Differentiation Rules

The process of differentiation is tedious for complicated functions. Therefore, rules for differentiating general functions have been developed, and can be proved with a little effort. Once sufficient rules have been proved, it will be fairly easy to differentiate a wide variety of functions. Some of the simplest rules involve the derivative of linear functions.

Derivative of a constant function

For any fixed real number $c$ ,

{\frac {d}{dx}}[c]=0

Intuition

The graph of the function $f(x)=c$ is a horizontal line, which has a constant slope of 0. Therefore, it should be expected that the derivative of this function is zero, regardless of the values of $x$ and $c$ .

Proof

The definition of a derivative is

\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}

Let $f(x)=c$ for all $x$ . (That is, $f$ is a constant function.) Then $f(x+\Delta x)=c$ . Therefore

{\frac {d}{dx}}[c]=\lim _{\Delta x\to 0}{\frac {c-c}{\Delta x}}=\lim _{\Delta x\to 0}{\frac {0}{\Delta x}}

Let $g(\Delta x)={\frac {0}{\Delta x}}$ . To prove that $\lim _{\Delta x\to 0}g(\Delta x)=0$ , we need to find a positive $\delta$ such that, for any given positive $\varepsilon$ , ${\Big |}g(\Delta x)-0{\Big |}<\varepsilon$ whenever $0<|\Delta x-0|<\delta$ . But ${\Big |}g(\Delta x)-0{\Big |}=0$ , so ${\Big |}g(\Delta x)-0{\Big |}<\varepsilon$ for any choice of $\delta$ .

Examples

${\frac {d}{dx}}[3]=0$
${\frac {d}{dx}}[z]=0$

Note that, in the second example, $z$ is just a constant.

Derivative of a linear function

For any fixed real numbers $m$ and $c$ ,

{\frac {d}{dx}}[mx+c]=m

The special case ${\frac {dx}{dx}}=1$ shows the advantage of the ${\frac {d}{dx}}$ notation—rules are intuitive by basic algebra, though this does not constitute a proof, and can lead to misconceptions to what exactly $dx$ and $dy$ actually are.

Intuition

The graph of $y=mx+c$ is a line with constant slope $m$ .

Proof

If $f(x)=mx+c$ , then $f(x+\Delta x)=m(x+\Delta x)+c$ . So,

$f'(x)$	$=\lim _{\Delta x\to 0}{\frac {m(x+\Delta x)+c-mx-c}{\Delta x}}$
	$=\lim _{\Delta x\to 0}{\frac {m(x+\Delta x)-mx}{\Delta x}}$
	$=\lim _{\Delta x\to 0}{\frac {mx+m\Delta x-mx}{\Delta x}}$
	$=\lim _{\Delta x\to 0}{\frac {m\Delta x}{\Delta x}}$
	$=m$

Constant multiple and addition rules

Since we already know the rules for some very basic functions, we would like to be able to take the derivative of more complex functions by breaking them up into simpler functions. Two tools that let us do this are the constant multiple rule and the addition rule.

The Constant Rule

For any fixed real number $c$ ,

{\frac {d}{dx}}{\big [}c\cdot f(x){\big ]}=c\cdot {\frac {d}{dx}}{\big [}f(x){\big ]}

The reason, of course, is that one can factor $c$ out of the numerator, and then of the entire limit, in the definition. The details are left as an exercise.

Example

We already know that

{\frac {d}{dx}}{\big [}x^{2}{\big ]}=2x

Suppose we want to find the derivative of $3x^{2}$

${\frac {d}{dx}}{\big [}3x^{2}{\big ]}$	$=3{\frac {d}{dx}}{\big [}x^{2}{\big ]}$
	$=3\cdot 2x$
	$=6x$

Another simple rule for breaking up functions is the addition rule.

The Addition and Subtraction Rules

{\frac {d}{dx}}{\big [}f(x)\pm g(x){\big ]}={\frac {d}{dx}}{\big [}f(x){\big ]}\pm {\frac {d}{dx}}{\big [}g(x){\big ]}

Proof

From the definition:

$\lim _{\Delta x\to 0}\left[{\frac {{\big [}f(x+\Delta x)\pm g(x+\Delta x){\big ]}-{\big [}f(x)\pm g(x){\big ]}}{\Delta x}}\right]$	$=\lim _{\Delta x\to 0}\left[{\frac {{\big [}f(x+\Delta x)-f(x){\big ]}\pm {\big [}g(x+\Delta x)-g(x){\big ]}}{\Delta x}}\right]$
	$=\lim _{\Delta x\to 0}\left[{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\right]\pm \lim _{\Delta x\to 0}\left[{\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right]$

By definition then, this last term is ${\frac {d}{dx}}\left[f(x)\right]\pm {\frac {d}{dx}}\left[g(x)\right]$

Example

What is the derivative of $3x^{2}+5x$ ?

${\frac {d}{dx}}{\big [}3x^{2}+5x{\big ]}$	$={\frac {d}{dx}}{\big [}3x^{2}+5x{\big ]}$
	$={\frac {d}{dx}}{\big [}3x^{2}{\big ]}+{\frac {d}{dx}}{\big [}5x{\big ]}$
	$=6x+{\frac {d}{dx}}{\big [}5x{\big ]}$
	$=6x+5$

The fact that both of these rules work is extremely significant mathematically because it means that differentiation is linear. You can take an equation, break it up into terms, figure out the derivative individually and build the answer back up, and nothing odd will happen.

We now need only one more piece of information before we can take the derivatives of any polynomial.

The Power Rule

{\frac {d}{dx}}{\big [}x^{n}{\big ]}=nx^{n-1}

For example, in the case of $x^{2}$ the derivative is $2x^{1}=2x$ as was established earlier. A special case of this rule is that ${\frac {dx}{dx}}={\frac {dx^{1}}{dx}}=1x^{0}=1$ .

Since polynomials are sums of monomials, using this rule and the addition rule lets you differentiate any polynomial. A relatively simple proof for this can be derived from the binomial expansion theorem.

This rule also applies to fractional and negative powers. Therefore

${\frac {d}{dx}}{\big [}{\sqrt {x}}{\big ]}$	$={\frac {d}{dx}}{\big [}x^{\frac {1}{2}}{\big ]}$
	$={\frac {1}{2}}x^{-{\frac {1}{2}}}$
	$={\frac {1}{2{\sqrt {x}}}}$

Derivatives of polynomials

With these rules in hand, you can now find the derivative of any polynomial you come across. Rather than write the general formula, let's go step by step through the process.