AP Calculus AB Ultimate Guide
Unit 1 - Limits and Continuity
A limit is the intended height of a graph, as x gets very close to a value. This means that when we follow a function, the limit will show what the height that the point is supposed to be, even if it doesn’t concur with the actual function
\lim_{x\to n}f\left(x\right)
Take this limit for example →\lim_{x\to2}\frac{x^2+5x+6}{x+2}
We can find this limit by plugging in x to the function → 4 + 10 + 6 / 4 → 5
Solving Limits
When you substitute the number and get \frac00, there are a few different ways you can solve it.
Factoring and eliminating the hole →\lim_{x\to2}\frac{t^2-3t+2}{t^2-4}=\frac00
The first option is factor the function so that we can eliminate the hole → \lim_{t\to2}\frac{\left(t-2\right)\left(t-1\right)}{\left(t+2\right)\left(t-2\right)}
Then, once you eliminate the hole, plug the value in again→\lim_{t\to2}\frac{t-1}{t+2}=\frac{2-1}{2+2}=\frac14
Making 1 fraction → \lim_{x\to0}\frac{\frac{1}{2+x}-\frac12}{x}=\frac00
First, to get the numerator sorted out, we need to find a common denominator for the two fractions in the numerator →\frac{1}{\left(2+x\right)}-\frac12\Rightarrow\frac{2}{2\left(2+x\right)}-\frac{2\left(2+x\right)}{2\left(2+x\right)}\Rightarrow\frac{2-2-x}{2\left(2+x\right)}\Rightarrow-\frac{x}{2\left(2+x\right)}
Now, we can simplify the fraction to get one fraction → -\frac{x}{2\left(2+x\right)}\cdot\frac{1}{x}\Rightarrow-\frac{1}{2\left(2+x\right)}\Rightarrow\lim_{x\to0}-\frac{1}{2\left(2+0\right)}=-\frac14
Multiplying by conjugate →\lim_{x\to0}\frac{x}{\sqrt{16+x}-4}=\frac00
In this situation, it would be best to try multiplying by the conjugate →\frac{x\left(\sqrt{16+x}+4\right)}{\left(\sqrt{16+x}-4\right)\left(\sqrt{16+x}+4\right)}
When we multiply everything out, we know that we only have to do the front terms and the back terms so we get →\frac{x\left(\sqrt{16+x}+4\right)}{16+x-16}=\frac{x\left(\sqrt{16+x}+4\right)}{x}=\sqrt{16+x}+4=\sqrt{16+0}+4=8
Here are some rules for limits as well! You can use these to help assist you in solving limits:
It is also helpful to remember → \lim_{x\to0}\frac{\sin x}{x}=1
x can be any number or symbol, but when it is the same on the bottom and top, it is equal to 1
I’ve also seen this limit rule →\lim_{x\to0}\frac{1-\cos x}{x}=0
Finding Limits Graphically / Numerically
Another way to find limits is by looking at a graph.
For a limit to exist, the function must approach the same height value from both sides
In this example, the limit does exist even though there is a hole because the function approaches the same height from both sides
When the function approaches two different heights, the function might have an actual value, but the limit does not exist (mean girls reference!)
Here is an example, where the f(2) = 4, but the limit DNE
Limits approaching from the left side will have a - sign above the number (like 2-) and limits approaching from the right has a + sign above the number (like 2+)
We can also find limits from the table
Don’t pay attention to the number it is approaching, as both sides may be approaching something different. In this example we can see that → \lim_{x\to3^{-}}y_1=5 and \lim_{x\to3^{+}}y_1=15
Because both limits approach something different, the limit does not exist → \lim_{x\to3}y_1=DNE
Squeeze Theorem
if g(x) <= f(x) <= h(x), while x does not equal c, in some interval about c, and \lim_{x\to c}g\left(x\right)=\lim_{x\to c}h\left(x\right)=L , then \lim_{x\to c}f\left(x\right)=L
Basically, the function f(x) is in between two functions (g(x) <= f(x) <= h(x)). By taking the limit of each function, and solving it, we can get the value of f(x).
