Let's prove the quotient rule by assuming we know the product rule and the chain rule, using the identity $f(x) = \frac{g(x)}{h(x)} = g(x) \cdot \left( h(x) \right)^{-1}$.

We'll differentiate $f(x) = g(x) \cdot \left( h(x) \right)^{-1}$ using the product and chain rules.

Step 1: Differentiate using the product rule

The product rule states that if $f(x) = u(x) \cdot v(x)$, then:

$f'(x) = u'(x) v(x) + u(x) v'(x)$

Here, let $u(x) = g(x)$ and $v(x) = \left( h(x) \right)^{-1}$. Differentiating $f(x) = g(x) \cdot \left( h(x) \right)^{-1}$ gives:

$f'(x) = g'(x) \cdot \left( h(x) \right)^{-1} + g(x) \cdot \frac{d}{dx} \left( h(x)^{-1} \right)$

Step 2: Differentiate $h(x)^{-1}$ using the chain rule

Next, apply the chain rule to differentiate $h(x)^{-1}$.

The chain rule says that if ( v(x) = \left( h