Calculating Derivatives

The following is a short summary of the key rules for calculating derivatives. Keep in mind that knowing how to calculate derivatives is useless unless you know what they are and how to use them.

There are two types of formulas for calculating derivatives, which we may classify as

formulas for calculating the derivatives of elementary functions
and
structural type formulas.

Formulas for Derivatives of Elementary Functions

Power Rule: d/du (uⁿ) = nu^n-1
d/du (ln u) = 1/u
d/du (e^u) = e^u
d/du (sin u) = cos u
d/du (cos u) = - sin u
d/du (tan u) = sec ² u

Note that these formulas may be used only to calculate derivatives of functions in precisely the form indicated. The name of the independent variable is irrelevant--it is the form that is important. For example, the power rule may be used to calculate that d/dx (x⁵) = 5x⁴ or that d/du (u⁵) = 5u⁴, but gives us no help at all with d/du (x⁵).

All other formulas, the structural type formulas, reduce derivatives of more complicated functions until we merely need to apply the above formulas a number of times.

Structural Type Formulas

These formulas may be divided into two groups; one group is used almost without thought, while the other group needs to be carefully memorized.

In the following, assume that x, y, u, v and t are variables, while c and k are constants.

The first group of formulas, which is used almost without thought, may be expressed as:

The derivative of a constant times a function equals the contant times the derivative of the function.
The derivative of a sum equals the sum of the derivatives.
The derivative of a difference equals the difference of the derivatives.

Symbolically, we write these rules as:

d/dx (cu) = c du/dx
d/dx (u+v) = du/dx + dv/du
d/dx (u-v) = du/dx - dv/du.

When we apply these rules, we say that we are differentiating "term by term." Using these rules along with the power rule, it is very easy to differentiate any polynomial. Some special cases such as the following come up so often that we tend to take them for granted:

d/du (c) = 0
d/du (cu) = c
d/du (au+b) = a

The last three rules are somewhat more difficult. They are called the product rule, the quotient rule and the chain rule. Of these, the product and quotient rules can be used routinely, since it is easy to recognize when you have a product or quotient, but it is more difficult and takes more practice to use the chain rule correctly.

The product rule may be thought of as the derivative of a product equals the first factor times the derivative of the second plus the second factor times the derivative of the first.

The quotient rule may be thought of as the derivative of a quotient equals the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

Symbolically, we express these rules as follows:

Product Rule: d/dx (uv) = u dv/dx + v du/dx
Quotient Rule: d/dx (u/v) = ( v du/dx - u dv/dx)/v²

However, we should remember the rules in words rather than trying to memorize jumbles of symbols.

The chain rule, as mentioned before, is a little trickier to use. Fortunately, its formula is easier to remember than some of the others.

Chain Rule: dy/dx = (dy/du)· (du/dx)

As usual, remember the meaning, especially since the symbols x,y and u are not necessarily the ones that will be used in the function you will be differentiating.

The chain rule is used for calculating the derivatives of composite functions. The easiest way to recognize that you are dealing with a composite function is by the process of elimination: if none of the other rules apply, then you have a composite function. This makes it very important to be totally familiar with the other rules for differentiation. The difficult part of using the chain rule is recognizing that you need it and then doing the algebra necessary to decompose the function. We have looked at two ways of doing that in class. Study them and, if you have difficulty, see your instructor. Once you have decomposed the function, so that you have written down something in the form

y = f(u)
u = g(x),

it is easy to actually carry out the rest of the chain rule. For example, given the function y=√(x² + 5) you would write something like

y = u^1/2
u = x² + 5

and then, since d/du(u^1/2) = (1/2)u^-1/2 and d/dx(x² + 5) = 2x, we get

dy/dx = (dy/du) · (du/dx) = (1/2)u^-1/2 · 2x = x/√(x² + 5).

Summary and Overall Strategy

The following is a general approach to calculating derivatives which, if followed, will enable you to routinely calculate every derivative we run across in the class. It depends on your being able to recognize individual terms, constant factors, products and quotients and also being able to rewrite functions in a more useful style, such as being able to recognize that something like x/50 can be rewritten as (1/50) ·x or that √x can be rewritten as x^1/2.

The Strategy

Differentiate term by term.: Deal with each term separately and, for each term, recognize any constant factor.
For each term, recognize whether it is one of the elementary functions - power, trigonometric, exponential or natural logarithm of the independent variable.: If it is, you can easily apply the appropriate formula and you will be done. If it's not, continue.
Decide whether the term is a product or a quotient.: If it is, use the appropriate formula. Note that the appropriate formula will have you calculating two other derivatives and you will have to go back to the beginning to deal with those.
If the term isn't a product or a quotient, you will need to use the Chain Rule. Try to use it correctly.

The Bottom Line

The process of differentiation is, for most of the functions encountered in a lower division calculus class, a very mechanical process. With sufficient practice, reliably calculating derivatives should and will take no more mental effort than doing ordinary arithmetic. It's important to reach that level of expertise so that you can expend your mental energy on more significant matters.

Decomposing Functions

If you have difficulty decomposing a composite function into components, consider the following two methods which, together, work on almost all composite functions ordinarily encountered in elementary calculus. Shown below is how each method works on the same function, y = sin(x²).

The Black Box Method

In this method, think of a function as a black box which takes an input, the value of the independent variable, and produces an output, the value of the function corresponding to that value of the independent variable.

To use this method, try to break up the black box into two separate black boxes, the second of which takes the output of the first as its input. This can generally be done by carefully considering how one would calculate the function for a particular value of the independent variable.

Consider how one would evaluate y = sin(x²) for a particular value of x, say for x = 3. Most people would say that the value would be sin(9), and might then take out a calculator (a prototypical black box) to find a decimal approximation to sin(9).

If one thinks about what was actually done, one took the input, 3, and came up with an output value of 9 by squaring 3. One then takes 9 as input and comes up with the output of sin(9).

One can see that, in general, one would calculate y by taking x, squaring it to get x², and then taking x² and calculating its sine. Symbolically,

x --> x² --> sin(x²) = y.

If one represents the initial output, x², by the symbol u, one can write

x --> x² = u --> sin(u) = y.

By reading right to left, one comes up with the decomposition

y = sin(u),
u = x².

The Walt Disney Method

The Walt Disney Method is based upon the idea of making one's wish come true. Often one will look at a composite function and, at least subliminally, wish it were something similar but somewhat simpler. With this method, one expediently but mathematically legally makes that wish come true - at least partially.

Consider again the function y = sin(x²). One might look at that and wish that it was the simpler function y = sin(x). One makes that wish almost come true by writing it as y = sin(u), which one can legally do by letting u = x².

Looking at that again, one has come up with the decomposition

y = sin(u),
u = x²,

exactly the same one generated through the use of the Black Box Method.

The Walt Disney Method makes use of the general principles that one can always assign a symbol to represent anything on wants, provided the symbol isn't already in use, and one can always replace any expression with anything else equal to it, the substitution principle.

In general, one implements the Walt Disney Method by looking at the form of a function y = f(x) and wistfully noticing that it would look like a much simpler function, say g(x), which would be much easier to differentiate were some part of it just x. It's also important that the independent variable x not appear outside of that part. One calls that part u and has the decomposition

y = g(u),
u = h(x),

where h(x) is the part that one wished was just x.

Last modified by Alan H. Stein.