When g(x) and h(x) are different, the limit of f(x) cannot be determined
Vertical Asymptotes
A vertical asymptote is defined as → the line x = a is a vertical asymptote of the graph of a function f(x) if either \lim_{x\to a^{\pm}}f\left(x\right)=\pm\infty
If you get confused with HA, remember that vertical asymptote limits will be approaching a NUMBER and result in ±infinity
You can find them by looking at a graph or looking at a table. The table will have values that get really big (in either pos or neg direction) really fast
You can find them without a calculator however
When direct substitution gives you \frac{n}{0}, you know it’s a vertical asymptote
The only possible one-sided limits are +∞ or -∞
Pick a value just to the left or right of your x on the side you are approaching from, and consider whether your height is positive or negative. This will give you the sign of the limit
Example →\lim_{x\to2^{-}}\frac{x^2+9}{x-2}\Rightarrow\lim_{x\to1.99}\frac{\left(1.99\right)^2+9}{1.99-2}\Rightarrow\frac{+}{-}\Rightarrow\lim_{x\to2^{-}}f\left(x\right)=-\infty
Horizontal Asymptotes
A horizontal asymptote is defined as → the line y = b is a horizontal asymptote of a function f(x) if either \lim_{x\to\pm\infty}f\left(x\right)=b
If you get confused with VA, remember that hortizontal asymptote limits will be approaching infinity and give 0/NA/#
It is also helpful to remember BOBO BOTNA EATSDC *remember from precalc?* (Bigger On Bottom = 0; Bigger On Top = No Asymptote; Exponents Are The Same = Divide Coefficients)
You can also find these by looking at a graph, but with horizontal asymptotes it takes some mental imagining (example function → f\left(x\right)=\frac{3x^2+2x-1}{5x^3+4x^2-x-2}
First, find the end behavior for the function, so just cancel out anything that isn’t the highest degree →EB=\frac{3x^2}{5x^3}=\frac{3}{5x}
Next, we need to think about both limits, so taking the limit of the end behavior as x approaches ± ∞.
If the denominator grows faster than the numerator, it gets smaller and smaller towards 0
If the numerator grows faster than the denominator, then the function will grow towards ± ∞, and therefore no HA
If the exponents are the same, then the limits will both approach the divided coefficients
In our case, 3/5x, the 5x will continue to grow further, while the numerator stays the same, meaning that the fraction will get closer and closer to 0
Continuity
Here are the types of discontinuity in graph form →
Removable
\lim_{x\to a^{-}}=\lim_{x\to a^{+}}\ne f\left(a\right)
The limits approaching the left or the right are equal, but the actual value is not
Jump
\lim_{x\to a^{+}}\ne\lim_{x\to a^{-}}
Here, the limits do not match, the function value may be the same as one of the limits, however
Infinite
\lim_{x\to a}=\text{DNE and/or }\lim_{x\to a^{+}}\ne\lim_{x\to a^{-}}
Here, the actual value does not exist and the limits approach infinity
Expanding the function to be continuous
When a function has a removable discontinuity, it is sometimes possible to extend the function. To do this, find the discontinuity and simply add the term to the existing piece wise function
Making functions continuous
Since a continuous function is \lim_{x\to c}f\left(x\right)=f\left(c\right) , we can use this in the following problem\begin{cases} x+2,x\ne4\\ k^{2}-2, x=4 \end{cases}
x=4 would be the actual value, and the other one would be the limit as x approaches 4. This means we just have to set them equal to each other and set x=4, meaning that we end up with 6 = k² - 2 →k = 2
Intermediate Value Theorem (IVT)
The IVT explains if a function is continuous between a and b, then it will pass through every value between f(a) and f(b)
for example, if we have the points (1, -1) and (2, 5) and we want to know about any real solutions where f(x) = x3 - x - 1. Here is how to justify →
check for continuity → f(x) is continuous, so IVT applies
sandwich your values → f(1) = -1 < 0 < 5 = f(2)
Answer (yes/no) by IVT → There is a real solution by IVT
Unit 2 - Differentiation: Definition and Fundamental Properties
Remember our average rate of change formula →\frac{f\left(a\right)-f\left(b\right)}{a-b}
But… what if we want to find the slope at one point? Finding that using our formula would lead to a number over zero
let’s say we have a function x². If we want the slope at point x=2, we take another point, for example (3, 9), and move that point closer and closer to x=2 until it gets infinitely close, and take the slope of that
That gives us the definition of a derivative
The definition of a derivative is this →f^{\prime}\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}
Basically it is \frac{rise}{run}=\frac{y_2-y_1}{x_2-x_1}=\frac{f\left(a\right)-f\left(b\right)}{a-b}=\frac{f\left(x+h\right)-f\left(x\right)}{x+h-x} , except we want to find the instantaneous rate of change (the rate of change of a single point). By differentiating a function, we get the derivative which is the rate of change at each point on a function
For example, find \frac{d}{dx}x^2=\lim_{h\to0}\frac{\left(x+h\right)^2-\left(x\right)^2}{h}
f^{\prime}\left(x\right)=\lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h} , now time to cancel terms!
