Difficulty: Foundational | Prerequisites: Calculus I and II fluency (these are reference formulas you should already know)
These differentiation and integration formulas are the mechanical toolkit of calculus. In Calculus III, you still need them constantly: computing velocity and acceleration from position functions, evaluating line integrals, working with partial derivatives, and setting up multiple integrals. This is a reference sheet, not new material. If any of these formulas feel unfamiliar, revisit your Calculus I and II notes before moving forward.
A compact reference of every derivative rule, trig derivative, inverse trig derivative, basic integral, and integration technique (u-substitution, integration by parts) you need to have at your fingertips. The unit circle values are included because they come up in nearly every integral involving trig functions.
Power rule
d/dx [xⁿ] = nxⁿ⁻¹. The workhorse of differentiation. In simple terms, bring the exponent down and reduce it by one.
Product rule
d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x). Used whenever two functions are multiplied together.
Quotient rule
d/dx [f(x)/g(x)] = [f'(x)g(x) − f(x)g'(x)] / [g(x)]². Think "low d-high minus high d-low, over low squared."
Chain rule
d/dx [f(g(x))] = f'(g(x)) · g'(x). Used whenever one function is nested inside another. In simple terms, differentiate the outside, then multiply by the derivative of the inside.
Antiderivative / indefinite integral
The reverse of differentiation. ∫f(x)dx = F(x) + C, where F'(x) = f(x). The +C accounts for the fact that many functions share the same derivative.
u-substitution
A technique for undoing the chain rule in integration. Choose u = g(x), compute du = g'(x)dx, and rewrite the integral in terms of u.
Integration by parts
∫u dv = uv − ∫v du. Used when the integrand is a product of two functions and substitution does not simplify it. Think of it as the integration version of the product rule.
Power rule: d/dx [xⁿ] = nxⁿ⁻¹
Product rule: d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
Quotient rule: d/dx [f(x)/g(x)] = [f'(x)g(x) − f(x)g'(x)] / [g(x)]²
Chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x)
Exponential: d/dx [aˣ] = aˣ ln(a); in particular, d/dx [eˣ] = eˣ
Logarithmic: d/dx [ln(x)] = 1/x; d/dx [log_a(x)] = 1/(x ln a)
d/dx [sin(x)] = cos(x)
d/dx [cos(x)] = −sin(x)
d/dx [tan(x)] = sec²(x)
d/dx [cot(x)] = −csc²(x)
d/dx [sec(x)] = sec(x)tan(x)
d/dx [csc(x)] = −csc(x)cot(x)
d/dx [sin⁻¹(x)] = 1/√(1 − x²)
d/dx [cos⁻¹(x)] = −1/√(1 − x²)
d/dx [tan⁻¹(x)] = 1/(1 + x²)
d/dx [cot⁻¹(x)] = −1/(1 + x²)
d/dx [sec⁻¹(x)] = 1/(|x|√(x² − 1))
d/dx [csc⁻¹(x)] = −1/(|x|√(x² − 1))
∫sin(x)dx = −cos(x) + C
∫cos(x)dx = sin(x) + C
∫tan(x)dx = ln|sec(x)| + C
∫ln(x)dx = x ln(x) − x + C
∫eᵃˣ dx = (1/a)eᵃˣ + C
u-substitution: choose u = g(x), find du = g'(x)dx, rewrite the integral as ∫f(u)du, integrate, then substitute back. The method for ∫f(g(x))g'(x)dx.
Integration by parts: ∫u dv = uv − ∫v du. Choose u as the function that simplifies when differentiated (often logarithmic or polynomial) and dv as the rest.
The standard angles and their sine/cosine values you need to know.
