5 Calculus

A derivative is a local rate of change.

The derivative is the limit of average change as the interval shrinks; it describes the tangent slope at a point.

Example

For f(x)=x², (f(1+h)−f(1))/h=2+h, so f′(1)=2 as h→0.

State the variable and point before differentiating, then interpret the sign and units of the result.

A derivative at one point is not the total change across an interval.

At SL, estimate $\lim_{x\to a}f(x)$ from values approaching $a$ on both sides or from the graph; formal analytic limit methods are not required. The derivative is the limiting gradient and may be written $dy/dx$, $f'(x)$, $dV/dr$ or $ds/dt$, with units of output per unit input. A limit can exist even when the displayed point is missing, because it describes nearby behaviour rather than only $f(a)$.

The first derivative reveals where a function rises or falls.

If f′(x)>0 the graph increases and if f′(x)<0 it decreases; sign changes identify possible stationary extrema.

Example

For f′(x)=x−2, the function decreases before x=2 and increases after it, so x=2 is a local minimum.

Make a sign chart around every critical value rather than relying on the derivative being zero alone.

A stationary point can be a saddle or inflection; f′=0 is a candidate, not a classification.

The power rule differentiates xⁿ by lowering the exponent.

For a constant n, d(xⁿ)/dx=nxⁿ⁻¹; the factor n records how rapidly the power changes.

Example

d(3x⁵−2x²)/dx=15x⁴−4x.

Apply $d(ax^n)/dx=anx^{n-1}$ term by term for integer exponents in this objective, retaining coefficients and differentiating constants to zero.

The derivative of $x^0$ is zero. Rational powers and chain-rule factors belong to SL 5.6, so do not import them into this integer-power step.

A tangent uses the derivative; a normal uses its negative reciprocal.

At x=a, the tangent slope is f′(a); a perpendicular normal has slope −1/f′(a) when the tangent is neither horizontal nor vertical.

Example

For y=x² at x=1, the tangent is y−1=2(x−1), while the normal is y−1=−(x−1)/2.

Find the point and tangent slope first, then use point–slope form for the requested line.

A vertical tangent has a horizontal normal, so the reciprocal-slope formula needs a separate case.

Integration reverses differentiation up to a constant.

An antiderivative F satisfies F′=f; all antiderivatives differ by a constant because differentiation removes constants.

Example

∫(6x²−4)dx=2x³−4x+C; differentiating checks the result.

Integrate each term, include +C for an indefinite integral and use a condition to determine C.

An indefinite integral is a family, not one function until a condition is supplied.

A boundary condition selects the constant: if $dy/dx=3x^2+x$ and $y(1)=10$, then $y=x^3+x^2/2+C$ gives $C=8.5$. A definite integral accumulates signed change; when $f(x)>0$, technology can evaluate $\int_a^b f(x)\,dx$ as the area between the curve and the $x$-axis. Write the correct integral before calculating.

Product, quotient and chain rules preserve how functions are built.

Differentiate the outer and inner structure rather than expanding blindly: (uv)′=u′v+uv′ and (f(g(x)))′=f′(g)g′.

Example

For y=(x²+1)³, y′=3(x²+1)²·2x=6x(x²+1)².

Name the outer operation and inner function before applying a rule.

The derivative of a product is not the product of derivatives.

Core derivatives: $d(x^n)/dx=nx^{n-1}$ for rational $n$ where defined, $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, $(e^x)'=e^x$, and $(\ln x)'=1/x$. Also $(u/v)'=(u'v-uv')/v^2$. Example: for $y=x^2e^x$, the product rule gives $y'=e^x(x^2+2x)$; for $y=\ln(3x+1)$, the chain rule gives $3/(3x+1)$.

The second derivative describes curvature and acceleration.

f″ measures how the slope changes; its sign indicates concavity and, in kinematics, acceleration.

Example

For f(x)=x³, f″=6x, so the graph is concave down for x<0 and up for x>0.

Use f′ for increasing/decreasing and f″ for curvature; test continuity when claiming an inflection.

A positive second derivative does not mean the function itself is positive.

Optimization compares critical points with the feasible boundary.

A maximum or minimum occurs at a critical point or endpoint of the allowed domain; the model and constraints decide which is meaningful.

Example

For a rectangle with perimeter 20, A=x(10−x) is largest at x=5, not at an unconstrained value outside 0≤x≤10.

Differentiate, solve candidates, then compare all endpoints and check units.

A local maximum need not be the global maximum when the domain is restricted.

Classification: if $f'$ changes from positive to negative, the stationary point is a local maximum; negative to positive gives a local minimum. Alternatively, at $f'(a)=0$, $f''(a)>0$ implies a local minimum and $f''(a)<0$ a local maximum. A point of inflexion requires a change in concavity, so $f''=0$ alone is not sufficient; for example $y=x^4$ has $f''(0)=0$ but no concavity change. Use 'concave-up' for $f''>0$ and 'concave-down' for $f''<0$.