\lim_{h\to0}\frac{2xh+h^2}{h}\Rightarrow\lim_{h\to0}\frac{h\left(2x+h\right)}{h}\Rightarrow\lim_{h\to0}2x+0\Rightarrow\frac{d}{dx}f\left(x\right)=2x
This fits within the power rule! Because →\frac{d}{dx}x^2=nx^{n-1}=2x^{2-1}=2x
Higher Order Notation
Let’s say we have the function f(x) = cos(x).
Finding the derivative of the derivative would be called the second derivative and would be written as so →\frac{d^2y}{dx^2},f^{\prime\prime}\left(x\right)=\frac{d}{dx}\left(\frac{d}{dx}\cos x\right)=\frac{d}{dx}-\sin x=-\cos x
Higher order derivative is noted by adding more primes or by this notation → \frac{d^{n}y}{dx^{n}}
Alternative Definition of Derivative
You might also see the derivative written these ways
f^{\prime}\left(a\right)=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ f^{\prime}\left(a\right)=\lim_{x\to a}\frac{f\left(x\right)-f\left(a\right)}{x-a}\\ f^{\prime}\left(a\right)=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a-h\right)}{2h}
In these cases, a is a plugged in value. You won’t have to manually solve these limits, but there is 3 things you should do
1 - find/spot the original function
2 - find the derivative of that function
3 - plug in any points (a) if you have to
Derivative Rules
Here are some more derivative rules:
Constant Rule → \frac{d}{dx}c=0
Coefficients → \frac{d}{dx}af\left(x\right)=a\cdot f^{\prime}\left(x\right)
Addition/Subtraction →\frac{d}{dx}\left(f\pm g\right)=f^{\prime}\pm g^{\prime}
Power Rule → \frac{d}{dx}\left(x^{n}\right)=nx^{n-1}
Product Rule → \frac{d}{dx}\left(f\cdot g\right)=\left(f\cdot g^{\prime}\right)+\left(g\cdot f^{\prime}\right)
You can remember this rule by simply remembering that it’s just multiplying by each others derivatives
If there is three or more terms, you do it like so → \frac{d}{dx}\left(fgh\right)=f^{\prime}gh+fg^{\prime}h+fgh^{\prime}
Quotient Rule → \frac{d}{dx}\left(\frac{f}{g}\right)=\frac{\left(g\cdot f^{\prime}\right)-\left(f\cdot g^{\prime}\right)}{g^2}
A way to remember this rule is that double ds come first (denominator x derivative) and then you switch (numerator x derivative). And then you just have to bring the denominator over and square it!
