Pretend that we didn't cover the quotient rule, but had covered the product and chain rules. Prove that the quotient rule holds using the identity 9(x) = g(x) (h(x)). (Hint: -1 h(x)
After applying the product and chain rules, you may need to simplify some fractions to get the form we're used to.)

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

Pretend that we didn't cover the quotient rule, but had covered the product and chain rules. Prove that the quotient rule holds using the identity g(x) = g(x) (h(x))^-1. (Hint: After applying the product and chain rules, you may need to simplify some fractions to get the form we're used to.)

Learn how to prove the quotient rule using the product and chain rules in calculus. This detailed step-by-step guide covers differentiation techniques, including simplifying fractions and applying key rules. Suitable for college-level or AP Calculus students.

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