$$\sum_{k=1}^n c = nc$$
$$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
$$\sum_{k=1}^n c+a = \sum_{k=1}^n c + \sum_{k=1}^n a$$
A summation is just a sum of all terms with the respective index plugged in. For instance,
$$\sum_{k=1}^3 n = n_1 + n_2 + n_3 = 3n$$
Pay attention to bounds, though, because
$$\sum_{k=2}^3 n = n_2 + n_3 = 2n$$
Remember that k is merely the index term in the sum unless it features in the equation/summation. Also remember that to find limits when the lower bound is not 1 and is instead y, you must subtract all indices up to y. For instance,
$$\sum_{k=2}^3 n = [\sum_{k=1}^3 n - \sum_{k=1}^1 n] = 3n - n = 2n$$
Also recall the following definition.
DEFINITION OF A SEQUENCE: any set of things in one-to-one correspondence with the natural numbers.
A limit describes a function as it approaches, but never fully reaches, a value. It would be more precise to say that the limit reaches a value as it approaches |∞|, and that can only be one ∞.
$$\lim_{n\to 4} n$$ As n reaches 4, what does n reach? The answer is obviously 4.
$$\lim_{n\to ∞} = ∞$$
$$\lim_{n\to 2} 3n = 3(2) = 6$$
The finite ones are as simple as plugging in. Unless...
$$\lim_{n\to 0} \frac{1}{n} = undefined = NL = \varnothing$$
or
$$\lim_{n\to -1} \frac{1}{\sqrt[2]{n}} = undefined = NL = \varnothing$$
Things get a bit more interesting when you have infinite limits, with multiple powers.
$$\lim_{n\to ∞} \frac{56n^2+96n+64}{3n^2}$$
In this case, you factor out the highest power to get:
$$\lim_{n\to ∞} \frac{56+\frac{96}{n}+\frac{64}{n^2}}{3}$$
We then apply the reasoning that since the denominators of $\frac{96}{n}$ and $\frac{64}{n^2}$ grow exponentially larger, their values approach 0. This leaves us with $\frac{56}{3}$ as our answer.
Derivatives are slopes of tangent lines. To find them, multiple strategies can be used. The long-form formula is:
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = f^\prime(x) = \frac{dy}{dx}$$
You can also use the power function rule, if x is raised to a power n.
$$f(x) = x^n, f^\prime(x) = nx^{n-1}$$
A higher-order derivative is simply taking the derivative of a single expression, x number of times. Roman numerals replace the prime symbol.
Sometimes you will encounter problems that would be troublesome to convert to one of the prelisted forms, such as quotients. There are a couple rules for such cases:
$$f(x) = \frac{T}{B}, f^{\prime}(x) = \frac{T^{\prime}B - B^{\prime}T}{B^2}$$
Where T is the top term and B is the bottom term, and:
$$f(x) = F\times S, f^{\prime}(x) = F^{\prime}S + S^{\prime}F$$
Where F is the first term and S is the second.
Occasionally, you will see these problems conjoined with trigonometric statements and formulas. The following govern most cases:
$$y = sin (x), y^{\prime} = cos (x)$$
$$y = cos (x), y^{\prime} = -sin (x)$$
$$y = tan (x), y^{\prime} = sec^2 (x)$$
$$y = cot (x), y^{\prime} = -csc^2 (x)$$
$$y = sec (x), y^{\prime} = sec (x) \times tan (x)$$
$$y = csc (x), y^{\prime} = -csc (x) \times cot (x)$$
$$y = e^{x}, y^{\prime} = e^{x}$$
$$y = ln (x), y^{\prime} = \frac{1}{x}$$
$$y = arcsin (x), y^{\prime} = \frac{1}{\sqrt[2]{1-x^2}}$$
$$y = arccos (x), y^{\prime} = \frac{-1}{\sqrt[2]{1-x^2}}$$
$$y = arctan (x), y^{\prime} = \frac{1}{1+x^2}$$
When derivatives have parenthetical expressions, one can apply the "chain rule." It is easiest to see with an example:
$$2(3x)^2 = 4(3x)(3) = 36x$$
$$a(b)^n = nab^{\prime}(b)^{n-1}$$
If c is a factor of f, it is also a factor of $y^{\prime}$ and can thus be factored out of the function until an arbitrary step in the process (such as the last).
$$\frac{3}{x} = 3(x)^{-1} = -3(x)^{-2} = \frac{-3}{x^2}$$
Note: make sure to keep your original trigonometric argument throughout this process, it can be easy to forget it.
$$y = sin^{4}(2x) = (sin (2x))^{4}, y^{\prime} = 4(sin (2x))^{3}(cos (2x))(2) = (8)(sin^{3}2x)(cos (2x))$$
Note: sometimes the argument itself can be raised to a power, don't mistake these situations for a different case.
$$y = sin(2)^2 = sin(4), y^{\prime} = cos(4)$$
Implicit derivatives mark the change within a derivative with respect to a variable, and are of the form $\frac{d}{dt}$ in related rates, where the variable of the denominator is time. Using implicit derivatives, we can find the relationship of one variable in relation to the time variable, such as $\frac{dx}{dt}$ or $\frac{dy}{dt}$. This sort is also known as a related rate: the rate of change of a variable with respect to time.
Extrema refers to the highest and lowest points on a graph. If the points given are the highest on a specific interval, but not necessarily the entire function, they are the relative maximum and the relative minimum. Conversely, if they are the extremes of the entire function, they are the absolute maximum and the absolute minimum.
On a specific interval, the extrema will occur either at the endpoints or at the "turning points." Turning points occur when the function reaches a horizontal tangent line (derivative with slope = 0), and then continues in the opposite direction (either increasing or decreasing).