Fundamentals
The core facts every aspirant should own — each a titled nugget with a real-world story, the concept in plain words, and a memory trick. Works even when the internet doesn't.
300 fundamentals
“A GPS route may point to a road that ends abruptly; the path’s direction approaches a point (limit) even though the road isn’t there (f(a) missing).”
Limit equals f(a) only if the function is continuous at a. If f is undefined or jumps, the limit can still exist, but f(a) differs or does not exist. Exams often give a limit value and ask for f(a); verify continuity first.
Memory trick: Limit = f(a) ⇔ continuity at a
“Seeing a speedometer reading (function value) and assuming it’s the acceleration (derivative) is the classic mix-up.”
A derivative measures the instantaneous rate of change, not the function’s value. It is the limit of Δy/Δx as Δx→0, giving the tangent slope. Trap: plugging the point into the original function instead of the derivative, or confusing units.
Memory trick: Derivative = Δy/Δx limit.
“Turning a blender’s speed knob (inner) changes the blade’s rotation speed (outer) – the total effect is the product of both.”
The chain rule states d/dx f(g(x)) = f'(g(x))·g'(x). Differentiate the outer function, then multiply by the inner derivative. Forgetting g'(x) (e.g., writing d/dx sin 2x = cos 2x) loses a factor and costs marks.
Memory trick: Chain: outer × inner – never drop the inner derivative.
“When two cyclists ride side by side, the combined acceleration equals each rider’s acceleration added, not multiplied.”
For u(x)v(x), (uv)' = u'v + uv'. For u/v, (u/v)' = (u'v – uv')/v². The plus sign in the product rule and the squared denominator in the quotient rule are frequent sources of error.
Memory trick: Product: +; Quotient: – on top, v² bottom.
“A chef tasting two sauces thinks the flavor change is just the product of each sauce’s change, but the real change adds both contributions.”
A common slip is writing (uv)' = u'v' or (u/v)' = u'v'/v². The correct formulas add the missing terms: (uv)' = u'v + uv' and (u/v)' = (u'v – uv')/v². Missing a term gives a wrong answer.
Memory trick: Never multiply derivatives – add the missing term.
“Whisking batter: volume ∝ x³, the rate grows as 3·x².”
Derivative of xⁿ is n·x^(n-1) for any real constant n, provided x lies in the function’s domain. Works for fractions like √x (n=½) but only where x>0. Mistake: using the rule when n varies with x.
d/dx(xⁿ) = n·x^(n−1)
Memory trick: Only constant n – if n varies, stop.
“If a river’s speed varies with depth and depth varies with distance downstream, the overall speed change along the bank equals both rates multiplied.”
When y = f(u) and u = g(x), dy/dx = (dy/du)·(du/dx). The derivative passes through each intermediate variable, multiplying the rates. Mixing the order or omitting a factor leads to an incorrect chain.
dy/dx = dy/du · du/dx
Memory trick: Chain link: dy/du × du/dx.
“A car’s speedometer reads km/h (derivative); the tiny distance traveled in a second dt is dy = speed·dt, the differential.”
dy/dx is the instantaneous rate (slope). The differential dy = f'(x)·dx gives the predicted small change for a tiny increment dx. Students often treat dy as the same as the derivative; remember dy includes the factor dx.
Memory trick: dy = f'(x)·dx – derivative times tiny step.
“When a rectangle’s length and width both expand with time, its area changes as dA/dt = L·dW/dt + W·dL/dt.”
For y = u·v, dy/dx = u'v + uv' (product rule). For y = u/v, dy/dx = (u'v – uv')/v² (quotient rule). The denominator is always v²; missing the square or flipping the sign of the second term is a frequent exam error.
Memory trick: Quotient: “minus, then square” → (u'v – uv')/v²
“Turning the volume knob (outer function) changes the loudness, but the effect also depends on how fast the knob itself is turned (inner function).”
For a composite function y = f(g(x)), dy/dx = f'(g(x))·g'(x). Forgetting g'(x) drops a crucial factor, especially when g(x) is a power, trig or exponential inside another function.
Memory trick: Chain rule: outer × inner – link the chain!
“Knowing a road’s slope tells you direction, but you need the starting altitude to locate the exact point.”
Indefinite integration reverses differentiation, producing a family F(x)+C. The constant C is lost when differentiating, so it must be added back. Dropping +C gives an incomplete solution, especially in initial-value problems.
Memory trick: ∫ → always write +C.
“A recipe without a pinch of salt tastes flat; the missing ingredient changes the final dish.”
When finding antiderivatives, forgetting the +C omits one degree of freedom, leading to a wrong general solution. In differential equations, the missing constant makes the answer fail the given condition.
