5 Calculus

A limit describes the value that $f(x)$ approaches as $x$ approaches a point. At SL, estimate it from a graph or from table values on both sides; formal analytic limit calculations are not required.

The derivative is the gradient function and an instantaneous rate of change. Notation identifies the changing quantities: $f'(x)$ or $dy/dx$ for a function, $dV/dr$ for volume changing with radius, and $ds/dt$ for displacement changing with time.

Example

If table values of $f(x)$ are 3.98 at $x=1.99$ and 4.02 at $x=2.01$, they support $\lim_{x\to2}f(x)\approx4$. If the tangent gradient there is 5 metres per second, the derivative gives the local rate and its units.

The function value and limit need not agree when there is a hole, and one-sided behaviour may disagree at a jump. Do not substitute $x=2$ blindly; inspect values approaching from both sides.

Where f'(x)>0, f is increasing locally; where f'(x)<0, f is decreasing locally. A stationary point has f'(x)=0, but the sign must be checked on either side to classify the change.

Find critical x-values, split the domain into intervals and test the derivative sign. This produces a variation table and keeps domain restrictions visible instead of relying on a sketch.

For f(x)=x³−3x, f'(x)=3x²−3. The derivative is positive outside x=−1 and x=1 and negative between them, so the function rises, falls, then rises again.

f'(x)=0 does not automatically mean a maximum or minimum. A horizontal inflection can have zero derivative without changing from increasing to decreasing.

For $f(x)=ax^n$ with integer $n$, $f'(x)=anx^{n-1}$. Differentiate a sum term by term and treat a constant as having derivative zero.

Negative integer powers are included where the original function is defined: $d(x^{-2})/dx=-2x^{-3}$. Keep coefficients, signs and domain restrictions visible before simplifying.

Example

For $f(x)=4x^3-2x^{-1}+7$, $f'(x)=12x^2+2x^{-2}=12x^2+2/x^2$, with $x\ne0$ inherited from the original function.

This SL rule is bounded to integer exponents. Rational powers, chain, product and quotient rules belong to AHL 5.9; do not import them into this Objective.

At x=a, the tangent gradient is f'(a). If that gradient is non-zero, the normal gradient is −1/f'(a), because perpendicular non-vertical lines have product of gradients −1.

Find the point (a,f(a)), calculate the tangent gradient, then use point–gradient form. Handle a horizontal tangent separately: its normal is vertical and cannot be written with a finite gradient.

For y=x² at x=1, the point is (1,1), tangent gradient 2 and tangent y−1=2(x−1). The normal gradient is −1/2, giving y−1=−(x−1)/2.

A normal is not the negative of the tangent gradient. It is the negative reciprocal, and the vertical/horizontal special cases must be stated.

For integer $n\ne-1$, $\int ax^n\,dx=ax^{n+1}/(n+1)+C$. An indefinite integral is a family of anti-derivatives; a boundary condition determines the constant $C$.

A definite integral $\int_a^b f(x)\,dx$ connects anti-derivatives with accumulated signed area. At SL, use technology after first writing the correct integral; for a region above the x-axis, the integral equals its geometric area.

Example

If $dy/dx=3x^2+x$ and $y=10$ when $x=1$, then $y=x^3+x^2/2+C$ and $10=1+1/2+C$, so $C=8.5$. Thus $y=x^3+x^2/2+8.5$.

Do not omit $C$ in an indefinite integral or apply the power formula to $x^{-1}$. A differential-equation growth model belongs to AHL, not this SL anti-differentiation Objective.

A stationary point occurs at an interior value where $f'(x)=0$. Use technology when appropriate to generate $f'(x)$ and solve for the x-values, then calculate the corresponding y-values.

Classify a local maximum when the function changes from increasing to decreasing and a local minimum when it changes from decreasing to increasing. Compare endpoints as well when the greatest or least value on a restricted domain is required.

Example

For $f(x)=x^3-3x$, $f'(x)=3x^2-3=0$ at $x=\pm1$. The derivative changes $+$ to $-$ at $-1$ and $-$ to $+$ at $1$, giving a local maximum and minimum respectively.

$f'(x)=0$ alone does not guarantee an extremum, and a local extremum need not be the absolute extremum on the stated domain. The second-derivative test belongs to AHL 5.10.

Optimization maximises or minimises a specified objective subject to constraints. Calculus can locate interior candidates where the derivative is zero, but endpoints, feasibility and the meaning of the variables decide the answer.

Translate the context into an objective function, state the domain, solve for critical points and compare all feasible candidates. A minimum cost or maximum area is a claim about the whole allowed interval, not just a local curve shape.

