How to prove the quotient rule using the product and chain rules?

Here is the derivation using only the product rule and chain rule. Write $\displaystyle \frac{u}{v} = u \cdot v^{-1}$. Apply the **product rule** $\frac{d}{dx}[f \cdot g] = f'g + fg'$ with $f = u$ and $g = v^{-1}$: $$\frac{d}{dx}\left(u \cdot v^{-1}\ ight) = \frac{du}{dx} \cdot v^{-1} + u \cdot \frac{d}{dx}\left(v^{-1}\ ight)$$ Now apply the **chain rule** to differentiate $v^{-1}$: $$\frac{d}{dx}\left(v^{-1}\ ight) = -1 \cdot v^{-2} \cdot \frac{dv}{dx} = -\frac{1}{v^2} \cdot \frac{dv}{dx}$$ Substitute back: \begin{align*} \frac{d}{dx}\left(\frac{u}{v}\ ight) &= \frac{1}{v} \cdot \frac{du}{dx} + u \cdot \left(-\frac{1}{v^2} \cdot \frac{dv}{dx}\ ight) \\ &= \frac{1}{v} \cdot \frac{du}{dx} - \frac{u}{v^2} \cdot \frac{dv}{dx} \\ &= \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \end{align*} This is exactly the quotient rule. The key insight is rewriting $u/v$ as a product $u \cdot v^{-1}$, which is a trick that appears frequently in calculus.