Difficulty: Core Calculus | Prerequisites: Algebra Foundations notes, Trigonometry notes, Geometry Formulas notes
This is the heart of Calculus 1. Derivatives measure rates of change; integrals measure accumulation. Together they are linked by the Fundamental Theorem of Calculus. Every exam question in this course uses material from this document, so treat it as your primary revision resource.
Derivatives measure instantaneous rates of change and give you the slope of a function at any point. Integrals measure accumulated quantities and give you the area under a curve. The Fundamental Theorem of Calculus links the two: differentiation and integration are inverse operations.
Derivative
The instantaneous rate of change of a function at a point. Formally, f'(x) = lim(h->0) [f(x+h) - f(x)] / h, provided the limit exists.
Think of it as the slope of the curve at a single point, or how fast the output is changing per unit of input.
Critical point
A value x = c where either f'(c) = 0 or f'(c) does not exist. Critical points are where the function might have a local maximum, local minimum, or a plateau.
In simple terms, these are the candidates for peaks and valleys.
Concavity
Describes whether a curve bends upward or downward. f''(x) > 0 means concave up (holds water, like a bowl). f''(x) < 0 means concave down (spills water, like a hill).
Think of it as: concave up = happy face curve, concave down = sad face curve.
Inflection point
A point where the concavity of f(x) changes (from concave up to concave down, or vice versa).
In simple terms, it is where the curve changes from bending one way to bending the other.
Integral (definite)
The definite integral of f(x) from a to b, written as the integral from a to b of f(x) dx, gives the net signed area between the curve and the x-axis on that interval.
Think of it as adding up infinitely many infinitely thin rectangles under the curve.
Antiderivative (indefinite integral)
A function F(x) whose derivative is f(x). Written as the integral of f(x) dx = F(x) + C, where C is an arbitrary constant.
In simple terms, it reverses differentiation. The "+C" accounts for the fact that many functions share the same derivative.
Fundamental Theorem of Calculus (FTC)
Part I: If g(x) = integral from a to x of f(t) dt, then g'(x) = f(x). Differentiation undoes integration. Part II: The integral from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f.
In simple terms, to evaluate a definite integral, find the antiderivative and plug in the endpoints.
f'(x) = lim(h->0) [f(x+h) - f(x)] / h
This is the formal definition. You must be able to use it when asked, especially in early exams. In practice, you will use the rules below.
m = f'(a) is the slope of the tangent line to y = f(x) at x = a
The equation of that tangent line is y = f(a) + f'(a)(x - a)
f'(a) is the instantaneous rate of change of f(x) at x = a
If f(x) is position, then f'(x) is velocity
x = c is a critical point if f'(c) = 0 or f'(c) does not exist
f'(x) > 0 on an interval means f(x) is increasing there
f'(x) < 0 on an interval means f(x) is decreasing there
f'(x) = 0 on an interval means f(x) is constant there
f''(x) > 0 means concave up; f''(x) < 0 means concave down
x = c is an inflection point if the concavity changes at x = c
Rule | Formula |
|---|---|
Constant | (c)' = 0 |
Sum/Difference | [f(x) +/- g(x)]' = f'(x) +/- g'(x) |
Constant multiple | [c·f(x)]' = c·f'(x) |
Product rule | [f(x)·g(x)]' = f(x)·g'(x) + f'(x)·g(x) |
Quotient rule | [f(x)/g(x)]' = [g(x)·f'(x) - f(x)·g'(x)] / [g(x)]^2 |
Chain rule | [f(g(x))]' = f'(g(x))·g'(x) |
Power rule | (x^n)' = n·x^(n-1) |
Function | Derivative |
|---|---|
e^x | e^x |
a^x | a^x · ln(a) |
ln(abs(x)) | 1/x |
log_a(x) | 1 / (x · ln(a)) |
Function | Derivative |
|---|---|
sin(x) | cos(x) |
cos(x) | -sin(x) |
tan(x) | sec^2(x) |
cot(x) | -csc^2(x) |
