Differentiation Rules

While it's possible to compute derivatives using the definition of the derivative as a limit, it is too much work if you're differentiating anything complicated. It's more efficient to develop a set of differentiation rules to use as shortcuts.

Several remarks are in order. First, it is silly to memorize the rules. You should learn them by doing examples.

Second, differentiation is a skill --- it is something you learn by practicing. You can't learn to differentiate by watching the instructor do examples. Since you will need to compute derivatives in virtually all the calculus courses you take, it is worth learning to do well now to avoid grief in the future.

On the other hand, you should keep things in perspective. Calculus is not just a matter of computing derivatives and integrals. The computations acquire their meaning when they are used to do things: Compute rates of change, find maxima and minima, determine areas or volumes, and so on.

The fact that calculators can do arithmetic does not mean that people don't need to learn arithmetic. A certain amount of computation is necessary in order to develop an intuition for numbers.

In the same way, the fact that computer programs like Mathematica or Derive can compute derivatives and integrals does not mean that people don't need to learn to differentiate or integrate. I don't think you can have a complete understanding of what a derivative is if you've never computed a few derivatives. However, at some point you may want to use a computer to do the ugly computations, in the same way that you use a calculator to do arithmetic.

I'll illustrate the differentiation rules with some examples. In these examples, I'll write

$$\der {} x (\hbox{junk})$$

for the derivative of $\hbox{junk}$ with respect to x.

  1. The derivative of a constant is 0.

This makes sense, because a constant doesn't change, and the derivative gives the rate of change.


Example.

$$\der {} x 17^{-3} = 0.$$

$$\der {} x \pi = 0.$$

$$\der {} t \left[(\sin t)^2 + (\cos t)^2\right] = \der {} t 1 = 0.\quad\halmos$$


  1. ( The Power Rule) $\displaystyle \der {} x x^n = nx^{n-1}$ , for $n \ne 0$ .

This rule is called the Power Rule, because it tells you how to differentiate a power of the variable. In words, you "bring the power down" and decrease the power by 1.


Example.

$$\der {} x x^{217} = 217 x^{216}.$$

$$\der {} x \sqrt{x} = \der {} x x^{1/2} = \dfrac{1}{2} x^{-1/2} = \dfrac{1}{2\sqrt{x}}.$$

$$\der {} x \dfrac{1}{x^7} = \der {} x x^{-7} = -7x^{-8} = \dfrac{-7}{x^8}.$$

Notice that n can be any nonzero real number:

$$\der {} x x^{\sqrt{2}} = \sqrt{2} x^{\sqrt{2}-1}.\quad\halmos$$


  1. $\displaystyle \der {} x (c\cdot
   \hbox{junk}) = c\cdot \der {} x (\hbox{junk})$ .

This rule says that constants may be "pulled out" when you differentiate. Another way to think of this is that the constant just "sits there", while the thing it multiplies gets differentiated.

Don't confuse this rule with the first rule --- the derivative of a constant is 0. In that situation, the constant was "by itself". In this situation, the constant multiplies something.


Example. The second derivative of f is the derivative of $\der f x$ with respect to x. It's denoted $\dfrac{d^2f}{dx^2}$ . Thus,

$$\dfrac{d^2f}{dx^2} = \der {} x \left(\der f x\right).$$

Likewise, the third derivative is the derivative of the second derivative, and so on:

$$\dfrac{d^3f}{dx^3} = \der {} x \left(\dfrac{d^2f}{dx^2}\right).$$

For example, suppose $f(x) = x^{100}$ . Here are the first few derivatives:

$$\der f x = 100 x^{99}, \quad \dfrac{d^2f}{dx^2} = 99\cdot 100 x^{98}, \quad \dfrac{d^3f}{dx^3} = 98\cdot 99\cdot 100 x^{97}.\quad\halmos$$


Example. A function $f(x)$ has a vertical tangent at a if f is continuous at a and

$$\lim_{x\to a} |f'(x)| = +\infty.$$

Geometrically, this means that the graph has a vertical tangent line at a.

For example, consider $f(x) =
   x^{1/5}$ . Then $f'(x) = \dfrac{1}{5}x^{-4/5}$ , so

$$\lim_{x\to 0} |f'(x)| = \lim_{x\to 0} \dfrac{1}{5}x^{-4/5} = \dfrac{1}{5} \lim_{x\to 0} \dfrac{1}{x^{4/5}} = +\infty.$$

The graph has a vertical tangent at $x
   = 0$ . Picture:

$$\hbox{\epsfysize=2.5in \epsffile{diffrul1.eps}}\quad\halmos$$


Example. A function $f(x)$ has a cusp at a if f is continuous at a and:

For example, consider $f(x) =
   x^{2/5}$ . Then $f'(x) = \dfrac{2}{5} x^{-3/5}$ , so

$$\lim_{x\to 0^+} f'(x) = +\infty \quad\hbox{while}\quad \lim_{x\to 0^-} f'(x) = -\infty.$$

Thus, $f(x)$ has a cusp at $x = 0$ . Picture:

$$\hbox{\epsfysize=2.5in \epsffile{diffrul2.eps}}\quad\halmos$$


Remark. If the power n is a positive integer, it's pretty easy to derive the Power Rule from the binomial formula. Let $f(x) = x^n$ . Then

$$f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \dfrac{(x + h)^n - x^n}{h}.$$

First, if $n = 1$ , this comes out to

$$\lim_{h \to 0} \dfrac{(x + h) - x}{h} = \lim_{h \to 0} \dfrac{h}{h} = 1,$$

which is indeed $1\cdot x^0$ . So assume that $n \ge 2$ .