Kinematics links position, velocity and acceleration by differentiation.

If s(t) is position, v=s′ and a=v′=s″; signs depend on the chosen positive direction.

Example

For s=4t²−t, v=8t−1 and a=8, so velocity changes while acceleration stays constant.

State the time interval and direction convention before interpreting motion.

Negative velocity means motion opposite the chosen direction, not necessarily slowing down.

From $t_1$ to $t_2$, displacement is $\int_{t_1}^{t_2}v(t)\,dt$, while total distance is $\int_{t_1}^{t_2}|v(t)|\,dt$. Split the distance integral wherever $v(t)=0$. Speed is $|v|$: an object slows when velocity and acceleration have opposite signs and speeds up when their signs agree.

Indefinite integration reconstructs a family of functions.

Reverse the derivative term by term and retain the arbitrary constant; the constant represents the unknown vertical shift.

Example

∫(2x+sin x)dx=x²−cos x+C.

Differentiate the answer and use any initial value to fix C.

Do not omit C when no initial condition has selected one antiderivative.

Formula set: $\int x^n\,dx=x^{n+1}/(n+1)+C$ for $n\ne-1$, $\int dx/x=\ln|x|+C$, $\int e^x\,dx=e^x+C$, $\int\sin x\,dx=-\cos x+C$, and $\int\cos x\,dx=\sin x+C$. For a linear composite, account for the inner gradient: $\int\cos(2x+3)\,dx=\tfrac12\sin(2x+3)+C$. Reverse-chain recognition gives $\int k g'(x)[f(g(x))]\,dx$ by substitution or inspection.

A definite integral accumulates signed change over an interval.

∫ₐᵇf(x)dx is net signed area; geometric area may require splitting where f changes sign.

Example

∫₋¹¹x dx=0 by cancellation, although the total geometric area is 1.

Find roots and compare the requested quantity—net area, total area or accumulated change.

A negative integral is possible; it does not mean an area has become physically negative.

Fundamental evaluation: if $F'=f$, then $\int_a^b f(x)\,dx=F(b)-F(a)$. For area between $y=f(x)$ and $y=g(x)$, first find intersections, then integrate top minus bottom on each interval: $A=\int_a^b|f(x)-g(x)|\,dx$, splitting wherever their order changes. This preserves geometric area even when the signed integral cancels.

Continuity and differentiability are local conditions.

Continuity requires the function value and both one-sided limits to agree; differentiability additionally requires matching finite slopes.

Example

|x| is continuous at 0 but not differentiable there because its left slope is −1 and right slope is 1.

Check the function, limit and derivative separately at the suspected point.

Differentiability implies continuity, but continuity alone does not imply differentiability.

A limit converges when nearby function values approach one finite value; it diverges when no finite common value exists. For a polynomial, first principles gives $f'(x)=\lim_{h\to0}[f(x+h)-f(x)]/h$; for $f(x)=x^2$, simplifying produces $2x+h\to2x$. Higher derivatives may be written $d^ny/dx^n$ or $f^{(n)}(x)$. At a point, differentiability implies continuity, but a corner, cusp, vertical tangent or discontinuity prevents a finite two-sided derivative.

L’Hôpital’s rule compares leading limiting behaviour.

For suitable 0/0 or ∞/∞ forms, the ratio limit can equal the ratio of derivatives; the indeterminate form must be identified first.

Example

limₓ→0 sin x/x=lim cos x/1=1.

Verify the form and conditions before differentiating, then simplify and re-check the limit.

The rule is not a general quotient shortcut and does not apply directly to every finite ratio.

At a finite point or at infinity, first substitute and confirm an indeterminate $0/0$ or $\infty/\infty$ form. Then $\lim f/g=\lim f'/g'$ when the rule's conditions and the derivative-ratio limit hold; if the new ratio remains indeterminate, repeat. Alternatively use leading Maclaurin terms, for example $\sin x=x-x^3/6+\cdots$, so $\sin x/x\to1$. Products, differences and powers must first be algebraically converted to an eligible quotient or series form.

Implicit differentiation keeps both variables changing.

Differentiate each term with respect to x, treating y as y(x), so every y term contributes a factor dy/dx.

Example

From x²+y²=25, 2x+2y y′=0, hence y′=−x/y where y≠0.

Collect dy/dx terms, then substitute the point only after differentiating.

A vertical tangent can make the solved slope undefined even though the curve is smooth.

Related-rate workflow: write one equation connecting all changing quantities, differentiate with respect to time and only then substitute the instant's values. For a circle, $A=\pi r^2$ gives $dA/dt=2\pi r\,dr/dt$; at $r=3$ and $dr/dt=2$, $dA/dt=12\pi$ square units per unit time. For optimization, include feasible endpoints as candidates when the optimum can occur on the boundary.

