Calculus Domain Guide

Every derivative statement is a statement about linear approximation. The product rule, the chain rule, and implicit differentiation are all bookkeeping for the same question: if each input moves a little, how much does the output move? Reading them that way is what makes multivariable calculus feel like a continuation rather than a new subject — the gradient is just the list of answers to that question, one input at a time.

Every integral theorem in the domain is the same sentence at a different dimension: the total of a derivative over a region equals a total of the original object over the boundary of that region. The fundamental theorem ($\int_a^b F' = F(b) - F(a)$, a boundary of two points), Green's theorem (boundary curve), Stokes' theorem (boundary of a surface), and the divergence theorem (boundary surface of a solid) are four readings of it. Memorising them as one pattern is far cheaper than memorising four formulas.

Series questions are almost never asked as computations. They are asked as classifications: does this converge, and if so absolutely or conditionally? That is why the useful memory object is not a proof but an ordered decision list, with the two benchmark families — geometric and $p$-series — at the top because everything else is compared to them.

Start with what the exam actually assumes at speed. Derivative rules (product, quotient, chain), the derivatives of the elementary functions, integration by parts and substitution, and the standard antiderivatives are treated as reflexes, not as content. The first real decision point in most items is the fundamental theorem with a variable limit: $\dfrac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))h'(x) - f(g(x))g'(x)$. The two chain-rule factors and the minus sign on the lower limit are the whole question in a surprising number of items.

Series is the second block. Fix the two benchmarks first: a geometric series $\sum ar^n$ converges exactly when $|r| < 1$, to $a/(1-r)$; a $p$-series $\sum n^{-p}$ converges exactly when $p > 1$. Then run a fixed order on anything else. Do the terms fail to approach zero? Divergent, stop. Are the terms positive and rational-looking? Compare (or limit-compare) with a $p$-series. Are there factorials or $n$-th powers? Ratio or root test. Alternating with decreasing terms tending to zero? Convergent by the alternating series test, at which point the remaining question is whether $\sum |a_n|$ also converges — if not, the convergence is conditional, and conditional convergence is exactly the situation in which rearranging the terms can change the sum.

Power series then convert the classification skill into a computational one. Once you know $e^x = \sum x^n/n!$, $\sin x = \sum (-1)^n x^{2n+1}/(2n+1)!$, $\cos x = \sum (-1)^n x^{2n}/(2n)!$, $\dfrac{1}{1-x} = \sum x^n$ for $|x| < 1$, and $\ln(1+x) = \sum (-1)^{n+1} x^n / n$ for $-1 < x \le 1$, whole categories of limit and integral problems collapse. A limit such as $\lim_{x \to 0} (e^x - 1 - x)/x^2$ is answered by reading off the leading surviving term ($x^2/2$ over $x^2$, so $1/2$) with no differentiation at all, and an integral such as $\int_0^1 \sin(x^2)\,dx$ — which has no elementary antiderivative — becomes a series you can truncate.

Multivariable calculus reuses the same machinery with more inputs. Partial derivatives hold the other variables fixed; the gradient collects them; the directional derivative in a unit direction $\vec{u}$ is $D_{\vec{u}}f = \nabla f \cdot \vec{u}$, which is largest when $\vec{u}$ points along $\nabla f$ and zero when $\vec{u}$ is tangent to the level set. Unconstrained critical points solve $\nabla f = \vec{0}$ and are classified by the second-derivative test with $D = f_{xx}f_{yy} - f_{xy}^2$: $D > 0$ with $f_{xx} > 0$ is a local minimum, $D > 0$ with $f_{xx} < 0$ a local maximum, $D < 0$ a saddle, and $D = 0$ no information. Constrained problems — maximise $f$ subject to $g = c$ — use Lagrange multipliers: at an extremum the level sets are tangent, so $\nabla f = \lambda \nabla g$, and you solve that system together with the constraint.

Multiple integrals are mostly a question of describing the region. Convert to polar when the region is a disc or a sector and remember that $dA = r\,dr\,d\theta$; the factor $r$ is the Jacobian of the change of variables, and dropping it is the single most common multiple-integral error. In three dimensions the corresponding factors are $r$ for cylindrical and $\rho^2 \sin \phi$ for spherical coordinates. When the integrand is impossible in the given order, changing the order of integration is usually the intended move, and it requires re-deriving the limits from a sketch rather than swapping them in place.

Vector calculus closes the domain. A line integral $\int_C \vec{F} \cdot d\vec{r}$ measures work along a path. If $\vec{F} = \nabla f$ the integral is just $f(\text{end}) - f(\text{start})$, so recognising a conservative field is worth more than any parametrisation. On a simply connected domain in the plane, $\vec{F} = (P, Q)$ is conservative exactly when $Q_x = P_y$ — and the simple connectivity is a genuine hypothesis, not a formality. For non-conservative fields, Green's theorem trades the loop integral for a double integral over the enclosed region, the divergence theorem trades a closed-surface flux for a triple integral of $\nabla \cdot \vec{F}$, and Stokes' theorem trades a loop integral for a flux of $\nabla \times \vec{F}$ through any surface with that loop as its boundary.