Recall that $n!$ ( n factorial) is the product of the integers from 1 to n:

$$n! = 1\cdot 2\cdot \cdots \cdot n.$$

The binomial formula says

$$(x + h)^n = x^n + \dfrac{n!}{1!(n - 1)!} x^{n-1} h + \dfrac{n!}{2!(n - 2)!} x^{n-2} h^2 + \dfrac{n!}{3!(n - 3)!} x^{n-3} h^3 + \cdots + \dfrac{n!}{(n - 1)!1!} x^1 h^{n-1} + h^n.$$

Therefore,

$$(x + h)^n - x^n = \dfrac{n!}{1!(n - 1)!} x^{n-1} h + \dfrac{n!}{2!(n - 2)!} x^{n-2} h^2 + \dfrac{n!}{3!(n - 3)!} x^{n-3} h^3 + \cdots + \dfrac{n!}{(n - 1)!1!} x^1 h^{n-1} + h^n.$$

Next, divide by h:

$$\dfrac{(x + h)^n - x^n}{h} = \dfrac{n!}{1!(n - 1)!} x^{n-1} + \dfrac{n!}{2!(n - 2)!} x^{n-2} h + \dfrac{n!}{3!(n - 3)!} x^{n-3} h^2 + \cdots + \dfrac{n!}{(n - 1)!1!} x^1 h^{n-2} + h^{n-1}.$$

As $h \to 0$ , all the terms but the first go to 0, because they contain positive powers of h. Therefore,

$$f'(x) = \dfrac{n!}{1!(n - 1)!} x^{n-1} = nx^{n-1},$$

because

$$\dfrac{n!}{1!(n - 1)!} = \dfrac{1\cdot 2\cdot \cdots \cdot (n - 1)\cdot n} {1\cdot 2\cdot \cdots \cdot (n - 1)} = n.$$

If n is not a positive integer, the derivation gets a little trickier. For one thing, it is not obvious what $x^n$ means if n is, say, $\sqrt{2}$ . Nevertheless, the technicalities can be resolved, and the formula does work for all nonzero real powers.


Example.

$$\der {} x 7x^{15} = 7\cdot \der {} x x^{15} = 7\cdot (15x^{14}) = 135x^{14}.$$

I'll discuss trig functions later on, and I'll show that

$$\der {} x \sin x = \cos x.$$

Hence,

$$\der {} 7 \sin x = 7 \cos x.\quad\halmos$$


  1. The derivative of a sum is the sum of the derivatives:

$$\der {} x \left(\hbox{blip} + \hbox{blap}\right) = \der {} x \hbox{blip} + \der {} x \hbox{blap}.$$


Example.

$$\der {} x \left(9\sqrt{x} - \dfrac{14}{x^{3/2}}\right) = \der {} x \left(9 x^{1/2} - 14x^{-3/2}\right) = \dfrac{9}{2} x^{-1/2} + 21 x^{-5/2} = \dfrac{9}{2} \cdot \dfrac{1}{\sqrt{x}} + \dfrac{21}{x^{5/2}}. \quad\halmos$$


  1. ( The Product Rule)

$$\der {} x (\hbox{hi}\cdot \hbox{ho}) = \hbox{hi}\cdot \der {} x (\hbox{ho}) + \der {} x (\hbox{hi})\cdot \hbox{ho}.$$


Example.

$$\der {} x (x^3 + 2x^2 + 7)(x^4 - x^3 - x + 2) = (3x^2 + 4x)(x^4 - x^3 - x + 2) + (x^3 + 2x^2 + 7)(4x^3 - 3x^2 - 1).\quad\halmos$$


The following picture shows why the Product Rule works.

$$\hbox{\epsfysize=2.5in \epsffile{diffrul3.eps}}$$

If I change $\hbox{hi}$ by a small amount $d(\hbox{hi})$ and $\hbox{ho}$ by a small amount $d(\hbox{ho})$ the change in the product $\hbox{hi}\cdot
   \hbox{ho}$ is

$$d(\hbox{hi}\cdot \hbox{ho}) = \hbox{hi}\cdot d(\hbox{ho}) + \hbox{ho}\cdot d(\hbox{hi}) + d(\hbox{hi})\cdot d(\hbox{ho}).$$

However, the last term is a product of two small numbers, which is {\it very} small compared to the other two terms. I may neglect it, and I obtain

$$d(\hbox{hi}\cdot \hbox{ho}) \approx \hbox{hi}\cdot d(\hbox{ho}) + \hbox{ho}\cdot d(\hbox{hi}).$$

Now divide by $dx$ to pass to rates of change.

