Rules of Derivatives

By now, the limit definition is the method needed to find the derivative of a function.   The purpose of this section is to provide some rules to make calculations of derivatives much more efficient and less tedious.

---

Constant Rule

Consider the function \(f(x)=c\), where \(c\) is any real number.   The graph of \(f\) here is a horizontal line, which always has a slope of 0.   It follows then that: $$\frac{d}{dx}\left(f(x)\right)=0\text{  or  }\frac{d}{dx}(c)=0$$ This is easily verified by using the limit definition of derivative from Section 3.1, click the button below to see the proof.   Click for Proof

---

Power Rule

Now consider a power function, \(f(x)=x^n\), where \(n\) is a positive integer.   Let's consider a few examples and see if a pattern emerges for this derivative.   First, consider \(f(x)=x^2\) and use the limit definition to find the derivative:

\(\large{\frac{d}{dx}}\normalsize{\left(f(x)\right)=\lim_{h\rightarrow0}\left(\large{\frac{f(x+h)-f(x)}{h}}\right)}\)	\(\implies\lim_{h\rightarrow0}\left(\large{\frac{(x+h)^2-x^2}{h}}\right)\)
	\(\implies\lim_{h\rightarrow0}\left(\large{\frac{x^2+2xh+h^2-x^2}{h}}\right)\)
	\(\implies\lim_{h\rightarrow0}\left(\large{\frac{2xh+h^2}{h}}\right)\)
	\(\implies\lim_{h\rightarrow0}\left(\large{\frac{2xh}{h}+\frac{h^2}{h}}\right)\)
	\(\implies\lim_{h\rightarrow0}\left(2x+h\right)\rightarrow 2x\;\color{green}{\checkmark}\)

If you follow the same process to the functions:   \(g(x)=x^3,\;h(x)=x^4\) and \(p(x)=x^5\) the solutions would produce the derivatives of \(g'(x)=3x^2,\;h'(x)=4x^3\) and \(p'(x)=5x^4\).   Can you see any pattern to these derivatives?   These examples appear to show the exponent of the power function becoming the multiplier, and then the new exponent is one less than the original.

---

Constant Multiple Rule

Now we know how to find the derivative of \(h(x)=x^{12}\), but what happens when \(g(x)=-8x^{12}\)?   The Constant Multiple Rule addresses this problem.
For any function \(f(x)\), and any constant multiplier \(a\), the derivative is obtained by multiplying the derivate by that constant:

$$\frac{d}{dx}\left(a\cdot f(x)\right)=a\cdot\frac{d}{dx}\left(f(x)\right)$$

Example

---

General Sum & Difference Rules

The goal now is to expand the rules already discussed and apply them.   The first such application is for a polynomial.   For instance, how would we find \(\frac{d}{dw}\left(2w^3+9w^2-6w+4\right)\)?   This is a sum and difference of Power functions.   Remember that derivatives are limits, and the Limit Laws back in Chapter 2 allow us to utilize the following rules:

$$\frac{d}{dx}\left(f(x)+g(x)+h(x)\right)=\frac{d}{dx}\left(f(x)\right)+\frac{d}{dx}\left(g(x)\right)+\frac{d}{dx}\left(h(x)\right)$$

Of course, if there was subtraction at some point instead, the rule applies as well:

$$\frac{d}{dx}\left(f(x)-g(x)\right)=f'(x)-g'(x)$$

Example

---

Higher Order Derivatives

Now that we have some examples to find \(f'(x)\) from a function \(f(x)\), what happens if we try to take the derivative of a derivative? In other words:   \(\frac{d}{dx}\left(f'(x)\right)=\)??.
It seems feasible then that we have \(\left(f'(x)\right)'\).   This notation is very clunky, so we adopt the following notation:   \(\frac{d}{dx}\left(f'(x)\right)=f^{\prime\prime}(x)\).
This is called the second derivative of \(f\), and is commonly referred to as \(f\) double prime.   We can repeat this process to find the third derivative, the fourth derivate, etc.   However, we only use the prime notation for the first, second and third derivative.   For higher order derivatives, the following notation is used: $$f^{(n)}(x)=\frac{d}{dx}\left(f^{(n-1)}(x)\right)$$ In the alternative notation, we use \(\frac{d^2}{dx^2}\left(f(x)\right),\;\frac{d^3}{dx^3}\left(f(x)\right),\;\frac{d^4}{dx^4}\left(f(x)\right)\) for the second, third and fourth derivative.   In general, \(\frac{d^n}{dx^n}\left(f(x)\right)\) would represent the \(n^{th}\) order derivative.

Examples

---