Differentiation Techniques

Section 2.2 Differentiation Techniques

Consider \(f(x) = c\) for a known constant \(c\text{.}\)

\begin{align*} f'(x) = \dfrac{d}{dx}(c) \amp = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\\ \amp = \lim_{h \to 0} \dfrac{c - c}{h}\\ \amp = \lim_{h \to 0} \dfrac{0}{h}\\ \amp = \lim_{h \to 0} 0 = 0 \end{align*}

so, \(\dfrac{d}{dx}(\text{constant}) = 0\)

Theorem 2.2.1.

If \(f\) is differentiable and \(c\) is a constant, \(\dfrac{d}{dx} \left( c \cdot f(x) \right) = c \cdot \left( \dfrac{d}{dx}f(x) \right)\text{.}\)

Theorem 2.2.2.

If \(f\) and \(g\) are differentiable, then \(\dfrac{d}{dx} \left( f(x) \pm g(x) \right) = \dfrac{d}{dx} \left( f(x) \right) \pm \dfrac{d}{dx} \left( g(x) \right)\text{.}\)

We can also calculate "higher derivatives" by reapplying these rules. We write these as follows:

find the second derivative \(f''(x)\) or \(\dfrac{d^2f}{dx^2}\) for \(f(x) = 6x^3 - x^{\frac{7}{6}} + \dfrac{\pi}{x^2}\)
we first find the first derivative:

\begin{equation*} f'(x) = \dfrac{d}{dx}\left( 6x^3 - x^{\frac{7}{6}} + \dfrac{\pi}{x^2} \right) = 18x^2 - \dfrac{7}{6}x^{\frac{1}{6}} - 2\pi x^{-3} \end{equation*}
now,

\begin{align*} f''(x) \amp = \dfrac{d}{dx} \left( 18x^2 - \dfrac{7}{6}x^{\frac{1}{6}} - 2\pi x^{-3} \right)\\ \amp = 18 \cdot 2x - \dfrac{7}{6} \cdot \dfrac{1}{6}x^{-\frac{5}{6}} - 2\pi \cdot -3x^{-4}\\ \amp = 36x - \dfrac{7}{36}x^{-\frac{5}{6}} + 6\pi x^{-4} \end{align*}

Example 2.2.1.

Find \(\dfrac{d}{dx}\) for \(f(x) = \sqrt{x} = x^\frac{1}{2}\text{:}\)

\begin{equation*} f'(x) = \dfrac{d}{dx}(x^{\frac{1}{2}}) = \dfrac{1}{2}x^{-\frac{1}{2}} \end{equation*}

Now, evaluate at \(x = 1\text{:}\)

\begin{align*} \dfrac{d}{dx}|_{x = 1} = f'(1) \amp = \dfrac{1}{2} \cdot 1^{-\frac{1}{2}}\\ \amp = \dfrac{1}{2\sqrt{1}} = \dfrac{1}{2} \end{align*}

Find the tangent line equation for \(f(x)\) at \(x = 1\text{:}\)

\begin{equation*} f_{\text{tan}}(x) = f(1) + f'(1)(x - 1) = 1 + \dfrac{1}{2}(x - 1) \end{equation*}

NOTE.

The derivative is the slope of the tangent line.

Figure 2.2.3.