A second (more elegant) proof of the Quotient Rule

6 February 2018 in MAE 113: Calculus for the Liberal Arts | No comments

The quotient rule proof can be quite tedious if we are only allowed to use the definition of the derivative. In this post, this presents an alternative way of proving the quotient rule if we are allowed to use the product rule.

The Quotient Rule: \left( \dfrac{f(x)}{g(x)}\right)' = \dfrac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}

Let h(x) = \dfrac{f(x)}{g(x)} be the quotient of differentiable functions. Then, we can rearrange so that we have

    \begin{align*} h(x) &= \frac{f(x)}{g(x)}\\ f(x) &= h(x)g(x). \end{align*}

Assuming that h(x) is differentiable, we can use the product rule on f(x) = h(x)g(x), which yields

    \[f'(x) = h'(x) g(x) + h(x) g'(x).\]

Then we isolate h'(x) on one side of the equation.

    \begin{align*} f'(x) &= h'(x) g(x) + h(x) g'(x)\\ h'(x) g(x) + h(x) g'(x) &= f'(x)\\ h'(x) g(x) &= f'(x) - h(x) g'(x)\\ h'(x) &= \frac{f'(x) - h(x) g'(x)}{g(x)} \end{align*}

Now, we substitute h(x) = \dfrac{f(x)}{g(x)}.

    \begin{align*} h'(x) &= \frac{f'(x) - h(x) g'(x)}{g(x)}\\ &= \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)}\\ &= \frac{f'(x)}{g(x)} - \frac{f(x) g'(x)}{[g(x)]^2}\\ &= \frac{f'(x)}{g(x)}\cdot\frac{g(x)}{g(x)} - \frac{f(x) g'(x)}{[g(x)]^2}\\ &= \frac{f'(x) g(x)}{[g(x)]^2} - \frac{f(x) g'(x)}{[g(x)]^2}\\ &= \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2}\\ \end{align*}

We have arrived at the desired result, and this completes our proof.

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