sec(x) | sec(x)·tan(x) |
csc(x) | -csc(x)·cot(x) |
Function | Derivative |
|---|---|
arcsin(x) | 1 / sqrt(1 - x^2) |
arccos(x) | -1 / sqrt(1 - x^2) |
arctan(x) | 1 / (1 + x^2) |
arccot(x) | -1 / (1 + x^2) |
arcsec(x) | 1 / (abs(x) · sqrt(x^2 - 1)) |
arccsc(x) | -1 / (abs(x) · sqrt(x^2 - 1)) |
Function | Derivative |
|---|---|
sinh(x) | cosh(x) |
cosh(x) | sinh(x) |
tanh(x) | sech^2(x) |
coth(x) | -csch^2(x) |
sech(x) | -sech(x)·tanh(x) |
csch(x) | -csch(x)·coth(x) |
Function | Derivative |
|---|---|
arcsinh(x) | 1 / sqrt(1 + x^2) |
arccosh(x) | 1 / sqrt(x^2 - 1) |
arctanh(x) | 1 / (1 - x^2) |
arccoth(x) | 1 / (1 - x^2) |
arcsech(x) | -1 / (abs(x) · sqrt(1 - x^2)) |
arccsch(x) | -1 / (abs(x) · sqrt(x^2 + 1)) |
Suppose f(x) is continuous on [a, b]. Divide [a, b] into n subintervals of width Delta x = (b-a)/n and choose a sample point x_i* from each. Then the integral from a to b of f(x) dx = lim(n->infinity) of the sum of f(x_i*)·Delta x.
Part I: If g(x) = integral from a to x of f(t) dt, then g'(x) = f(x) for all x in [a, b]. Differentiation undoes integration.
Part II: The integral from a to b of f(x) dx = F(b) - F(a), where F(x) is any antiderivative of f(x) (meaning F' = f).
Area under a curve: A = integral from a to b of f(x) dx
Area between curves (vertical slices): A = integral from a to b of [upper function - lower function] dx
Area between curves (horizontal slices): A = integral from a to b of [right function - left function] dy
Volume by cross-sections: V = integral from a to b of A(x) dx, where A(x) is the cross-sectional area
Volume by discs/washers: V = integral from a to b of pi·(outer^2 - inner^2) dx
Volume by cylindrical shells: V = integral from a to b of 2·pi·(radius)·(height) dx
Work: W = integral from a to b of F(x) dx, where F(x) is the force function
Average value: f_avg = [1/(b-a)] · integral from a to b of f(x) dx
Integral | Result |
|---|---|
integral of u^n du (n ≠ -1) | u^(n+1)/(n+1) + C |
integral of du/u | ln(abs(u)) + C |
integral of e^u du | e^u + C |
integral of a^u du | a^u / ln(a) + C |
integral of ln(u) du | u·ln(u) - u + C |
integral of (1/(u·ln u)) du | ln(abs(ln u)) + C |
Integral | Result |
|---|---|
integral of sin(u) du | -cos(u) + C |
integral of cos(u) du | sin(u) + C |
integral of tan(u) du | ln(abs(sec u)) + C |
integral of cot(u) du | ln(abs(sin u)) + C |
integral of sec(u) du | ln(abs(sec u + tan u)) + C |
integral of csc(u) du | ln(abs(csc u - cot u)) + C |
integral of sec^2(u) du | tan(u) + C |
integral of csc^2(u) du | -cot(u) + C |
integral of sec(u)·tan(u) du | sec(u) + C |
integral of csc(u)·cot(u) du | -csc(u) + C |
Integral | Result |
|---|---|
integral of du / sqrt(a^2 - u^2) | arcsin(u/a) + C |
integral of du / (a^2 + u^2) | (1/a)·arctan(u/a) + C |
integral of du / (u·sqrt(u^2 - a^2)) | (1/a)·arcsec(u/a) + C |
integral of du / (a^2 - u^2) | (1/(2a))·ln(abs((u+a)/(u-a))) + C |
integral of du / (u^2 - a^2) | (1/(2a))·ln(abs((u-a)/(u+a))) + C |
Integral | Result |
|---|---|
integral of arcsin(u) du | u·arcsin(u) + sqrt(1 - u^2) + C |
integral of arccos(u) du | u·arccos(u) + sqrt(1 - u^2) + C |
integral of arctan(u) du | u·arctan(u) - (1/2)·ln(1 + u^2) + C |
Integral | Result |
|---|---|
integral of sinh(u) du | cosh(u) + C |
integral of cosh(u) du | sinh(u) + C |
integral of tanh(u) du | ln(cosh(u)) + C |
integral of coth(u) du | ln(abs(sinh(u))) + C |
integral of sech(u) du | arctan(abs(sinh(u))) + C |
integral of csch(u) du | ln(abs(tanh(u/2))) + C |
integral of sech^2(u) du | tanh(u) + C |
integral of csch^2(u) du | -coth(u) + C |
integral of sech(u)·tanh(u) du | -sech(u) + C |
integral of csch(u)·coth(u) du | -csch(u) + C |
integral of u dv = u·v - integral of v du
Choose u and dv using the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trig, Exponential): pick u from whichever category comes first in LIATE.
