Basic Laws for Differentiation

# Basic Laws for Differentiation

Before we look at some other methods of differentiation, let's first note some important properties of differentiation:

 Theorem 1 (Addition Law): If $f = g + j$ where $g$ and $j$ are differentiable, then $\frac{d}{dx} f(x) = \frac{d}{dx} g(x) + \frac{d}{dx} j(x)$.
• Proof: From the definition of a derivative we know that $\frac{d}{dx} f(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}$. Now suppose that $f = g + j$. Then:
(1)
\begin{align} \frac{d}{dx} f(x) &= \frac{d}{dx} (g(x) + j(x)) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) + j(x + h) - g(x) - j(x)}{h} \right ) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) - g(x)}{h} + \frac{j(x + h) - j(x)}{h} \right ) \\ \: \frac{d}{dx} f(x) &= \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} + \lim_{h \to 0} \frac{j(x + h) - j(x)}{h} \\ \frac{d}{dx} f(x) &= \frac{d}{dx} g(x) + \frac{d}{dx} j(x) \quad \blacksquare \end{align}
 Theorem 2 (Subtraction Law): If $f = g - j$ where $g$ and $j$ are differentiable, then $\frac{d}{dx} f(x) = \frac{d}{dx} g(x) - \frac{d}{dx} j(x)$.
• Proof: Like property 1, from the definition of a derivative we know that $\frac{d}{dx} f(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}$, and suppose $f = g - j$, then:
(2)
\begin{align} \frac{d}{dx} f(x) &= \frac{d}{dx} (g(x) - j(x)) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) - j(x + h) - g(x) + j(x)}{h} \right ) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) - g(x)}{h} - \frac{j(x + h) - j(x)}{h} \right ) \\ \: \frac{d}{dx} f(x) &= \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} - \lim_{h \to 0} \frac{j(x + h) - j(x)}{h} \\ \frac{d}{dx} f(x) &= \frac{d}{dx} g(x) - \frac{d}{dx} j(x) \quad \blacksquare \end{align}
 Theorem 3 (Constant Multiplication Law): If $f$ is differentiable and $k \in \mathbb{R}$, then $\frac{d}{dx} kf(x) = k\frac{d}{dx} f(x)$.
• Proof: By the definition of a derivative we get that $\frac{d}{dx} f(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}$. For some $k \in \mathbb{R}$, we will apply the definition of the derivative to get:
(3)
\begin{align} \frac{d}{dx} kf(x) &= \lim_{h \to 0} \left ( k \cdot \frac{f(x + h) - f(x)}{h} \right ) \\ \frac{d}{dx} kf(x) &= \lim_{h \to 0} k \cdot \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ \frac{d}{dx} kf(x) &= k \cdot \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ \frac{d}{dx} kf(x) &= k \frac{d}{dx} f(x) \quad \blacksquare \end{align}