Single-Variable Derivatives & Integrals

A derivative compresses nearby behavior into a local linear model; an integral assembles many small contributions into a net total. Both are limits, and the Fundamental Theorem of Calculus connects them: continuous rates accumulate into a function whose derivative recovers the rate, while an antiderivative evaluates the corresponding accumulation between endpoints.

One physical quantity, two different questions #

A derivative answers “how fast is this changing now?” An integral answers “how much change accumulated over an interval?” They operate on the same model but return different kinds of information, with different units.

This unit check is a fast conceptual test: a derivative divides output units by input units, while an integral multiplies by the integration variable's units.

The derivative is local sensitivity #

Geometrically, secant slopes approach the tangent slope. In a model, the units matter: if $s(t)$ is measured in metres and $t$ in seconds, then $s'(t)$ has units metres per second. A derivative is not “just a formula”—it is a conversion factor from a small input change to a first-order output change.

If $f$ is differentiable at $a$, then

$$f(a+h)=f(a)+f'(a)h+r(h), \qquad \frac{r(h)}{h}\to0.$$

The linear term $f'(a)h$ dominates the remainder at sufficiently small scale. This is the precise reason tangent-line approximation works.

A cusp, corner, jump, or sufficiently wild oscillation can destroy differentiability. A vertical tangent is also not an ordinary real derivative: if difference quotients diverge to an infinite magnitude, report a vertical tangent or an infinite-slope limit, not the equation $f'(a)=\infty$ as though infinity were a real number.

The rules come from the limit #

For an integer $n\ge1$, the binomial theorem gives

$$(x+h)^n-x^n =nx^{n-1}h+\binom{n}{2}x^{n-2}h^2+\cdots+h^n.$$

After division by $h$, every term except $nx^{n-1}$ still contains a positive power of $h$ and vanishes as $h\to0$. Hence

$$\frac{d}{dx}x^n=nx^{n-1}.$$

The same difference-quotient algebra produces the sum, product, and quotient rules. Their compact forms are:

$$(af+bg)'=af'+bg',$$

$$(fg)'=f'g+fg',$$

$$\left(\frac{f}{g}\right)' =\frac{f'g-fg'}{g^2}, \qquad g\ne0.$$

The denominator condition belongs to the rule. A symbolic derivative outside the original real domain is not a derivative of the original function there.

The chain rule tracks nested change #

If $y=f(u)$ and $u=g(x)$, differentiability of $g$ at $x$ and of $f$ at $g(x)$ gives

$$\frac{dy}{dx} =f'(g(x))g'(x).$$

The outer function sees a small change $\Delta u\approx g'(x)\Delta x$, then converts it to $\Delta y\approx f'(u)\Delta u$. Multiplying the two local conversion factors yields the chain rule.

Useful elementary derivatives include

$$(e^x)'=e^x, \qquad (\ln x)'=\frac1x\quad(x>0),$$

$$(\sin x)'=\cos x, \qquad (\cos x)'=-\sin x.$$

For $x\ne0$, $(\ln\lvert x\rvert)'=1/x$. Logarithmic differentiation should therefore use $\ln\lvert y\rvert$ when the differentiated expression may be negative and should state where the original expression is nonzero. For a real power $x^\alpha$, the largest real domain depends on $\alpha$; the familiar formula $\alpha x^{\alpha-1}$ must be applied only where the chosen real power is differentiable.

Implicit, inverse, and parametric derivatives #

Suppose a curve is specified by $F(x,y)=0$. Differentiating along a differentiable branch $y(x)$ gives

$$F_x(x,y)+F_y(x,y)y'=0,$$

so

$$y'=-\frac{F_x}{F_y}$$

where $F_y\ne0$. The denominator is not an optional algebraic detail: it is the local condition that lets the curve be represented as a graph $y(x)$.

For a parametric curve $x=x(t)$, $y=y(t)$,

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

when $dx/dt\ne0$. If both numerator and denominator vanish, the displayed ratio is indeterminate; simplify the parametrization or analyze the limiting tangent rather than cancelling blindly.

For an inverse $f^{-1}$, if $f$ is locally one-to-one, differentiable, and $f'(x)\ne0$, then

$$(f^{-1})'(f(x))=\frac{1}{f'(x)}.$$

All three formulas express the same principle: local invertibility fails when the relevant rate in the denominator vanishes.

From derivatives to shape and bounds #

The mean value theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in(a,b)$ satisfies

$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

This theorem turns local derivative information into interval-wide conclusions:

• if $f'>0$ throughout an interval, $f$ is strictly increasing there;
• if $f'=0$ throughout an interval, $f$ is constant there;
• if $\lvert f'\rvert\le M$, then $\lvert f(b)-f(a)\rvert\le M\lvert b-a\rvert$;
• critical points are candidates for extrema, but endpoints and nondifferentiable domain points must also be checked.

The second derivative tracks change in slope. Where $f''>0$, slopes increase and the graph is convex (often called concave up in elementary calculus); where $f''<0$, slopes decrease. A sign change of $f''$ can certify an inflection point, but merely solving $f''=0$ cannot: $f(x)=x^4$ has $f''(0)=0$ without changing convexity.

