Defining and Taking Derivatives | AP Calculus AB/BC Class Notes

At this point, you should be able to recognize quite a few ways to write the derivative:

Some common notations are $f'(x)$ , $\dfrac{dy}{dx}$ , $\dfrac{d}{dx}[f(x)]$ , and $y'$ .

...and so on. However, on the AP exam, you need to know more than just how to write or take the derivative of a function. In this article, I'll be going over the definitions of the derivative and some common derivatives that you'll need to know for the AP exam.

Defining the Derivative

Both definitions of the derivative start from the slope of a secant line through two nearby points on the graph. The derivative at a point is the limit of those secant slopes as the second point approaches the first point. In other words, the derivative represents the instantaneous rate of change of the function at that point, or equivalently, the slope of the tangent line there.

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The Limit Definition

This is an important definition for the AP Calculus exam. You may be asked to recognize it, use it to evaluate a derivative at a point, or connect it to the slope of a tangent line and instantaneous rate of change. It is as follows:

Text version:
$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

You may also see it written as
$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Since the slope of the secant line between two points is the change in y divided by the change in x, the expression without the limit compares the outputs of the function at two nearby x-values.
Once you add the limit to the expression, the formula becomes the definition of the derivative: it gives the exact instantaneous rate of change at a point, provided the limit exists. It is the limit of the secant-line slopes as Δx approaches 0.
The expression without the limit, such as $(f(x+h)-f(x))/h$ , is called the difference quotient. The derivative is the limit of that difference quotient as $h \to 0$ .

Quick example in text: if $f(x)=x^2$ , then $f'(x)=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h} =\lim_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h} =\lim_{h\to 0}(2x+h)=2x$

The Alternate Definition

The second definition of the derivative that you should be able to recognize is the alternate definition (I know, it's a really creative name). It looks like this:

Text version:
$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

This definition also shows the slope of a secant line, written using function values at $x$ and $c$ . By taking the limit as $x$ approaches $c$ , the two points get closer together, and the expression gives the exact derivative at $x=c$ , provided the limit exists.

Quick example in text: if $f(x)=x^2$ , then $f'(3)=\lim_{x\to 3}\frac{x^2-9}{x-3} =\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3} =\lim_{x\to 3}(x+3)=6$

You should still be able to recognize and use both of these forms.

Calculating Basic Derivatives

With taking a derivative, there are a couple of rules that you'll need to know

Power Rule

The formula for the power rule looks something like this

Text version:
$\frac{d}{dx}(x^n)=nx^{n-1}$

With this formula, 'x' is the variable, and 'n' is the exponent
First, you have to look at the exponent. It's also important to remember your exponent rules here too (anything to the power of 0 is 1, negative exponents, etc.)
Next, you need to multiply the coefficient of the variable by the exponent
The last step is to subtract one from the original exponent. This is now the new exponent for the derivative!

Let's go over a quick example then:

Text version: let $f(x)=3x^2$ .

We can see that the we have an exponent of two and a coefficient of 3.
Now that we've noted this, we have to multiply the coefficient by the exponent.

Text version: $3\cdot 2=6$

Last but not least, we have to subtract one from the exponent value

Text version: $2-1=1$ , so $f'(x)=6x$

And voila! You've taken a derivative using the power rule!

Constant and Constant Multiple Rules

Two basic derivative rules on the AP exam are: $\frac{d}{dx}(c)=0$ for any constant $c$ , and $\frac{d}{dx}[c\,f(x)]=c\,f'(x)$ . Example: $\frac{d}{dx}(7x^3)=7\cdot 3x^2=21x^2$ , and $\frac{d}{dx}(5)=0$ .

Sum and Difference Rule

This rule is pretty straight forward and is used when you're taken a derivative of a function that has multiple terms.

The rule in its formal form looks like this

Text version:
$\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)$ and $\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)$

The first one shows the sum rule while the second one shows the difference rule (That apostrophe means 'the derivative of')

Quick example in text: $\frac{d}{dx}(x^3+4x-7)=3x^2+4$

This one's a pretty easy one to pick up on so we're gonna go ahead and go into the next one! If you need additional practice with this one there'll be some links to some great resources for practice problems.

Product Rule

This rule is used when two parts of a function are being multiplied by each other (this is different from chain rule!)

Text version:
$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

This looks super confusing but the saying my teacher taught me really helped: "first times the derivative of the second plus second times the derivative of the first"
In this case, the 'first' would be the variable 'f' while the 'second' is the variable is 'g'
First, you'll need to identify your two variables, pay attention to their coefficients and exponents
Next, identify which term will be your 'first' and which will be your 'second'
Now you're going to take the 'first' and multiply it by the derivative of the second and vice versa.

This can get pretty confusing through just words so we'll go over a quick example to help see how it should be done!