For a rectangle with fixed perimeter 20, A=x(10−x) on 0<x<10. A'(x)=10−2x gives x=5, and comparing the endpoints' limiting values confirms the square gives the largest area.

A stationary point is not automatically the global optimum. Check the full feasible domain and endpoints, state units, and interpret the best value in context. Kinematics questions are not set in SL examinations.

For equally spaced x-values with width h, the trapezoidal rule approximates ∫ₐᵇ f(x)dx by h/2[f(x₀)+2f(x₁)+⋯+2f(xₙ₋₁)+f(xₙ)]. It estimates signed area, not automatically total geometric area.

The endpoint ordinates have weight one and interior ordinates weight two. More strips usually reduce error for a smooth curve, but curvature and the number of strips still determine the approximation quality.

Using two strips for f(x)=x² on [0,2] gives h=1 and (1/2)[0+2(1)+4]=3, while the exact integral is 8/3. The difference is the discretisation error.

Do not use the rule with the wrong h or omit the doubled interior values. If the curve crosses the axis, signed integral and total area require different treatment.

Know $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, $(\tan x)'=\sec^2x$, $(e^x)'=e^x$, $(\ln x)'=1/x$ and $(x^n)'=nx^{n-1}$ for rational $n$ on the appropriate domain.

Use chain $(f(g(x)))'=f'(g(x))g'(x)$, product $(uv)'=u'v+uv'$ and quotient $(u/v)'=(u'v-uv')/v^2$. Identify the outer structure before choosing a rule, and combine rules when functions are nested.

Example

For $y=x^2e^{3x}$, $dy/dx=2xe^{3x}+3x^2e^{3x}=xe^{3x}(2+3x)$. In related rates, differentiate the connecting equation with respect to time, then substitute values with units only after differentiating.

Do not confuse this Objective with second derivatives in AHL 5.10. Product and quotient rules require both terms, and the chain rule requires the derivative of the inner function.

Concavity describes whether a curve bends above or below its tangent trend. The sign of the second derivative gives this local curvature when the function is twice differentiable.

An inflection point is where concavity changes, not merely where f''=0. Check the sign on both sides and include points where f'' is undefined if the original function is defined there.

For f(x)=x³, f''(x)=6x changes from negative to positive at x=0, so the origin is a stationary inflection: f'(0)=0 but the function keeps increasing through it.

A zero second derivative alone does not prove an inflection or extremum. Use neighbouring intervals and the domain, not one substituted value.

Use $\int x^n\,dx=x^{n+1}/(n+1)+C$ for rational $n\ne-1$, $\int x^{-1}\,dx=\ln|x|+C$, $\int\sin x\,dx=-\cos x+C$, $\int\cos x\,dx=\sin x+C$, $\int\sec^2x\,dx=\tan x+C$ and $\int e^x\,dx=e^x+C$.

For $\int f(g(x))g'(x)\,dx$, let $u=g(x)$ or recognize the reverse chain rule. Constant factors must be adjusted so the derivative of the inner function is present.

Example

$\int4x\sin(x^2)\,dx$: take $u=x^2$, so $du=2x\,dx$ and the integral becomes $2\int\sin u\,du=-2\cos(x^2)+C$.

Differentiate the result to verify it. Do not use the ordinary power rule for $x^{-1}$, omit absolute values in $\ln|x|$, or forget the inner derivative adjustment.

Rotating a region about an axis creates thin discs, washers or shells. About the x-axis, a disc method uses V=π∫[radius]²dx; washers subtract the inner radius squared from the outer.

Choose the variable and axis so the radius and bounds are explicit. If the region crosses the axis or the radii change order, split the integral rather than silently using one formula.

Rotating y=x on 0≤x≤2 about the x-axis gives V=π∫₀²x²dx=8π/3. The radius is y=x, not the horizontal coordinate itself by definition.

An area integral is not automatically a volume. Square the radius, include π, and check whether the axis and region create overlapping solids.

For area between a curve and an axis, split at intersections and use absolute geometric pieces when the question asks for area because an integral below the axis is negative. For revolution about the y-axis, $V=\pi\int_a^b x^2\,dy$; about the x-axis, $V=\pi\int_a^b y^2\,dx$. Express both bounds and radius in the integration variable.

For position s(t), velocity v(t)=ds/dt and acceleration a(t)=dv/dt=d²s/dt². Integrating velocity or acceleration needs initial conditions to recover the earlier quantity.

The sign depends on the chosen positive direction. A negative velocity means motion opposite that direction; a negative acceleration does not necessarily mean the object is slowing if velocity is also negative. Keep the sign convention fixed throughout the interval.