Another way to remember this is “low d high - high d low over low squared”
You also sometimes might be able to work out problems without the quotient rule like so →\frac{d}{dx}\left\lbrack\frac{x^3-4x+8}{x}\right\rbrack=\frac{x^3}{x}-\frac{4x}{x}+\frac{8}{x}=x^2-4+8x^{-1}=2x-8x^{-2}
Tangent & Normal Lines
A tangent line is the line at that point tangent to the function. To find the tangent line, you will need to use point slope formula (y2 - y1 = m(x2 - x1)), but instead of m, we plug in the slope from the derivative
For example, find the tangent line at point x=2 for the function f(x) = 2x²
First, we need to plug in the point to the original function f(2) = 8
Then, find the derivative of f →f’(x) = 4x
Then, we need to plug in 2 in the derivative → f’(2) = 8
Then using these values, we get the tangent line equation →y-8=8\left(x-2\right)
The slope for a normal line is just perpendicular, so the normal line equation would be → y - 8 = (-1/8)(x-2)
Linear Approximation
linear approximation is using a tangent line to approximate a value
Example, we want the value of x = 8.1 when y = √x
First, we write a tangent line close to the point, at x = 8
y^{\prime}\left(8\right)=\frac12x^{-\frac12}=\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt8}
Now that we have slope, the y-coordinate of the point would be √8, so our tangent line would be →y-\sqrt8=\frac{1}{2\sqrt8}\left(x-8\right)
Using this tangent line, we can plug in the real value of x, 8.1, and get out our y-value →y-\sqrt8=\frac{1}{2\sqrt8}\left(8.1-8\right)\Rightarrow\frac{1}{2\sqrt8}\left(0.1\right)+\sqrt8\Rightarrow2.846
That is very close to the actual value y(8.1)!
Points Parallel to Lines
At what point on the graph y = 3x + 1 is the tangent line parallel to the line y = 5x -1
We already know what slope we should be looking for and that is 5, because it is parallel, so we need to set the derivative equal to the slope → 3^{x}\ln3=5
Now, we need to solve for the x-value → 3^{x}=\frac{5}{\ln3}\Rightarrow\log_3\frac{5}{\ln3}=x
Now that we have our x-value, we can solve for the y value, by simply plugging that x-value back into our undifferentiated function → 3^{\log_3\frac{5}{\ln3}}+1=y_1
That gives us our messy looking point \left(\log_3\frac{5}{\ln3},3^{\log_3\frac{5}{\ln3}}+1\right)
Differentiability
In order to be differentiable, a function must have equal one-sided derivatives and be continuous where the rule for the function changes
This means that → DIFFERENTIABILITY implies CONTINUITY
if f’(a) exists, then f(a) is continuous
However, the inverse is not true… just because something is continuous doesn’t imply differentiability
How to check if a piecewise function is differentiable
Given this piecewise function: f(x)=\begin{cases}3x^2,x\le1\\ 3x,x>1\end{cases}
First, we need to check continuity
f\left(1\right)=\lim_{x\to1^{-}}3x^2=\lim_{x\to1^{+}}3x=3
So, continuity is good!
Now, we need to check the derivatives → f^{\prime}(x)=\begin{cases}6x,x\le1\\ 3,x>1\end{cases}
\lim_{x\to1^{-}}6x=6\ne3=\lim_{x\to1^{+}}3
Because our left and right derivatives are not equal, the function isn’t differentiable @ x=1 and f’(1) DNE. However, this function is still continuous
Derivatives of Trig Functions
Ways to remember
the functions with c are always negative (cos=-sin, csc=-csccot, cot=-csc²)
sin loves cos, but cos doesn’t love sin (sin=cos, cos=-sin)
the tangents are always squared and love their partners more (sec², -csc²)
the sec and csc are self absorbed and always with their tangent partners (-csccot & sectan)
Derivatives of Exponents & Logarithms
Unit 3 - Differentiation: Composite, Implicit, and Inverse Functions
The Chain Rule → \frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime}\left(g\left(x\right)\right)\cdot g^{\prime}\left(x\right)
The chain rule is used for functions inside of functions. It works like Russian nesting dolls the derivative of the outside function with all the normal dolls inside, then multiplied by the derivative of the next doll inside with the rest of the dolls inside
Example → \frac{d}{dx}\sec\left(x^3-2\right)=\left(\sec\left(x^3-2\right)\tan\left(x^3-2\right)\right)\left(3x^2\right)=3x^2\sec\left(x^3-2\right)\tan\left(x^3-2\right)
Implicit Differentiation
Sometimes, the functions aren’t always explicitly defined and we have both x and y → x² + y² = 1
Remember that taking the d/dx of x is just 1! Think of it like the chain rule → d/dx(f(x)) = f’(x) * 1. However, when we take the derivative of y with respect to x, that is dy/dx or y’.
