14. Calculus

Learning objectives

14.1Understand derived functions• Understand the idea of a derived function.• Only an informal understanding of limits is expected; differentiation from first principles is not required.14.2Use differentiation notation• Use notation for derivatives and increments, including f'(x), f''(x), dy/dx, d^2y/dx^2, delta y / delta x and the limiting notation as delta x -> 0.14.3Differentiate standard functions• Know and use derivatives of x^n for any rational n, sin x, cos x, tan x, e^x and ln x.• Includes constant multiples, sums and composite functions using the chain rule.• For trigonometric functions, angles are always in radians.14.4Differentiate products and quotients• Differentiate products and quotients of functions.14.5Find gradients, tangents and normals• Use differentiation to find gradients, tangents and normals.14.6Find stationary points• Use differentiation to find stationary points.• Points of inflexion are not included.14.7Apply differentiation to rates and approximations• Apply differentiation to connected rates of change, small increments and approximations.14.8Apply differentiation to maxima and minima• Apply differentiation to practical problems involving maxima and minima.14.9Use derivative tests for maxima and minima• Use first and second derivative tests to discriminate between maxima and minima.• Points of inflexion are not included.• Give full justification of conclusions.• Explain how to distinguish between a maximum point and a minimum point when required.• Unless specified otherwise, any valid method is allowed.14.10Understand integration as reverse differentiation• Understand integration as the reverse process of differentiation.• Solutions for indefinite integrals should include an arbitrary constant.14.11Integrate powers of x• Integrate sums of terms in powers of x, including 1/x and 1/(ax + b).• Solutions for indefinite integrals should include an arbitrary constant.14.12Integrate standard composite functions• Integrate functions of the form (ax + b)^n for any rational n, sin(ax + b), cos(ax + b), sec^2(ax + b) and e^(ax+b).• Includes the case where n = -1.• For trigonometric functions, angles are always in radians.• Solutions for indefinite integrals should include an arbitrary constant.14.13Evaluate definite integrals and areas• Evaluate definite integrals and apply integration to plane areas.• Plane areas include areas between a line and a curve, between two curves, and a sum of two areas.14.14Apply calculus to kinematics• Apply differentiation and integration to kinematics problems involving displacement, velocity and acceleration of a particle moving in a straight line with variable or constant acceleration.14.15Use kinematics graphs• Use the relationships in 14.14 to draw and use displacement-time, distance-time, velocity-time, speed-time and acceleration-time graphs.

The derived function $f'(x)$ gives the gradient of $y=f(x)$ at each value of $x$ . It turns a curve into a new function describing how quickly and in which direction the original function is changing.

Between $x$ and $x+\Delta x$ , the average gradient is $\Delta y/\Delta x$ . As the second point moves closer and $\Delta x$ tends to $0$ , the secant gradient approaches the tangent gradient. That limiting gradient is the derivative at $x$ .

For $f(x)=x^2$ , the average gradient from $x=3$ to $x=3.1$ is $(3.1^2-3^2)/0.1=6.1$ . Moving the second point still closer makes this value approach $6$ , which is $f'(3)$ because the derived function is $f'(x)=2x$ .

A derivative is not one fixed gradient unless the original graph is a straight line. Only an informal limit idea is required here; deriving rules from first principles is outside this syllabus.

Differentiation notation tells you both the order of change and the variables involved. Several symbols can name the same first derivative, but finite changes and limiting change must not be confused.

Notation	Meaning
$f'(x)$ or $\dfrac{dy}{dx}$	first derivative; gradient/rate of change
$f''(x)$ or $\dfrac{d^2y}{dx^2}$	derivative of the first derivative
$\dfrac{\Delta y}{\Delta x}$	average change over a non-zero interval
$\displaystyle\lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}$	instantaneous derivative $\dfrac{dy}{dx}$

If $y$ depends on $x$ , $dy/dx$ means differentiate $y$ with respect to $x$ . Differentiating again gives $d/dx(dy/dx)=d^2y/dx^2$ ; the square belongs to the derivative order, not to the value $dy/dx$ .

