KindiakMath

Differentiating Products

Previously, we explored the mighty chain rule:

\displaystyle \frac{\mathrm d}{\mathrm dx}( f( g(x) ) ) = f'( g(x) ) \cdot g'(x).

By considering f(x) = x^2 and g(x) = h(x), we have

\displaystyle \frac{\mathrm d}{\mathrm dx}( h(x)^2 ) = 2 \cdot h(x) \cdot h'(x).

Using this result, we showed that

\begin{aligned} \frac{\mathrm d}{\mathrm dx}( ( f(x) + g(x) )^2 ) &= \frac{\mathrm d}{\mathrm dx}(f(x)^2) + \frac{\mathrm d}{\mathrm dx}(g(x)^2) \\ &\phantom{==} + 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ).\end{aligned}.

This result is responsible for the product rule.

Theorem 1 (Product Rule). For functions f(x), g(x) with derivatives f'(x), g'(x),

\displaystyle \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) = f'(x) \cdot g(x) + g'(x) \cdot f(x).

Proof. We have the result

\begin{aligned} \frac{\mathrm d}{\mathrm dx}\left((f(x) + g(x))^2 \right) &= \frac{\mathrm d}{\mathrm dx}(f(x)^2 ) + \frac{\mathrm d}{\mathrm dx}(g(x)^2 ) \\ &\phantom{==} + 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ).\end{aligned}

On the other hand, using vanilla algebra,

(f(x) + g(x))^2 = f(x)^2 + 2 \cdot f(x) \cdot g(x) + g(x)^2.

Hence,

2 \cdot f(x) \cdot g(x) = (f(x) + g(x))^2 - f(x)^2 - g(x)^2.

Using linearity,

\begin{aligned} \frac{\mathrm d}{\mathrm dx} ( 2 \cdot f(x) \cdot g(x) ) &= \frac{\mathrm d}{\mathrm dx}\left( (f(x) + g(x))^2 - (f(x))^2 - (g(x))^2 \right) \\ 2 \cdot \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) &= \frac{\mathrm d}{\mathrm dx}\left( (f(x) + g(x))^2 \right) - \frac{\mathrm d}{\mathrm dx}(f(x)^2 ) - \frac{\mathrm d}{\mathrm dx}(g(x)^2 ) \\ 2 \cdot \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) &= 2 \cdot (f'(x) \cdot g(x) + g'(x) \cdot f(x) ) \\ \frac{\mathrm d}{\mathrm dx} ( f(x) \cdot g(x) ) &= f'(x) \cdot g(x) + g'(x) \cdot f(x). \end{aligned}

Previously, we have also proven that

\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \frac 1{h(x)} \right) = -\frac{h'(x)}{h(x)^2}.

Together with the product rule, we can prove the quotient rule.

Theorem 2 (Quotient Rule). For g(x) \neq 0,

\displaystyle \frac{ \mathrm d }{ \mathrm dx } \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}.

Proof. Using the product rule,

\begin{aligned} \frac{ \mathrm d }{ \mathrm dx } \left( \frac{f(x)}{g(x)} \right) &= \frac{ \mathrm d }{ \mathrm dx } \left( f(x) \cdot \frac 1{g(x)} \right) \\ &= f'(x) \cdot \frac 1{g(x)} + \frac{\mathrm d}{\mathrm dx} \left( \frac 1{g(x)} \right) \cdot f(x) \\ &= f'(x) \cdot \frac 1{g(x)} + \left( -\frac{g'(x)}{g(x)^2} \right) \cdot f(x) \\ &= \frac {f'(x) \cdot g(x)}{g(x)^2} - \frac{g'(x) \cdot f(x)}{g(x)^2} \\ &= \frac {f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2} .\end{aligned}

Most differentiation problems becomes simply applying these results one after another, ensuring very careful algebraic calculations.

Example 1. Given that f(x) \neq 0, g(x) \neq 0, and \displaystyle y = \frac{f(x)}{g(x)}, show that

\displaystyle \frac{ \mathrm dy }{ \mathrm dx } = 0 \quad \iff \quad \frac{f'(x)}{ f(x) } = \frac{g'(x)}{g(x)}.

Solution. Using the quotient rule,

\displaystyle \begin{aligned} \frac{ \mathrm dy }{ \mathrm dx } &= \frac {f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}. \end{aligned}

Since g(x) \neq 0,

\displaystyle \frac{ \mathrm dy }{ \mathrm dx } = 0 \quad \iff \quad f'(x) \cdot g(x) - g'(x) \cdot f(x) = 0.

Since f(x) \neq 0,

\displaystyle f'(x) \cdot g(x) - g'(x) \cdot f(x) = 0 \quad \iff \quad \frac{f'(x)}{ f(x) } = \frac{g'(x)}{g(x)}.

The equation \displaystyle \frac{ \mathrm dy }{ \mathrm dx } = 0 is called the zero derivative condition, which plays a vital role in optimisation applications.

Oddly enough, when compared with the chain rule, there is really not much else to be discussed about these two rules until we start differentiating other kinds of functions, like the trigonometric functions, exponential functions, and logarithm functions.

So for now, we switch gears and discuss turning points, applying it in the context of optimisation.

—Joel Kindiak, 8 Jan 26, 2036H

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