The integral is signed accumulation #

For a partition $a=x_0<x_1<\cdots<x_n=b$, choose a sample $x_i^*$ in each subinterval and form

$$\sum_{i=1}^n f(x_i^*)\Delta x_i, \qquad \Delta x_i=x_i-x_{i-1}.$$

If these sums approach one value as the maximum subinterval width tends to zero, that value is the definite integral:

$$\int_a^b f(x)\,dx.$$

Continuous functions on a closed interval are Riemann integrable. The integral is signed: regions where $f<0$ subtract from the total. The geometric area between the graph and the axis is instead $\int_a^b\lvert f(x)\rvert\,dx$, usually split at sign changes.

Units multiply. If $v(t)$ is metres per second, then $\int_{t_0}^{t_1}v(t)\,dt$ is net displacement in metres. Integrating speed $\lvert v(t)\rvert$ gives total distance.

Do not conflate two notations:

$$\int_a^b f(x)\,dx$$

is a number, while

$$\int f(x)\,dx=F(x)+C$$

denotes the family of antiderivatives satisfying $F'=f$ on an interval. The constant is required because differentiation erases additive constants.

Why differentiation and integration invert one another #

Define an accumulation function for continuous $f$:

$$A(x)=\int_a^x f(t)\,dt.$$

Then

$$\frac{A(x+h)-A(x)}{h} =\frac1h\int_x^{x+h}f(t)\,dt.$$

The right-hand side is the average value of $f$ on the shrinking interval. Continuity forces that average toward $f(x)$, so

$$A'(x)=f(x).$$

This is the first part of the Fundamental Theorem of Calculus. The evaluation part says that if $F'=f$ on $[a,b]$ and $f$ is continuous, then

$$\int_a^b f(x)\,dx=F(b)-F(a).$$

The continuity hypotheses are a clean sufficient condition. More advanced integration theories weaken them, but the elementary FTC should not be applied across a singularity without first treating the integral as improper and proving convergence.

Integration techniques reverse derivative rules #

Not every elementary integrand has an elementary antiderivative, so a finite table of reverse rules cannot cover every case. The main hand techniques are structured reversals.

Substitution reverses the chain rule #

If $u=g(x)$ and the needed continuity conditions hold,

$$\int f(g(x))g'(x)\,dx =\int f(u)\,du.$$

For definite integrals, either transform the bounds to $u$-values or substitute back before evaluating; mixing the two coordinate systems is an error.

Integration by parts reverses the product rule #

From $(uv)'=u'v+uv'$,

$$\int u\,dv=uv-\int v\,du.$$

For example,

$$\int xe^x\,dx =xe^x-\int e^x\,dx =e^x(x-1)+C.$$

Differentiate the result to verify it. In repeated integration by parts, track signs and boundary terms explicitly rather than trusting a memorized pattern.

Rational functions require algebra first #

For $P(x)/Q(x)$:

1. perform polynomial division when $\deg P\ge\deg Q$;
2. factor $Q$ over the chosen field;
3. include a full partial-fraction template—one term for every power of a repeated linear factor, and a linear numerator for every power of an irreducible quadratic factor;
4. solve the coefficients and integrate term by term.

The decomposition is valid away from the zeros of $Q$. An antiderivative containing $\ln\lvert x-a\rvert$ is defined separately on intervals that do not cross the pole $x=a$; one constant $C$ need not serve disconnected domain components.

Improper integrals are convergence limits #

An infinite interval or an unbounded integrand is not covered by the ordinary Riemann integral on a closed bounded interval. The notation is defined through a limit. For example,

$$\int_a^\infty f(x)\,dx =\lim_{R\to\infty}\int_a^R f(x)\,dx,$$

provided that the limit exists and is finite. If $f$ is unbounded near the right endpoint instead, then

$$\int_a^b f(x)\,dx =\lim_{t\to b^-}\int_a^t f(x)\,dx.$$

A singularity inside $(a,b)$ requires two integrals split at the singular point, and both must converge. Opposite divergences may cancel in a Cauchy principal value, but that is a different definition and does not make the ordinary improper integral converge.

The power family gives the two reference thresholds. For $p\ne1$,

$$\int_1^R x^{-p}\,dx=\frac{R^{1-p}-1}{1-p},$$

so $\int_1^\infty x^{-p}\,dx$ converges exactly when $p>1$. At the other end, $\int_0^1x^{-p}\,dx$ converges exactly when $p<1$. When $p=1$, the antiderivative is $\ln x$, and the relevant endpoint limit diverges in both cases. The existence of an antiderivative formula therefore does not by itself establish a finite integral.

Practical verification and failure modes #

For a derivative, compare units, domain, and a finite-difference estimate away from singular points. For an indefinite integral, differentiate the candidate. For a definite integral, also check sign, scale, and any symmetry.

A good workflow preserves meaning at every step: state the domain, identify the local or accumulated quantity and its units, choose the rule whose hypotheses match the structure, and verify the result by reversing the operation whenever possible.