$\Delta y/\Delta x$ uses actual finite increments and is generally only an average. The symbols $dy$ and $dx$ belong to derivative notation and should not be replaced casually by finite changes.

Differentiate each standard outer function, then multiply by the derivative of its inner expression. This chain-rule factor is what adjusts the rate for a composite function.

$f(x)$	$f'(x)$
$x^n$	$nx^{n-1}$ for rational $n$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2x$
$e^x$	$e^x$
$\ln x$	$1/x$

\frac{d}{dx}f(g(x))=f'(g(x))g'(x)

For $y=(3x^2+4)^{1/3}$ , differentiate the outer power and multiply by $6x$ : $dy/dx=\frac13(3x^2+4)^{-2/3}(6x)=2x(3x^2+4)^{-2/3}$ .

Trigonometric derivative rules require angles in radians. Preserve the domain of the original function, especially for rational powers and $\ln x$ , and do not omit the inner derivative.

When two changing functions are multiplied or divided, differentiating each part separately is not enough. Keep the product or quotient structure visible until every derivative is in place.

(uv)'=u'v+uv',\qquad \left(\frac uv\right)'=\frac{u'v-uv'}{v^2}

For $y=x^2e^{3x}$ , take $u=x^2$ and $v=e^{3x}$ . Then $dy/dx=2xe^{3x}+x^2(3e^{3x})=xe^{3x}(2+3x)$ . The chain rule is still needed inside $v'$ .

For a quotient, write the numerator in the order $u'v-uv'$ before simplifying. Reversing those terms changes the sign. A quotient may instead be rewritten as a product with a negative power when that makes the structure clearer.

The derivative of $uv$ is not $u'v'$ , and the derivative of $u/v$ is not $u'/v'$ . Factor only after the full rule has been applied so that terms are not lost.

At $x=a$ , the derivative $f'(a)$ is the tangent gradient. A normal is perpendicular to the tangent, so its gradient is the negative reciprocal when the tangent gradient is non-zero.

m_{\text{tan}}=f'(a),\qquad m_{\text{normal}}=-\frac1{m_{\text{tan}}},\qquad y-f(a)=m(x-a)

For $y=\ln(4x-3)$ at $x=2$ , the point is $(2,\ln5)$ and $dy/dx=4/(4x-3)$ , so $m_{\text{tan}}=4/5$ . The tangent is $y-\ln5=\frac45(x-2)$ and the normal is $y-\ln5=-\frac54(x-2)$ .

Use the coordinates of the point on the curve, not just the derivative. If the tangent gradient is $0$ , the tangent is horizontal and the normal is vertical, so the negative-reciprocal formula is not written as a finite number.

A stationary point has a horizontal tangent, so solve $dy/dx=0$ for its $x$ -coordinate and then substitute into the original curve to find its $y$ -coordinate.

Differentiate, factor or otherwise solve the derivative equation, reject values outside the stated domain, and recover every corresponding point from the original function. A derivative equation gives candidate $x$ -values, not complete coordinates.

For $y=x^3-2x^2-4x+8$ , $dy/dx=3x^2-4x-4=(3x+2)(x-2)$ . Hence $x=-2/3$ or $x=2$ , giving stationary points $(-2/3,256/27)$ and $(2,0)$ .

This step locates stationary points but does not by itself determine whether each is a maximum or minimum. Points of inflexion are outside this syllabus, so do not add them as a third required category.

A derivative converts a small change or rate in one variable into the corresponding change or rate in another. The same chain relationship controls connected rates and small-increment approximations.

Situation	Relationship
connected rates	$\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}$
small increment	$\Delta y\approx\dfrac{dy}{dx}\Delta x$ at the starting value of $x$

If $V=\frac43\pi r^3$ and $dr/dt=0.5\,\mathrm{cm\,s^{-1}}$ , then $dV/dt=4\pi r^2(dr/dt)$ . At $r=0.25\,\mathrm{cm}$ this is $\pi/8\,\mathrm{cm^3\,s^{-1}}$ . Units confirm that a volume rate was found.