Further derivative pairs unlock advanced integrals.

Key derivatives are $(\tan x)'=\sec^2x$, $(\sec x)'=\sec x\tan x$, $(\cosec x)'=-\cosec x\cot x$, $(\cot x)'=-\cosec^2x$, $(a^x)'=a^x\ln a$, $(\log_a x)'=1/(x\ln a)$, $(\arcsin x)'=1/\sqrt{1-x^2}$, $(\arccos x)'=-1/\sqrt{1-x^2}$ and $(\arctan x)'=1/(1+x^2)$.

Example

Because $1/[(x+1)(x+2)]=1/(x+1)-1/(x+2)$, partial fractions give $\int dx/[(x+1)(x+2)]=\ln|x+1|-\ln|x+2|+C$. For $\int dx/[1+(2x+1)^2]$, the linear inner derivative gives $\tfrac12\arctan(2x+1)+C$.

Match an integrand to a derivative pair, include the reciprocal inner-gradient factor for a linear composite, and use partial fractions before integrating a rational expression when required.

Inverse-trig derivative domains and logarithmic absolute values matter; a memorised form without its domain or inner-gradient factor is incomplete.

Advanced integration techniques expose hidden structure.

Substitution reverses a chain rule. Integration by parts reverses a product rule: $\int u\,dv=uv-\int v\,du$; repeat it when the remaining integral still contains a product suited to the same process.

Example

For $\int xe^x\,dx$, choose $u=x$ and $dv=e^x dx$. Then $du=dx$, $v=e^x$, so the integral is $xe^x-\int e^x dx=e^x(x-1)+C$.

Use substitution when one factor is the derivative of an inner function; use parts when differentiating one factor simplifies it. Transform limits as well as the differential in a definite substitution.

Partial fractions belong to AHL 5.15. In this objective, do not mix them into the choice between substitution and integration by parts.

Volumes of revolution depend on slices and axis choice.

Rotating a region creates disks, washers or shells; the radius and thickness must match the axis and variable of integration.

Example

Rotating y=x from 0 to 1 about the x-axis gives V=π∫₀¹x²dx=π/3.

Sketch the region, identify outer/inner radius or shell height, and state the limits.

Using a radius measured from the wrong axis can produce a plausible but incorrect volume.

Area between $x=f(y)$ and the $y$-axis from $y=c$ to $y=d$ is $A=\int_c^d|f(y)|\,dy$, splitting where the curve crosses the axis. Rotation about the $x$-axis gives $V=\pi\int_a^b(R(x)^2-r(x)^2)\,dx$; about the $y$-axis use radii expressed in $y$, $V=\pi\int_c^d(R(y)^2-r(y)^2)\,dy$. Sketch the region so the radius is measured perpendicular to the chosen axis.

A differential equation becomes a model after a condition selects a solution.

The equation states a rate relationship; solving it gives a family, and an initial or boundary condition chooses the member that fits the system.

Example

dy/dx=2y with y(0)=3 gives y=3e^(2x), not the whole family Ce^(2x).

Separate variables or use the appropriate method, then substitute the condition and check the derivative.

A mathematical solution can still be physically invalid if it violates domain or sign constraints.

Method map: separable equations rearrange to $g(y)dy=f(x)dx$; homogeneous $dy/dx=F(y/x)$ uses $y=vx$; linear $y'+P(x)y=Q(x)$ uses integrating factor $\mu=e^{\int P(x)dx}$ so $(\mu y)'=\mu Q$. Euler's method for $y'=f(x,y)$ uses $x_{n+1}=x_n+h$ and $y_{n+1}=y_n+h f(x_n,y_n)$. Apply the initial condition after integration and check the solution in the original equation.

A Maclaurin series approximates a function near zero.

The series uses derivatives at 0: f(x)=f(0)+f′(0)x+f″(0)x²/2!+…; truncation creates an approximation error.

Example

e^x≈1+x+x²/2 for small x, so e^.1≈1.105 using three terms.

State the expansion point and order, then judge whether the input lies in a range where the truncation is useful.

A convergent series can still be inaccurate when too few terms are used far from the expansion point.

Standard series: $e^x=1+x+x^2/2!+\cdots$, $\sin x=x-x^3/3!+x^5/5!-\cdots$, $\cos x=1-x^2/2!+x^4/4!-\cdots$, $\ln(1+x)=x-x^2/2+x^3/3-\cdots$, and $(1+x)^p=1+px+p(p-1)x^2/2!+\cdots$. Obtain related series by substituting, multiplying, differentiating or integrating within the valid interval; a differential equation can determine successive coefficients after substituting $y=\sum a_nx^n$.