As a previous student of AP calculus,
derivatives, I know, require quite a bit of dedication to get down.
With all the different rules for deriving functions, it can be quite
difficult keeping all of them straight in your mind. This guide will
act as a resource containing all of these rules for differentiation.
This guide assumes you have a basic
understanding of the derivative and its meaning as the rate of change
of a function.
Derivative of Constants and Variables with Coefficients
Deriving a constant in calculus is easy: it's 0. Based on the fact that a horizontal line never changes, this ought to be understandable (i.e. the derivative of 5 is 0).
Deriving a variable with a coefficient is just about as easy: it's the coefficient. Based on the fact that a line has a constant rate of change, this is also quite intuitive (i.e. the derivative of 10x is 10).
Sum and Difference Rules
Differentiation of sums and differences is incredibly simple. The image below demonstrates the formula:
All that means is that if you need to find the derivative of a sum, the formula is just to derive each of the parts of the expression individually, and then add them together. For instance, the derivative of 3x + 6 is the same as the derivative of 3x plus the derivative of 6, which is 3.
Please note that deriving the individual parts of the expression does NOT work for multiplication or division!
in order to find the derivative of an expression raised to a
power, take the power down and bring it to the front of the
expression as a coefficient, and then reduce the power by one.
instance, let us say you wanted to solve for the derivative of x^3.
The answer would be 3x^2 â€“ you bring the 3 down as a coefficient,
then reduce by the power from 3 to 2.
In my opinion, the trickiest part of the power rule are negative powers. If an expression is being raised to a negative power, you still must subtract 1 from the power. Thus, the derivative of 2x^-3 is -6x^-4.
If there has been any rule that has cost me the most points on a test, it is, without a doubt, the rule for deriving products. As I stated with great emphasis in the section about addition and subtraction, you cannot derive the individual factors and then multiply them together.
Rather, you must follow the formula below:
The mnemonic device I like to use to remember the product rule is "the first times the derivative of the second plus the second times the derivative of the first." The key thing is simply to remember not to take the product apart and try to use a tactic similar to deriving sums: it does not work.
is an example: To find the derivative with respect to x of 3ab, you
would take the first (3a) times the derivative of the second (db/dx),
plus the second (b) times the derivative of the first (3 * da/dx), to
get 3a * (db/dx) + 3b * (da/dx).
However, you may run into instances where deriving a product of three is necessary. This task is not much different from the basic two-factor product rule:
for any number of factors in a product you must find the derivative
of, multiply all but one of the factors together, and multiply by the
derivative of the remaining one, then add all possible combinations
to each other.
This one looks a bit ickier, but a little catchy phrase can help immensely with getting this one down pat: low d high minus high d low over low low. Try saying it a few times, and then let's look at how it breaks down through an example. Say you want to find the derivative of 3/x with respect to x.
Low d high, or, x * d/dx(3) minus high d low, or (3 * d/dx(x)), over low low (x^2).
Put it all together and you get (0 â€“ 3)/x^2, or -3/x^2.
As a sidenote, this particular problem could have been solved with the power rule, as well, by rewriting it as 3x^-1. See if you can get the same result.
The tricky thing about the quotient rule is to remember to distribute the negative. Right when you plop that minus sign down on your paper, place an opening parenthesis so you do not forget.
Chain Rule for DerivativesIt's time I address the beast of all rules: the chain rule. Every differentiation you perform must obey the chain rule, which addresses deriving a composition of functions. The formula for this is below:
Basically, the chain rule states to derive each "layer" of a composition and multiply all of the derivatives together. The chain rule is much more commonly seen when transcendental expressions (logarithms, exponents, trigonometric functions and their inverses) are involved. Some examples of the chain rule in action will be shown as we go along.
For logarithms, there is, thankfully, no temptation to separate anything, like there is in multiplication and division. However, you must remember the following formula:
Put into English, the derivative of a logarithm is 1 over the product of the thing you are log-ing times the natural log of the base of the logarithm. Finally, due to the chain rule, you must multiply by the derivative of what you are log-ing, as it is the next "layer" of this expression. For example, the derivative of log6 2x is (1/(2x * ln 6)) * 2, which simplifies to 2/(2x * ln 6).
It should be noted that, for natural logarithms, the formula is simplified because a is constant e, whose natural logarithm is simply 1. The simplified formula for the derivative of a natural logarithm is:
example, the derivative of ln 5x is just (1/5x) * 5, which simplifies
to just 1/x.
I like to remember the rule for logarithms by thinking of the bone "ulna."
Exponential RuleThe formula for solving the derivative of exponential expressions is quite similar to that of solving logarithmic expressions. Since the two go hand-in-hand, "ulna" will assist you in remembering the exponential rule, as well. It is:
Essentially, take the expression and multiply it by the natural log of what is being powered. For instance, the derivative of 9x is 9x * ln 9.
Once again, constant e is special and makes the process simpler. If your expression is eu, the derivative is just eu*(du/dx). y = ex is a rare instance in calculus where the rate at which the function increases is the function itself.
Trigonometric FunctionsThe formulas for all trigonometric functions' derivatives are in the below image. They require no explanation in and of themselves:
Also, bear in mind how to handle powered trigonometric functions. sin2x means sin(x)2, and you would thus employ the chain rule for calculus derivatives by taking the derivative of the outside layer (which is 2 sin(x) by the power rule) times the derivative of the inside layer (cos(x)), to get the answer: 2 sin(x) cos(x).
Inverse Trigonometric Functions
The last derivative formulas we are going to cover are the ones for inverse trigonometric functions. Although there are six inverse trigonometric functions, there are essentially only three formulas you must remember. Arccosine's, arccotangent's, and arccosecant's derivatives are just the negative derivatives of arcsine, arctangent, and arcsecant, respectively. These get tricky because of their similarity:
Arctangent can be remembered as the one that has addition instead of subtraction; however, arcsine's and arcsecant's order is what trips me up. Try to remember that the absolute value of u and the u^2 term are next to each other (in arcsecant).
Let's do an example. The derivative of arccosine(2x) would be deduced by following these steps:
With the tips provided above, along with some good study time with the above calculus formulas, you should be on your way to success at calculus, and all of these differentiation tasks will become second nature to you. Good luck with your studies!