A big part of calculus involves rates of change. Rates of changes involves the ratio of the differences in one variable when there is a change in another variable.
Calculus derivatives focus on the instantaneous rate of change at a point on the function’s domain. Given a step size or a small increment of \(h\), as \(h\) approaches zero, the derivative at a point is obtained. This gives the limit definition of the derivative at a point \(a\) (in the function \(f(x)\)).
Limit Definition
\[f'(a) = \displaystyle{\lim_{h \to 0}}\dfrac{f(a + h) - f(a)}{h}\]
\(\textbf{Example One}\)
With the limit definition of the derivative, find the derivative of \(g(x) = 3x\) at the point \(x = 2\).
\[g'(2) = \displaystyle{\lim_{h \to 0}} \dfrac{g(2 + h) - g(2)}{h}\] \[g'(2) = \displaystyle{\lim_{h \to 0}} \dfrac{3(2 + h) - 3(2)}{h}\]
\[g'(2) = \displaystyle{\lim_{h \to 0}} \dfrac{6 + 3h - 6}{h}\] \[g'(2) = \displaystyle{\lim_{h \to 0}} \dfrac{3h}{h}\]
\[g'(2) = \displaystyle{\lim_{h \to 0}} 3\]
\[g'(2) = 3\]
The derivative of \(g(x) = 3x\) at \(x = 2\) is simply 3. This three represents the slope at the point \(x = 2\). Furthermore, this slope of three represents the slope for all x-values in \(g(x)\).
\(\textbf{Example Two}\)
With the limit definition of the derivative, find the derivative of \(f(x) = -2x^2\) for any value of \(x\).
\[f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{f(x + h) - f(x)}{h}\]
\[f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{-2(x + h)^2 - -2x^2}{h}\]
\[f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{-2(x^2 + 2hx + h^2) +2x^2}{h}\]
\[f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{-2x^2 - 4hx - 2h^2 + 2x^2}{h}\]
\[f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{-4hx - 2h^2}{h}\]
\[f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{-2h(2x - h)}{h}\]
\[f'(x) = \displaystyle{\lim_{h \to 0}} -2(2x - h)\]
\[f'(x) = \displaystyle{\lim_{h \to 0}} (-4x) + \lim_{h \to 0} 2h\]
\[f'(x) = -4x + 0\]
The derivative of \(f(x) = -2x^2\) is \(f'(x) = -4x\). In this second example, there is more algebra involved. The main goal is to use algebra and factoring to eliminate the h on the bottom. Once the h on the bottom of the fraction is removed, you can apply the limit as h approaches zero.
The limit definition of the derivative helps in understanding where the derivative comes from but it is not the fastest. A more efficient way and common way of obtaining derivatives is through the power rule for polynomials.
Consider the polynomial function \(f(x)\) where:
\[f(x) = a_{n}x^{n} + a_{n - 1}x^{n - 1} + ... + a_{3}x^3 + a_{2}x^2 + a_{1}x + a_{0}\]
where \(a_{n}\) to \(a_{0}\) are numeric constant coefficients.
The derivative \(f'(x)\) of the above polynomial function would be:
\[f'(x) = a_{n}n x^{n - 1} + a_{n - 1} (n - 1) x^{n - 2} + ... + 3 a_{3}x^2 + 2 a_{2}x^1 + a_{1}\]
Notice that the \(a_{0}\) intercept term goes away. In addition, the terms are multiplied by the exponent and the exponent decreases by one. The examples below will give a better idea on how this all works.
For calculus derivatives, they can operate with addition and subtraction signs. Here are some basic rules.
If \(h(x) = f(x) + g(x)\) then \(h'(x) = f'(x) + g'(x)\).
If \(h(x) = f(x) - g(x)\) then \(h'(x) = f'(x) - g'(x)\).
If \(g(x) = f_1(x) + f_2(x) + f_3(x) + \dots + f_n(x)\) then \(g'(x) = f'_1(x) + f'_2(x) + f'_3(x) + \dots + f'_n(x)\). This also holds true for the subtraction case.
Here are some examples of finding derivatives of functions. Note that \(f'(x)\) refers to the same derivative as \(\dfrac{d}{dx} f(x)\).
\(\textbf{Example One}\)
Find the derivative of the function \(f(x) = 2x\).
\[\begin{array}{lcl} f'(x) & = & \dfrac{d}{dx} 2x^1 \\ & = & 2 \dfrac{d}{dx} x^1 \\ & = & 2 \times 1 \times x^{1 - 1} \\ & = & 2 \times 1 \times 1 \\ & = & 2 \end{array}\]
\(\textbf{Example Two}\)
Given \(f(x) = 3x^3\) what is \(f'(x)\)?
\[\begin{array}{lcl} f'(x) & = & \dfrac{d}{dx} 3x^3 \\ & = & 3 \dfrac{d}{dx} x^3 \\ & = & 3 \times 3x^2 \\ & = & 9x^2 \end{array}\]
\(\textbf{Example Three}\)
What is the derivative of \(f(x) = x^{\pi}\)?
\[ f'(x) = \pi x^{\pi - 1} \]
Remember that \(\pi \approx 3.14\) is a number.
\(\textbf{Example Four}\)
What is the derivative of \(g(x) = x(x^2 - 4x + 2)\)?
This particular example looks scary but it is actually not too bad. The distributive law is applied first and derivatives can be taken separately.
Rewrite \(g(x)\) as:
\[g(x) = x(x^2 - 4x + 2) = x^3 - 4x^2 + 2x\]
Then you can take the derivatives of each term.
\[g'(x) = \dfrac{d}{dx}(x^3 - 4x^2 + 2x)\]
\[g'(x) = \dfrac{d}{dx} x^3 - \dfrac{d}{dx} 4x^2 + \dfrac{d}{dx} 2x\]
\[g'(x) = 3x^2 - 8x + 2\]