Here's how the proof looks if I use limits.

$$\der {} x \left[f(x) g(x)\right] = \lim_{h\to 0} \dfrac{f(x + h)g(x + h) - f(x)g(x)}{h} =$$

$$\lim_{h\to 0} \dfrac{f(x + h)g(x + h) - f(x)g(x + h) + f(x)g(x + h) - f(x)g(x)}{h} =$$

$$\lim_{h\to 0} \dfrac{f(x + h)g(x + h) - f(x)g(x + h)}{h} + \lim_{h\to 0} \dfrac{f(x)g(x + h) - f(x)g(x)}{h} =$$

$$\lim_{h\to 0} \dfrac{f(x + h) - f(x)}{h}\cdot g(x + h) + \lim_{h\to 0} f(x)\cdot \dfrac{g(x + h) - g(x)}{h} =$$

$$\left(\der {} x f(x)\right)\cdot g(x) + f(x)\cdot \left(\der {} x g(x)\right).$$

  1. ( The Quotient Rule)

$$\der {} x \dfrac{\hbox{hi}}{\hbox{ho}} = \dfrac{\hbox{ho}\cdot \der {} x (\hbox{hi}) - \hbox{hi}\cdot \der {} x (\hbox{ho})}{\hbox{ho ho}}.$$

(That's $\hbox{ho ho } = \hbox{ho}^2$ on the bottom!)


Example.

$$\der {} x \dfrac{x^2 + 1}{x^2 - 3x - 1} = \dfrac{(x^2 - 3x - 1)(2x) - (x^2 + 1)(2x - 3)} {(x^2 - 3x - 1)^2}.$$

Note, however, that if either the top or the bottom is just a {\it number}, it is better not to use the Quotient Rule. If the number is on the bottom, you can divide:

$$\der {} x \dfrac{x^2 - 4x + 6}{2} = \der {} x \left(\dfrac{1}{2}x^2 - 2x + 3\right) = x - 2.$$

If the number is on the top, you can use

$$\der {} x \dfrac{\hbox{number}}{\hbox{junk}} = -\dfrac{\hbox{number}\cdot \der {} x \hbox{junk}}{\hbox{junk}^2}.$$

You shouldn't memorize this formula; it's a special case of the Chain Rule, which I'll talk about later. Here's an example:

$$\der {} x \dfrac{5}{x^2 + 3} = -\dfrac{5\cdot (2x)}{(x^2 + 3)^2}. \quad\halmos$$


I obtained the derivative by trying to find the slope of the tangent line to a curve. Here are some examples which apply the differentiation rules to problems on tangent lines.


Example. Find the equation of the tangent lines to $y = x^2$ which pass through the point $(2,0)$ .

$$\hbox{\epsfysize=2.5in \epsffile{diffrul4.eps}}$$

The x-axis is clearly one such tangent line. Its equation is $y = 0$ .

To find the other tangent, let $(c,c^2)$ be the point of tangency. The slope of the tangent line is $y' = 2x$ ; when $x = c$ , $y' = 2c$ .

On the other hand, the line passes through $(2,0)$ and $(c,c^2)$ , so its slope is $\dfrac{c^2 - 0}{c - 2}$ . equate the two expressions for the slope and solve for c:

$$2c = \dfrac{c^2}{c - 2}, \quad 2c^2 - 4c = c^2, \quad c^2 - 4c = 0, \quad c = 0 \quad\hbox{or}\quad c = 4.$$

$c = 0$ gives the x-axis, which I know about already. $c = 4$ gives $y' = 8$ ; since the line passes through $(2,0)$ , its equation is

$$y = 8(x - 2).\quad\halmos$$


Example. At what point(s) does the graph of $y = \dfrac{x}{(x - 1)^2}$ have a horizontal tangent?

A horizontal line has slope 0. Since the derivative gives the slope of the tangent, I'll find the derivative and set it equal to 0:

$$y' = \dfrac{(x - 1)^2(1) - (x)(2)(x - 1)}{(x - 1)^4} = \dfrac{-(x + 1)}{(x - 1)^3}.$$

Clearly, $y' = 0$ when $x = -1$ . $y(-1) = -\dfrac{1}{4}$ , so there is a horizontal tangent at the point$\left(-1,-\dfrac{1}{4}\right)$ .


Example. Find the equation of the line which is {\it perpendicular} to the graph of $y = x^2 - 3x - 4$ at the point $(3,-4)$ .

To be perpendicular to the graph is the same as being perpendicular to the tangent line. The slope of the tangent is given by the derivative, which is $y' = 2x - 3$ . Hence, at $x = 3$ , the tangent has slope $y'(3) = 3$ .

Perpendicular line have slopes which are negative reciprocals of each other. Therefore, the slope of the perpendicular line is $-\dfrac{1}{3}$ . Since the line passes through the point $(3,-4)$ , its equation is

$$-\dfrac{1}{3}(x - 3) = y + 4, \quad\hbox{or}\quad -\dfrac{1}{3}x - 3 = y.\quad\halmos$$


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