Application of Derivatives QRT – Quick Revision Tables bring the important concepts, formulas and results of the chapter together in one place. Use these tables for a quick recap while revising, practising questions or preparing for your exams.
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Table 1: Application of Derivatives QRT – Rate of Change of Quantities
| Concept | Formula / Result |
|---|---|
| Rate of Change | If \displaystyle y=f(x), then \displaystyle \frac{dy}{dx} gives the rate of change of \displaystyle y with respect to \displaystyle x. |
| Rate at a Point | At \displaystyle x=c, the rate of change is \displaystyle \left.\frac{dy}{dx}\right|_{x=c}=f'(c) |
| Rate with Time | If a quantity \displaystyle y changes with time \displaystyle t, its rate of change is \displaystyle \frac{dy}{dt}. |
| Related Rates | If \displaystyle x and \displaystyle y both depend on \displaystyle t, then \displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}, provided \displaystyle \frac{dx}{dt}\ne0. |
| Chain Rule in Rate Problems | If \displaystyle y=f(x) and \displaystyle x varies with time, then \displaystyle \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt} |
| Sign of Rate | \displaystyle \frac{dy}{dx}\gt0 means \displaystyle y increases as \displaystyle x increases; \displaystyle \frac{dy}{dx}\lt0 means \displaystyle y decreases. |
| Circle | Area: \displaystyle A=\pi r^2, so \displaystyle \frac{dA}{dt}=2\pi r\frac{dr}{dt} |
| Rectangle — Perimeter | \displaystyle P=2(x+y), so \displaystyle \frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right) |
| Rectangle — Area | \displaystyle A=xy, so \displaystyle \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt} |
| Cube — Volume | \displaystyle V=x^3, so \displaystyle \frac{dV}{dt}=3x^2\frac{dx}{dt} |
| Cube — Surface Area | \displaystyle S=6x^2, so \displaystyle \frac{dS}{dt}=12x\frac{dx}{dt} |
| Marginal Cost | If \displaystyle C(x) is the total cost of producing \displaystyle x units, then \displaystyle MC=\frac{dC}{dx} |
| Marginal Revenue | If \displaystyle R(x) is the total revenue from selling \displaystyle x units, then \displaystyle MR=\frac{dR}{dx} |
| Working Rule — Related Rates |
1. Identify the quantities that vary with time. 2. Write the relation connecting them. 3. Differentiate with respect to \displaystyle t. 4. Substitute the given values and rates. 5. State the answer with correct units. |
Maths Better Tip… In related-rate problems, differentiate the required relation with respect to time first, then substitute the given values. Keep the signs carefully: “increasing” gives a positive rate, while “decreasing” gives a negative rate.
Table 2: Application of Derivatives QRT – Increasing & Decreasing Functions
| Concept | Formula / Result |
|---|---|
| Increasing Function | On an interval \displaystyle I, \displaystyle x_1\lt x_2\Rightarrow f(x_1)\lt f(x_2). As \displaystyle x increases, \displaystyle f(x) also increases. |
| Decreasing Function | On an interval \displaystyle I, \displaystyle x_1\lt x_2\Rightarrow f(x_1)\gt f(x_2). As \displaystyle x increases, \displaystyle f(x) decreases. |
| Constant Function | \displaystyle f(x)=k, where \displaystyle k is constant, throughout the interval. |
| Increasing at a Point | \displaystyle f is increasing at \displaystyle x_0 if it is increasing in some open interval containing \displaystyle x_0. |
| Decreasing at a Point | \displaystyle f is decreasing at \displaystyle x_0 if it is decreasing in some open interval containing \displaystyle x_0. |
| Derivative Test — Increasing | If \displaystyle f is continuous on \displaystyle [a,b], differentiable on \displaystyle (a,b) and \displaystyle f'(x)\gt0 for every \displaystyle x\in(a,b), then \displaystyle f is increasing on \displaystyle [a,b]. |
| Derivative Test — Decreasing | If \displaystyle f is continuous on \displaystyle [a,b], differentiable on \displaystyle (a,b) and \displaystyle f'(x)\lt0 for every \displaystyle x\in(a,b), then \displaystyle f is decreasing on \displaystyle [a,b]. |
| Derivative Test — Constant | If \displaystyle f'(x)=0 for every \displaystyle x\in(a,b), then \displaystyle f is constant on \displaystyle [a,b]. |
| General Sign Condition | For a continuous function, \displaystyle f'(x)\ge0 throughout an interval implies \displaystyle f is increasing; similarly, \displaystyle f'(x)\le0 implies \displaystyle f is decreasing. |
| Where to Check | To find intervals of increase/decrease, consider the points where \displaystyle f'(x)=0 and where \displaystyle f'(x) is not defined, and use them to divide the domain into disjoint intervals. |
| Sign-Chart Method | Find \displaystyle f'(x) → solve \displaystyle f'(x)=0 → mark the critical points → divide the domain into disjoint intervals → determine the sign of \displaystyle f'(x) in each interval → conclude increasing where \displaystyle f'(x)\gt0 and decreasing where \displaystyle f'(x)\lt0. |
| Sign of Derivative | \displaystyle f'(x)\gt0 indicates increasing behaviour; \displaystyle f'(x)\lt0 indicates decreasing behaviour; \displaystyle f'(x)=0 alone does not determine whether the function is increasing or decreasing. |
Maths Better Tip… For interval questions, never stop after finding \displaystyle f'(x)=0. These points only divide the domain. The sign of \displaystyle f'(x) in each resulting interval tells you where the function is increasing or decreasing.