For $y=\tan x$ , $dy/dx=\sec^2x$ . Increasing $x$ from $-\pi/4$ by a small amount $h$ gives $\Delta y\approx\sec^2(-\pi/4)h=2h$ .

Evaluate the derivative at the relevant instant or starting value. The small-change relation is an approximation, whereas a connected-rate equation is an instantaneous equality; do not discard signs or units.

A practical maximum or minimum is found only after the constraints have reduced the target quantity to a function of one variable on a realistic domain.

Write the constraint, eliminate extra variables, state the physical domain, differentiate the target function, solve $f'(x)=0$ , and verify that the candidate gives the required maximum or minimum. Finish by reporting the requested dimensions or quantity, not just $x$ .

An open tank with square base $x$ and volume $5$ has height $h=5/x^2$ and metal area $A=x^2+4xh=x^2+20/x$ for $x>0$ . Since $A'=2x-20/x^2$ , the stationary value satisfies $x^3=10$ . Also $A''=2+40/x^3>0$ , so the tank has minimum area at $x=\sqrt[3]{10}$ and $h=5/(\sqrt[3]{10})^2$ .

A stationary value outside the physical domain is invalid. A calculus candidate is not automatically the requested optimum: justify its nature and check any relevant endpoint or feasibility condition.

A maximum or minimum conclusion needs evidence about how the gradient changes, not merely the equation $f'(x)=0$ .

Test at $x=a$	Maximum	Minimum
first derivative	$f'$ changes $+\to-$	$f'$ changes $-\to+$
second derivative, given $f'(a)=0$	$f''(a)<0$	$f''(a)>0$

If stationary points occur at $x=0.5$ and $x=3$ , with $f''(x)=4x-7$ , then $f''(0.5)=-5<0$ , so the first is a maximum; $f''(3)=5>0$ , so the second is a minimum.

If $f''(a)=0$ , the second derivative test is inconclusive, so use the first derivative test or another valid argument. State the derivative value or sign change and the resulting conclusion; points of inflexion are not included.

Integration reverses differentiation. If $F'(x)=f(x)$ , then every function $F(x)+C$ has derivative $f(x)$ because the derivative of a constant is zero.

\int f(x),dx=F(x)+C\quad\Longleftrightarrow\quad F'(x)=f(x)

Because $d(2x^3)/dx=6x^2$ , $\int6x^2\,dx=2x^3+C$ . The constant is essential: $2x^3+1$ and $2x^3-7$ have the same derivative. If a point on the curve is given, substitute it after integrating to determine $C$ .

Sometimes a known derivative exposes a less obvious integral. For example, differentiating $xe^{4x}$ gives $e^{4x}+4xe^{4x}$ , so the relationship can be rearranged to recover an antiderivative involving $xe^{4x}$ .

An indefinite integral is a family of functions and must include $C$ . A definite integral has numerical limits and therefore does not carry an arbitrary constant in its final value.

For powers of $x$ , increase the exponent by $1$ and divide by the new exponent. The exponent $-1$ is the exception and produces a logarithm.

\int x^n,dx=\frac{x^{n+1}}{n+1}+C\ (n\ne-1),\qquad \int\frac1x,dx=\ln|x|+C

\int\frac1{ax+b},dx=\frac1a\ln|ax+b|+C\quad(a\ne0)

Rewrite before integrating. For example, $\int\left(4x+5-\frac1{2x+3} ight)dx=2x^2+5x-\frac12\ln|2x+3|+C$ . Differentiating the result recovers the integrand.

Do not apply the power formula with $n=-1$ , because it would divide by zero. The factor $1/a$ compensates for the derivative of $ax+b$ ; absolute-value bars may be omitted only when the domain guarantees $ax+b>0$ .

For a standard function of $ax+b$ , use its usual antiderivative and divide by the inner derivative $a$ . This is the reverse chain rule.