Memory trick: No +C = incomplete answer.
“Measuring water flowing between two bridges is a definite integral.”
The definite integral ∫_aᵇ f(x)dx equals the signed area between the curve and the x-axis from a to b. Area above the axis is positive, below is negative, and swapping limits reverses the sign. For total (unsigned) area split the interval at the zeros of f(x).
Memory trick: Upper-lower → sign; swap limits → opposite sign.
“A cyclist’s net displacement counts downhill as negative, just like a signed integral.”
A frequent slip is to assume the integral always gives a positive area. The definite integral actually yields signed area, so portions of the curve below the x-axis subtract from the total. To get the total (unsigned) area, break the interval at each root and add the absolute values.
Memory trick: Below axis = negative; add |area| for total.
“When you see ∫f(x)·f'(x)dx, it's like a lock and key—substitution opens it instantly.”
If the integrand contains a function and its exact derivative, let u = that function. Then du = f'(x)dx and the integral becomes ∫u du, which is trivial. The trap: forcing substitution when the derivative isn’t present wastes time; switch to parts or other tricks then.
Memory trick: f·f' → SUB first
“When navigating a campus, you first look for the straight path before wandering through the garden, just like trying substitution before by-parts.”
A simple u-substitution often turns the integral into a basic form; by-parts adds extra terms and can complicate it. The exam trap is using by-parts on ∫f(g(x))g'(x)dx where u = g(x) works instantly. Always check if the integrand contains a function and its derivative first.
Memory trick: Derivative of inner present? → try SUB first
“Climbing stairs where each step height follows xⁿ gives total height x^(n+1)/(n+1).”
The power rule ∫xⁿdx = xⁿ⁺¹/(n+1)+C holds for any real n except n = –1. It follows from reversing the differentiation of xⁿ⁺¹. Remember that n = –1 leads to the logarithmic case, not division by zero.
∫xⁿ dx = x^(n+1)/(n+1) + C
Memory trick: n≠-1 → add 1, divide.
“Continuous 100% interest growth over time integrates to ln|x|.”
∫(1/x)dx = ln|x|+C for x≠0. The absolute value ensures the antiderivative works for both positive and negative x because d/dx[ln|x|]=1/x. Omitting the | | gives the wrong sign on the negative side.
∫(1/x) dx = ln|x| + C
Memory trick: 1/x → ln|x|, keep absolute.
“When you measure water collected in a bucket you get one volume; writing the formula for how fast the tap fills gives a family of curves.”
A definite integral ∫_aᵇ f(x)dx has upper and lower limits and evaluates to a single number, the net area. An indefinite integral ∫ f(x)dx has no limits; its result is a general antiderivative F(x)+C, a family of functions. In JEE, missing the limits or adding +C to a definite integral loses marks.
Memory trick: Definite = ‘Done’ → number; Indefinite = ‘Infinite’ → family (+C).
“A speedometer shows the car’s instant speed (derivative); the odometer shows total distance travelled over time (integral).”
Differentiation turns y=f(x) into its instantaneous rate dy/dx, the slope at each point. Integration reverses this, accumulating infinitesimal pieces to recover the original function (plus C) or total area. Trap: treating ∫ f'(x)dx as f(x) without the +C.
Memory trick: Diff = slope, Int = sum – think ‘D’ for ‘Derivative’, ‘I’ for ‘Integral’ = ‘Add up’.
“Finding a particle’s position from its velocity requires adding the unknown starting point, otherwise the path is incomplete.”
Every indefinite integral ∫ f(x)dx yields an antiderivative plus an arbitrary constant C because differentiation removes constants. Omitting +C makes the answer incomplete and can cost marks, especially when later applying initial conditions. Trap: writing ∫ x dx = x²/2 without +C.
Memory trick: Indefinite Integral → Include C.
“Rolling a fair die, the chance of a 4 is 1 favourable outcome out of 6 total outcomes.”
Probability = favourable outcomes ÷ total outcomes, so 0 ≤ P ≤ 1. Always list all equally likely outcomes before dividing. If the result exceeds 1, the sample space was counted incorrectly.
Memory trick: P>1? Re-count the sample space.
“Thinking the chance of drawing a red card is 14/13 shows you counted extra cards.”
A probability greater than 1 is impossible; it means the numerator or denominator is wrong—usually double-counted favourable cases or omitted equally likely outcomes. Verify the sample space first.
Memory trick: P>1 → recount outcomes.
“Two light switches that can never be on together act like mutually exclusive events.”
Independent events satisfy P(A∩B)=P(A)P(B); mutually exclusive events have P(A∩B)=0 and are therefore dependent because one occurring stops the other. Do not treat exclusive events as independent.
Memory trick: Exclusive → zero overlap.