to derive the quotient rule in a similar fashion to what was done here: Product rule intuition
I'm trying to get the change in $\frac{f}{g}$.
$$\frac{f + df}{g + dg} - \frac{f}{g} =$$
$$=\frac{gf + gdf} {g * (g + dg)} - \frac{fg + fdg}{g * (g + dg)} =$$
$$=\frac{gdf - fdg}{g * (g + dg)}$$
which is pretty close to the quotient rule. But I can't figure out why the denominator is $g * (g + dg)$ rather than $g^2$. Where am I going wrong?
I'm not interested in deriving the quotient rule in terms of another rule (like the product rule/chain rule). Thanks.
 A: $$f(x+\triangle x)=f(x)+f'(x)\triangle x+O(\triangle x^2)$$
$$g(x+\triangle x)=g(x)+g'(x)\triangle x+O(\triangle x^2)$$
$$h(x)=\frac{f(x)}{g(x)}$$
$$h(x+\triangle x)=h(x)+h'(x)\triangle x+O(\triangle x^2)$$
where
$$h'(x)=\frac{f'g-fg'}{g^2}$$
so
$$\frac {f(x+\triangle x)}{g(x+\triangle x)} = \frac{f(x)}{g(x)}+\frac{f'g + fg'}{g^2} \triangle x + O(\triangle x^2)$$
and
$$\frac {f(x+\triangle x)}{g(x+\triangle x)} \ne \frac{f + f' \triangle x}{g + g' \triangle x} $$
A: Your denominator is $g^2 + g \, dg$, but the second term is negligible compared with $g^2$.
Let $h = f/g$.  It may seem more clear if we write your result like this (using infinitesimal intuition):
\begin{equation}
dh = \left(\frac{g \, \frac{df}{dx} - f \, \frac{dg}{dx}}{g^2 + g \, dg}\right) \, dx.
\end{equation}
Now the only infinitesimal term in the expression in parentheses is $g \, dg$, so that term is negligible.
Of course, this is just infinitesimal intuition, not a rigorous proof.
A: My proof of quotient rule goes as following:
$$\frac{\mathrm{d} g}{\mathrm{d} x}=\frac{\mathrm{d}g}{\mathrm{d} z}.\frac{\mathrm{d}z }{\mathrm{d} x}$$
Let: $$g=\frac{1}{f}|z=f$$
So: $$\frac{\mathrm{d} \frac{1}{f}}{\mathrm{d} x}=\frac{\mathrm{d}\frac{1}{f} }{\mathrm{d} f}.\frac{\mathrm{d}f }{\mathrm{d} x}=-\frac{1}{f^{2}}.\frac{\mathrm{d} f}{\mathrm{d} x}=-\frac{f'}{f^{2}}$$
Using the product rule: $$\frac{\mathrm{d} \frac{f}{g}}{\mathrm{d} x}=f'\frac{1}{g}+f(\frac{1}{g})'=\frac{f'}{g}-\frac{fg'}{g^{2}}$$
A: I don't know if that's formally correct. I will propose another derivation instead. It builds upon the product rule, which you said you have already proved. It also uses the chain rule.
$$\begin{align*}
\left(\frac{f(x)}{g(x)}\right)' &= (f(x)g^{-1}(x))' = f'(x)g^{-1}(x) + f(x)(g^{-1}(x))' =\\[0.5ex]
&= f'(x)g(x)g^{-2}(x) + f(x)(-g^{-2}(x)g'(x)) =\\[2ex]
&= \frac{f'(x)g(x) - f(x)g'(x)}{g^{2}(x)}
\end{align*}$$
Note that here $g^{-1}$ does not represent the inverse function of $g$, but its multiplicative inverse.