\frac{d}{dx}\left(x\right)=\frac{dx}{dx}=1\\ \frac{d}{dx}\left(y\right)=\frac{dy}{dx}=y^{\prime}
This just means that whenever we are taking the derivative of y, we have to multiply it by dy/dx. A more complicated explanation is because of the chain rule, treating the singular function f(x) as a “f(g(x))” composite function, where the outside is “f( )” and inside is “x”
f’(x) * x’, but x’ (or dx/dx) is 1 so we normally leave it out, but with y → f^{\prime}\left(y\right)\cdot\left(\frac{dy}{dx}\text{ or } y^{\prime}\right)
This is why we have to multiply by dy/dx when taking the derivative of a non-x variable with respect to x
But you don’t really need to remember that, just remember that when taking the derivative of y, you have to multiply by dy/dx
Here is the steps to implicit differentiation
Differentiate both sides of the equation with respect to x
x^2+y^2=1\Rightarrow2x+2y\left(\frac{dy}{dx}\right)=0
Collect the terms with dy/dx on one side of the equation
2y\left(\frac{dy}{dx}\right)=-2x
Factor out dy/dx and/or solve for dy/dx
\frac{dy}{dx}=\frac{-2x}{2y}=-\frac{x}{y}
Tangent Lines
Let’s say we wanted to find the tangent line to the curve of x² - xy + y² = 7 at the point (-1, 2)
We’re already given the point, so we just need to find the derivative. For this one, we have -xy so we use product rule (xy’ + yx’)
2x-\left(\left(x\right)\left(\frac{dy}{dx}\right)+\left(y\right)\left(1\right)\right)+2y\left(\frac{dy}{dx}\right)=0
2x-x\frac{dy}{dx}-y+2y\frac{dy}{dx}=0\Rightarrow x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x+y
\frac{dy}{dx}\left(-x+2y\right)=-2x+y\Rightarrow\frac{dy}{dx}=\frac{-2x+y}{-x+2y}
Now, we need to plug it into the derivative equation
\frac{dy}{dx}\vert_{\left(-1,2\right)}=\frac{-2\left(-1\right)+2}{-\left(-1\right)+2\left(2\right)}=\frac45
Now, we can use this and the point to find the tangent and normal lines
y-2=\frac45\left(x+1\right)
Higher Order Derivatives
When we do higher order derivatives with implicit differentiation, we solve for the first derivative like normal, but whenever we start solving for the second one, we substitute dy/dx with our first derivative. Example → what is the second derivative of 5 = x² - cos(y)
Find the first derivative → 0=2x+\sin\left(y\right)\left(\frac{dy}{dx}\right)\Rightarrow\frac{dy}{dx}=-\frac{2x}{\sin y}
Then we need to use the quotient rule to find the second derivative (substituting dy/dx with our first derivative) →\frac{d^2y}{dx^2}=\frac{\left(\sin y\right)\left(-2\right)-\left(-2x\right)\left(\cos\left(y\right)\frac{dy}{dx}\right)}{\sin^2y}\Rightarrow\frac{-2\sin\left(y\right)+2x\cos\left(y\right)\left(-\frac{2x}{\sin y}\right)}{\sin^2y}
Then, we can get rid of that numerator fraction by multiplying a sin(y) to the top and bottom → \frac{d^2y}{dx^2}=\frac{-2\sin^2y-4x^2\cos y}{\sin^3y}
Here’s implicit differentiation with a harder example: x² = 2ycos(xy)
2x=\left(2y\left(-\sin xy\right)\left(x\frac{dy}{dx}+y\right)+\cos xy\left(2\frac{dy}{dx}\right)\right)\Rightarrow2x=-2yx\sin xy\frac{dy}{dx}-2y^2\sin xy+2\cos xy\frac{dy}{dx}\Rightarrow2x+2y^2\sin xy=\frac{dy}{dx}\left(-2yx\sin xy+2\cos xy\right)\Rightarrow\frac{dy}{dx}=\frac{2x+2y^2\sin xy}{-2yx\sin xy+2\cos xy}=\frac{x+y^2\sin xy}{-yx\sin xy+2\cos xy}
Inverse Function Derivatives