0° (0 rad): sin = 0, cos = 1
30° (π/6): sin = 1/2, cos = √3/2
45° (π/4): sin = √2/2, cos = √2/2
60° (π/3): sin = √3/2, cos = 1/2
90° (π/2): sin = 1, cos = 0
180° (π): sin = 0, cos = −1
270° (3π/2): sin = −1, cos = 0
All other standard angles follow by symmetry across the axes. In quadrant II, cosine is negative. In quadrant III, both are negative. In quadrant IV, sine is negative.
Function | Derivative |
|---|---|
xⁿ | nxⁿ⁻¹ |
sin(x) | cos(x) |
cos(x) | −sin(x) |
tan(x) | sec²(x) |
cot(x) | −csc²(x) |
sec(x) | sec(x)tan(x) |
csc(x) | −csc(x)cot(x) |
eˣ | eˣ |
aˣ | aˣ ln(a) |
ln(x) | 1/x |
log_a(x) | 1/(x ln a) |
sin⁻¹(x) | 1/√(1 − x²) |
cos⁻¹(x) | −1/√(1 − x²) |
tan⁻¹(x) | 1/(1 + x²) |
Integral | Result |
|---|---|
∫sin(x)dx | −cos(x) + C |
∫cos(x)dx | sin(x) + C |
∫tan(x)dx | ln |
∫ln(x)dx | x ln(x) − x + C |
∫eᵃˣ dx | (1/a)eᵃˣ + C |
Students often forget the negative sign in d/dx [cos(x)] = −sin(x). The derivative of cosine has a minus sign; the derivative of sine does not.
Students mix up the quotient rule order. It is "low d-high minus high d-low," not the other way round. Getting the subtraction backwards flips the sign of the entire derivative.
Students apply the power rule to eˣ and write xeˣ⁻¹. The power rule is for xⁿ (variable base, constant exponent). For aˣ (constant base, variable exponent), the derivative is aˣ ln(a).
Students forget +C on indefinite integrals. Without the constant of integration, the answer is incomplete.
Students confuse sin⁻¹(x) (the inverse sine function, aka arcsin) with 1/sin(x) (the cosecant). They are entirely different functions.
⚠️ You will not be given a formula sheet for most Calculus III exams. These derivatives and integrals need to be memorised.
⚠️ The chain rule appears in nearly every partial derivative problem. If you cannot apply it automatically, partial derivatives will be very slow.
⚠️ u-substitution is the single most commonly tested integration technique. Practise choosing u until it becomes instinctive.
⚠️ Unit circle values appear whenever you evaluate a definite integral with trig functions. Know at minimum 0, π/6, π/4, π/3, π/2 and their sine/cosine values.
Fill in the blank: d/dx [tan(x)] = ______.
True or false: d/dx [eˣ] = xeˣ⁻¹.
Fill in the blank: ∫cos(x)dx = ______.
True or false: sin(π/4) = 1/2.
Fill in the blank: in integration by parts, ∫u dv = ______.
Q: Differentiate f(x) = x³ sin(x).
A: Use the product rule. f'(x) = 3x² sin(x) + x³ cos(x).
Q: Find d/dx [ln(cos(x))].
A: Use the chain rule. d/dx [ln(cos(x))] = (1/cos(x)) · (−sin(x)) = −sin(x)/cos(x) = −tan(x).
Q: Evaluate ∫2x cos(x²) dx using u-substitution.
A: Let u = x². Then du = 2x dx. The integral becomes ∫cos(u) du = sin(u) + C = sin(x²) + C.
Q: What is cos(5π/6)?
A: 5π/6 is in quadrant II (reference angle π/6). Cosine is negative in quadrant II, so cos(5π/6) = −√3/2.
Q: Evaluate ∫x eˣ dx using integration by parts.
A: Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. By parts: ∫x eˣ dx = xeˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x − 1) + C.
Every one of these formulas carries into multivariable calculus. Partial derivatives are just single-variable derivatives taken one variable at a time, so the same rules apply. Multiple integrals stack single-variable integrals, so your integration techniques are used in every double and triple integral. The unit circle values return whenever you convert between Cartesian and polar coordinates.
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