Integrand	Antiderivative
$(ax+b)^n$ , $n\ne-1$	$\dfrac{(ax+b)^{n+1}}{a(n+1)}+C$
$1/(ax+b)$	$\dfrac1a\ln\|ax+b\|+C$
$\sin(ax+b)$	$-\dfrac1a\cos(ax+b)+C$
$\cos(ax+b)$	$\dfrac1a\sin(ax+b)+C$
$\sec^2(ax+b)$	$\dfrac1a\tan(ax+b)+C$
$e^{ax+b}$	$\dfrac1a e^{ax+b}+C$

$\int\sin((\phi+\pi)/3)\,d\phi=-3\cos((\phi+\pi)/3)+C$ , because differentiating the cosine contributes an extra factor $1/3$ as well as the negative sign.

Trigonometric integration assumes radians. Check the result by differentiating: a missing reciprocal inner factor is revealed immediately. The case $n=-1$ uses the logarithmic row, not the power row.

A definite integral gives signed accumulation. A geometric area must be non-negative, so identify intersections and integrate upper function minus lower function on every interval where their order is fixed.

\int_a^b f(x),dx=F(b)-F(a),\qquad A=\int_a^b(\text{upper}-\text{lower}),dx

Solve for all relevant intersection or axis-crossing values, sketch the ordering mentally from the equations, split the interval when the upper curve changes or when a curve crosses the axis, then add the positive area pieces.

Between $y=6x-x^2$ and $y=x$ from $x=0$ to $x=5$ , the curve is above the line. The area is $\int_0^5(5x-x^2)dx=[\frac52x^2-\frac13x^3]_0^5=125/6$ .

A negative definite integral is signed area, not a negative physical area. For a region made from two pieces or two curves that exchange order, split at the crossing and add magnitudes rather than allowing cancellation.

For straight-line motion, differentiation moves from displacement to velocity to acceleration; integration moves back and introduces constants fixed by the particle's stated conditions.

v=\frac{ds}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2},\qquad s=\int v,dt,\qquad v=\int a,dt

If $v=6t-12$ and $s=5$ at $t=0$ , then $a=6$ and $s=3t^2-12t+C$ . The initial condition gives $C=5$ , so $s=3t^2-12t+5$ . The particle is instantaneously at rest when $v=0$ , at $t=2$ .

The sign of $v$ gives direction. To find total distance over an interval, find every time at which $v=0$ , split the interval there, and add the magnitudes of the displacement changes. Directly integrating velocity gives net displacement, not necessarily distance.

Keep displacement, distance, velocity, speed and acceleration distinct and include units. A zero acceleration means constant velocity, not necessarily zero velocity; a zero velocity at one instant does not mean the particle remains at rest.

A kinematics graph carries information through its slope and signed area. Identify the vertical quantity first: displacement and velocity may be negative, while distance and speed are non-negative.

Graph	Slope	Area under graph
displacement-time	velocity	not used here as a standard kinematics quantity
distance-time	speed; never negative	not used here as a standard kinematics quantity
velocity-time	acceleration	displacement change
speed-time	rate of change of speed	distance travelled
acceleration-time	not required here	velocity change

A horizontal displacement-time segment means rest; a straight segment means constant velocity. On a velocity-time graph, crossing the time axis marks an instant of rest and usually a direction change; the gradient gives acceleration, so a horizontal velocity segment has zero acceleration.

Areas below the time axis on a velocity-time graph count negatively towards displacement but positively towards total distance after taking their magnitudes. A distance-time graph cannot decrease, even when the particle reverses direction.

Do not confuse a graph's height with its slope. For example, a velocity of zero is read from the height of a velocity-time graph, whereas zero acceleration is shown by a horizontal tangent or segment on that graph.

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—Cambridge IGCSE Additional Mathematics (0606)Read

14. Calculus

Learning objectives

See a derivative as changing gradient

Read differentiation notation correctly

Differentiate standard and composite functions

Choose the product or quotient rule

Build tangents and normals from a derivative

Locate stationary points exactly

Use a derivative as a local conversion factor

Turn a practical constraint into an optimum

Justify maxima and minima with derivative tests

Reverse differentiation with an arbitrary constant

Integrate powers and logarithmic exceptions

Integrate standard composite functions

Use definite integrals to measure plane area

Connect displacement, velocity and acceleration

Translate between kinematics graphs

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