📂Analysis

Differentiable Function Properties

Theorem¹

Let’s say $f, g : [a,b] \to \mathbb{R}$. If $f,g$ is differentiable at $x\in [a,b]$, then $f+g$, $fg$, and $f/g$ are also differentiable at $x$ and the following equation holds.

$$ \begin{align} (f+g)^{\prime}(x) &=f^{\prime}(x)+g^{\prime}(x) \\ (fg)^{\prime}(x) &= f^{\prime}(x)g(x)+f(x)g^{\prime}(x) \\ \left( \frac{f}{g} \right)^{\prime}(x) &= \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)} \end{align} $$

However, $(3)$ holds when $g(x)\ne 0$.

Description

$(2)$ is commonly referred to as the product rule of differentiation.

Proof

$(1)$

By the definition of differentiation and the properties of the limit of functions, the following is true.

$$ \begin{align*} (f+g)^{\prime}(x) &=\lim \limits_{t \to x} \frac{(f+g)(x)-(f+g)(t)}{x-t} \\ &= \lim \limits_{t \to x} \frac{(f(x)+g(x))-(f(t)+g(t))}{x-t} \\ &= \lim \limits_{t \to x} \frac{(f(x)-f(t))+(g(x)+g(t))}{x-t} \\ &= \lim \limits_{t \to x} \left[ \frac{f(x)-f(t)}{x-t}+\frac{g(x)+g(t)}{x-t} \right] \\ &= \lim \limits_{t \to x} \frac{f(x)-f(t)}{x-t}+ \lim \limits_{t \to x}\frac{g(x)+g(t)}{x-t} \\ &= f^{\prime}(x)+g^{\prime}(x) \end{align*} $$

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$(2)$

The following holds by the definition of differentiation and the properties of the limit of functions.

$$ \begin{align*} (fg)^{\prime}(x) &= \lim \limits_{t \to x} \frac{(fg)(x)-(fg)(t)}{x-t} \\ &= \lim \limits_{t \to x}\frac{f(x)g(x)-f(t)g(t)}{x-t} \\ &= \lim \limits_{t \to x}\frac{f(x)g(x) {\color{blue}-f(t)g(x)+f(t)g(x)}-f(t)g(t)}{x-t} \\ &= \lim \limits_{t \to x}\left[ \frac{f(x)g(x) -f(t)g(x)}{x-t} + \frac{f(t)g(x)-f(t)g(t)}{x-t} \right] \\ &= \lim \limits_{t \to x}\left[ \frac{f(x) -f(t)}{x-t}g(x) + f(t)\frac{g(x)-g(t)}{x-t} \right] \\ &= \lim \limits_{t \to x} \left[\frac{f(x) -f(t)}{x-t}g(x)\right] + \lim \limits_{t \to x} \left[ f(t)\frac{g(x)-g(t)}{x-t} \right] \\ &= \lim \limits_{t \to x}\frac{f(x) -f(t)}{x-t}\lim \limits_{t \to x}g(x) + \lim \limits_{t \to x} f(t)\lim \limits_{t \to x}\frac{g(x)-g(t)}{x-t} \\ &= f^{\prime}(x)g(x)+f(x)g^{\prime}(x) \end{align*} $$

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$(3)$

Proved by a similar method to $(2)$.

$$ \begin{align*} \left( \frac{f}{g} \right)^{\prime}(x) &= \lim \limits_{ t \to x } \frac{(f/g)(x) -(f/g)(t)}{x-t} \\ &= \lim \limits_{ t \to x } \frac{f(x)/g(x) -f(t)/g(t)}{x-t} \\ &= \lim \limits_{ t \to x } \frac{f(x)/g(x) {\color{blue}-f(x)/g(t)+f(x)/g(t)}- f(t)/g(t)}{x-t} \\ &= \lim \limits_{ t \to x } \left[ \frac{f(x)/g(x) - f(x)/g(t) }{x-t}+\frac{f(x)/g(t)-f(t)/g(t)}{x-t} \right] \\ &= \lim \limits_{ t \to x } \left[ \frac{\frac{f(x){\color{blue}g(t)}}{g(x){\color{blue}g(t)}} -\frac{f(x){\color{blue}g(x)}}{g(t){\color{blue}g(x)}} }{x-t}+\frac{\frac{f(x){\color{blue}g(x)}}{g(t){\color{blue}g(x)}}-\frac{f(t){\color{blue}g(x)}}{g(t){\color{blue}g(x)}}}{x-t} \right] \\ &= \lim \limits_{ t \to x } \frac{1}{g(x)g(t)} \left[ {\color{red}\frac{f(x)g(t)-f(x)g(x) }{x-t}}+\frac{f(x)g(x)-f(t)g(x)}{x-t} \right] \\ &= \lim \limits_{ t \to x }\frac{1}{g(x)g(t)} \left[\frac{f(x)g(x)-f(t)g(x)}{x-t}{\color{red}-\frac{f(x)g(x)-f(x)g(t) }{x-t}} \right] \\ &= \lim \limits_{ t \to x }\frac{1}{g(x)g(t)} \left[\frac{f(x)-f(t)}{x-t}g(x)-f(x)\frac{g(x)-g(t) }{x-t} \right] \\ &= \lim \limits_{ t \to x }\frac{1}{g(x)g(t)} \lim \limits_{ t \to x } \left[\frac{f(x)-f(t)}{x-t}g(x)-f(x)\frac{g(x)-g(t) }{x-t} \right] \\ &= \frac{1}{g^{2}(x)}\left[ f^{\prime}(x)g(x)-f(x)g^{\prime}(x) \right] \\ &= \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)} \end{align*} $$

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