Students often think the derivative of a product is the product of the derivatives: (fg)' = f'g'. It is not. You must use the product rule: (fg)' = fg' + f'g.
Students forget the minus sign in the derivative of cos(x). The derivative of cos(x) is -sin(x), not sin(x).
Students confuse the quotient rule's numerator. It is g(x)f'(x) - f(x)g'(x), not the other way around. The denominator function times the derivative of the numerator comes first.
Students forget the "+C" on indefinite integrals. Without it, the answer is incomplete.
Students treat the integral from a to b of f(x)·g(x) dx as the product of two separate integrals. There is no such rule. Integration does not distribute over multiplication.
⚠️ The product rule and chain rule are used in nearly every derivative problem. Know them cold.
⚠️ The FTC Part II is how you evaluate every definite integral. It will appear on every exam.
⚠️ Volumes of revolution (disc/washer and shell methods) are a major topic on the Calculus 1 final. Know when to use each method.
⚠️ The integral of sec(x) and csc(x) are commonly tested and easy to forget. Memorise them.
⚠️ Integration by parts appears on most Calculus 1 finals. Remember LIATE for choosing u.
What is the derivative of x^5? Answer: 5x^4.
True or false: the derivative of e^x is x·e^(x-1). Answer: False. The derivative of e^x is e^x. The power rule does not apply to exponential functions.
What is the integral of cos(x) dx? Answer: sin(x) + C.
Fill in the blank: the integral from a to b of f(x) dx = F() - F(). Answer: F(b) - F(a).
True or false: if f'(x) > 0 on an interval, then f(x) is decreasing. Answer: False. f'(x) > 0 means f(x) is increasing.
Q: Find the derivative of f(x) = x^3 · sin(x).
A: Use the product rule: f'(x) = x^3 · cos(x) + 3x^2 · sin(x).
Q: Find the derivative of f(x) = e^(x^2).
A: Use the chain rule: f'(x) = e^(x^2) · 2x = 2x · e^(x^2).
Q: Evaluate the integral from 0 to pi of sin(x) dx.
A: The antiderivative of sin(x) is -cos(x). So: [-cos(pi)] - [-cos(0)] = [-(-1)] - [-(1)] = 1 + 1 = 2.
Q: Find the area between y = x^2 and y = x on [0, 1].
A: On [0, 1], x > x^2, so A = integral from 0 to 1 of (x - x^2) dx = [x^2/2 - x^3/3] from 0 to 1 = 1/2 - 1/3 = 1/6.
Q: Use integration by parts to find the integral of x · e^x dx.
A: Let u = x, dv = e^x dx. Then du = dx, v = e^x. The integral = x·e^x - integral of e^x dx = x·e^x - e^x + C = e^x(x - 1) + C.
Derivatives connect to limits (the derivative is defined as a limit), to optimisation (finding maxima and minima), and to related rates (connecting the rates of change of linked quantities). Integration connects to area, volume, work, and probability (probability density functions are integrated to find probabilities). Both derivatives and integrals extend into multivariable calculus, where partial derivatives and multiple integrals generalise these single-variable concepts.
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