Text version: let $y=x^2e^x$

This first step is to identify our terms. We have 'x^2' and 'e^x' and we can see that they are being multiplied by each other. Now, we'll assume that 'x^2' is going to be our 'first' and 'e^x' is going to be 'second'

Text version: $f(x)=x^2$ , $g(x)=e^x$

Now, we have to follow through with the formula and saying: "first times the derivative of the second plus second times the derivative of the first"

Text version: $y'=x^2(e^x)+e^x(2x)=x^2e^x+2xe^x$

Quotient Rule

Text version:
$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

Just like the product rule, this one's a little bit of a doozy and so my teacher taught us a song to help remember this (it's meant to go with the Snow White and the Seven Dwarfs HeIgh-Ho song): "You take the low, d hi, minus the high, d low, square the bottom and away we go, heigh-ho, heigh-ho!"

Quick example in text: if $y=\frac{x^2+1}{x},$ then $y'=\frac{x(2x)-(x^2+1)(1)}{x^2} =\frac{2x^2-x^2-1}{x^2} =\frac{x^2-1}{x^2}$

This rule is more about identifying when you should use it and following the formula.
With certain problems, especially since this is such a long process, you could change the denominator to a negative exponent (1/x = x^-1) and then use product rule

Chain Rule

Last but not least, we have chain rule. This is definitely the one that most people have the hardest time with so it's important that you can get the basics of it first.

Text version:
$\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)$

This would sound something like "the derivative of f of g of x equals to the f prime of g of x times g prime of x"
This is used when a function is composed of another one (kinda confusing, but it'll make sense)
It's sometimes called the "outside inside rule", since it is the derivative of the outside ignoring the inside times the derivative of the inside
Say you have a problem like this:

Text version: let $y=(3x+1)^2$

First, we need to identify our inner and outer functions. If we treat what's in the inner parentheses as one function and replace the whole with the variable 'x' for now.
What we have now is

Text version: the outside function is $u^2$ , where $u=3x+1$

We know how to take the derivative of this (power rule!) so we have the first part of this problem done
Now, we have to take the derivative of the 'inside' part of the original function

Text version: derivative of the outside is $2(3x+1)$ , and derivative of the inside is $3$

The last step is to multiply these two parts together

Text version: $y'=2(3x+1)\cdot 3=6(3x+1)=18x+6$

And that's it! These were the basic rules for taking a derivative that you'll need to know in order to get that five! If you need some additional practice, check out the resources linked below!

What Should I Memorize?

Below are several standard derivatives you should know for AP Calculus; this is not the full derivative toolkit, so also include inverse trig derivatives, inverse-function differentiation, and implicit differentiation procedures. Remember, these are listed in their most basic forms, so be prepared to see them with negative signs and coefficients in front of them and adjust your derivative accordingly. You should also know how to use these derivatives in conjunction with other functions (such as when using the product rule, quotient rule, or chain rule). Finally, you may notice that these functions seem quite familiar. This is because they are all either parents functions or trig functions to have memorized!

Know these standard derivatives:

$\dfrac{d}{dx}(x^n)=nx^{n-1}$
$\dfrac{d}{dx}(\sin x)=\cos x$
$\dfrac{d}{dx}(\cos x)=-\sin x$
$\dfrac{d}{dx}(e^x)=e^x$
$\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$
$\dfrac{d}{dx}(\tan x)=\sec^2 x$
$\dfrac{d}{dx}(\sec x)=\sec x\tan x$
$\dfrac{d}{dx}(\cot x)=-\csc^2 x$
$\dfrac{d}{dx}(\csc x)=-\csc x\cot x$

You should also be able to combine these with the product, quotient, and chain rules.

For the AP Calculus AB and BC exams, you are expected to know the standard derivative rules and the derivatives of common elementary functions, including power, trigonometric, exponential, and logarithmic functions. You may not always be directly asked to "memorize" them, but you should know them well enough to use them quickly and accurately on exam questions.

Additional AP Calculus Derivatives and Procedures

You should also know derivatives of inverse trigonometric functions that appear in AP Calculus: $\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}}$ , $\frac{d}{dx}(\arccos x)=-\frac{1}{\sqrt{1-x^2}}$ , and $\frac{d}{dx}(\arctan x)=\frac{1}{1+x^2}$ . You should also understand differentiation of inverse functions in general: if $y=f^{-1}(x)$ , then $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$ , when the derivative exists.

Other Derivative Skills You Should Know

A few more derivative ideas show up a lot and are absolutely worth knowing:

Differentiability and continuity: if a function is differentiable at a point, then it is continuous there, but a continuous function need not be differentiable.
Higher-order derivatives: things like $f''(x)$ describe rates of change of rates of change.
Derivatives of inverse functions and inverse trig functions: these are part of the derivative toolkit too, especially once you move past the basic rules.
Choosing the right procedure: make sure you can tell when to use product rule, quotient rule, chain rule, and implicit differentiation. For implicit differentiation, differentiate both sides with respect to $x$ , remembering to multiply by $y'$ when differentiating terms involving $y$ . Example: if $x^2+y^2=25$ , then $2x+2y\,y'=0$ , so $y'=-\frac{x}{y}$ ."

If you're familiar with these common derivatives, rules, and definitions, you'll have a strong start on AP Calculus differentiation—but be sure to also practice implicit differentiation, inverse and inverse trig derivatives, higher-order derivatives, and choosing efficient differentiation methods.

How do you defining and taking derivatives?