If s(t)=t³−6t²+9t, then v=3t²−12t+9 and a=6t−12. At t=1, v=0 but a=−6, so the object is instantaneously stationary while its velocity is changing.

Distance is not signed displacement. Use |v| or integrate speed for total distance, and use the stated time interval when locating turning points.

Also use $a=v\,dv/ds$ when velocity is given as a function of displacement. Over $[t_1,t_2]$, displacement is $\int_{t_1}^{t_2}v(t)\,dt$ while total distance is $\int_{t_1}^{t_2}|v(t)|\,dt$; split at zeros of $v$. Dot notation $\dot x$ and $\ddot x$ denotes first and second time derivatives.

Build a differential equation by translating the stated rate relationship and units. A separable equation $dy/dx=g(x)h(y)$ can be rearranged as $dy/h(y)=g(x)\,dx$ and integrated on both sides.

The integration constant gives the general solution; an initial condition selects a particular solution. For proportional change $dy/dx=ky$, separation gives $\ln|y|=kx+C$ and hence $y=Ae^{kx}$ on the relevant branch.

Example

If $dP/dt=0.2P$ and $P(0)=100$, then the general non-zero solution is $P=Ae^{0.2t}$ and the condition gives $A=100$, so $P(t)=100e^{0.2t}$.

Do not introduce a logistic capacity term unless the context states it. Keep equilibrium solutions that may be lost by division, and distinguish a general solution from the one satisfying initial data.

A slope field places a small line segment with slope dy/dx=f(x,y) at many points. A solution curve follows the local directions and cannot cross itself where the differential equation gives a unique slope.

Equilibria appear where f(x,y)=0, because the segments are horizontal there. Draw trajectories from an initial condition by following the field, while remembering that the sketch is qualitative unless a scale is stated.

For dy/dx=y−x, the line y=x has zero slope segments. A solution starting above that line initially rises more steeply; the field helps predict the direction before numerical work.

A slope field is not the solution curve and its segments are not vectors showing speed. Read slope, position and equilibrium separately.

For dy/dx=f(x,y), Euler's method uses y_(n+1)=y_n+h f(x_n,y_n). It replaces the curve on each short interval with the tangent at the current point.

Choose a step h, state the starting point and keep the same update rule. Smaller h usually reduces local error for a smooth problem, but accumulated error and model uncertainty remain.

For dy/dx=y, y(0)=1 and h=0.1, the first Euler step gives y₁=1+0.1(1)=1.1, while the exact value e^{0.1}≈1.105. The approximation is close but not identical.

Use the slope at the current point, not the next point, and do not confuse a numerical estimate with an exact solution. Check units of h and the direction of the step.

For a coupled system, update every component from the same current state: $x_{n+1}=x_n+h f_1(x_n,y_n,t_n)$ and $y_{n+1}=y_n+h f_2(x_n,y_n,t_n)$. Do not use the newly updated $x_{n+1}$ inside the same Euler step for $y$ unless a different method is explicitly specified. Technology or a spreadsheet may generate the table.

A phase portrait plots state variables against one another rather than against time. Arrows or trajectories show how the system evolves, and fixed points are states where all derivatives are zero.

Classify a fixed point by nearby trajectories: convergence suggests stability, while divergence suggests instability. Nullclines where one derivative is zero help locate and interpret the flow without solving every trajectory.

In a predator–prey portrait, a closed orbit can indicate repeating population states; it does not mean the two populations are constant. The axes and direction arrows are essential evidence.

A phase portrait is not a time-series graph. It may hide the speed of motion, and a visually central point is not automatically an equilibrium or stable state.

For $\dot x=ax+by$, $\dot y=cx+dy$ with distinct non-zero eigenvalues: positive real parts move away from the origin; negative real parts move toward it; complex values spiral; purely imaginary values give circles or ellipses; real values of opposite signs give a saddle. Exact solutions are required only for distinct real eigenvalues, using eigenvector modes.

For $d^2x/dt^2=f(x,dx/dt,t)$, define $y=dx/dt$. Then rewrite the equation as the coupled system $dx/dt=y$ and $dy/dt=f(x,y,t)$.

With step $h$, update simultaneously from the current state: $x_{n+1}=x_n+hy_n$ and $y_{n+1}=y_n+h f(x_n,y_n,t_n)$. Initial position and velocity provide the two starting values.

Example

For $x''=-x$, $x(0)=1$, $x'(0)=0$ and $h=0.1$, set $y=x'$. One Euler step gives $x_1=1+0.1(0)=1$ and $y_1=0+0.1(-1)=-0.1$.

The syllabus requires numerical Euler and phase-portrait treatment, not a general characteristic-equation solution. Use both current values in one update and do not confuse velocity $y$ with position $x$.