Table 3: Application of Derivatives QRT – Maxima & Minima Basics
| Concept | Formula / Result |
|---|---|
| Maximum Value | A function \displaystyle f has a maximum value at \displaystyle x=c if \displaystyle f(x)\le f(c) for all \displaystyle x sufficiently close to \displaystyle c. |
| Minimum Value | A function \displaystyle f has a minimum value at \displaystyle x=c if \displaystyle f(x)\ge f(c) for all \displaystyle x sufficiently close to \displaystyle c. |
| Local Maxima | \displaystyle x=c is a point of local maxima (or maximum) if \displaystyle f(c) is the greatest value of \displaystyle f in some open interval containing \displaystyle c. |
| Local Minima | \displaystyle x=c is a point of local minima (or minimum) if \displaystyle f(c) is the least value of \displaystyle f in some open interval containing \displaystyle c. |
| Local Maximum Value | If \displaystyle c is a point of local maximum, then \displaystyle f(c) is the local maximum value of \displaystyle f. |
| Local Minimum Value | If \displaystyle c is a point of local minimum, then \displaystyle f(c) is the local minimum value of \displaystyle f. |
| Turning Point | A local maximum or local minimum is a turning point where the function changes from increasing to decreasing or from decreasing to increasing. |
| Absolute Maximum | \displaystyle f(c) is the absolute maximum value on a domain if \displaystyle f(c)\ge f(x) for every \displaystyle x in the domain. |
| Absolute Minimum | \displaystyle f(c) is the absolute minimum value on a domain if \displaystyle f(c)\le f(x) for every \displaystyle x in the domain. |
| Local vs Absolute | Local: compare with nearby points. Absolute: compare with every point in the given domain or interval. |
| Point vs Value | The point is the value of \displaystyle x where the extremum occurs; the corresponding value is \displaystyle f(c). |
| Open Interval | A function on an open interval need not have an absolute maximum or minimum. For example, \displaystyle f(x)=x+2 on \displaystyle (0,1) has neither. |
| Closed Interval | A continuous function on a closed interval \displaystyle [a,b] has both an absolute maximum and an absolute minimum. |
Maths Better Tip… Do not confuse the point with the value: if the extremum occurs at \displaystyle x=c, the point is \displaystyle c, while the corresponding maximum or minimum value is \displaystyle f(c). For absolute extrema, consider the entire given domain or interval.
Table 4: Application of Derivatives QRT – Critical Points & Necessary Condition for Maxima/Minima
| Concept | Formula / Result |
|---|---|
| Critical Point | A point \displaystyle x=c in the domain of \displaystyle f is a critical point if \displaystyle f'(c)=0 or \displaystyle f is not differentiable at \displaystyle c. |
| Stationary Point | A point \displaystyle x=c where \displaystyle f'(c)=0 is called a stationary point. |
| Necessary Condition for Local Extremum | If \displaystyle f has a local maximum or minimum at \displaystyle x=c, then either \displaystyle f'(c)=0 or \displaystyle f is not differentiable at \displaystyle c. |
| Important Warning | \displaystyle f'(c)=0 is not sufficient to conclude that \displaystyle x=c is a local maximum or minimum. |
| Example — No Extremum | For \displaystyle f(x)=x^3, \displaystyle f'(0)=0, but \displaystyle x=0 is neither a local maximum nor a local minimum. |
| Critical Points | Critical points are points where a local maxima or minima may occur. They must be tested further using the first or second derivative test. |
| Derivative Does Not Exist | A point where \displaystyle f'(c) does not exist can be a critical point and may be a local maximum or minimum. |
| Point of Inflexion — Key Case | If \displaystyle f'(x) does not change sign through a critical point, the point is neither a local maximum nor a local minimum; it is called a point of inflexion. |
| What to Check First | Find the critical points by solving \displaystyle f'(x)=0 and identifying points where \displaystyle f is not differentiable, provided the points lie in the domain. |
| Working Rule | Find \displaystyle f'(x) → find critical points → test each critical point. Do not declare a maxima or minima merely because \displaystyle f'(c)=0. |
Maths Better Tip… Think of a critical point as a “possibility”, not an answer. A critical point only tells you where to look; the first or second derivative test tells you whether a local maxima or minima actually occurs.
Table 5: Application of Derivatives QRT – First Derivative Test
| Concept | Formula / Result |
|---|---|
| First Derivative Test | Let \displaystyle f be continuous at a critical point \displaystyle x=c. Examine the sign of \displaystyle f'(x) immediately to the left and right of \displaystyle c. |
| Local Maximum | If \displaystyle f'(x) changes from positive to negative as \displaystyle x moves from left to right of \displaystyle c, then \displaystyle c is a point of local maxima and \displaystyle f(c) is the local maximum value. |
| Local Minimum | If \displaystyle f'(x) changes from negative to positive as \displaystyle x moves from left to right of \displaystyle c, then \displaystyle c is a point of local minimum and \displaystyle f(c) is the local minimum value. |
| No Sign Change | If \displaystyle f'(x) does not change sign through \displaystyle c, then \displaystyle c is neither a local maxima nor a local minima. |
| Sign Change: \displaystyle +\to- | Function changes from increasing to decreasing → local maxima. |
| Sign Change: \displaystyle -\to+ | Function changes from decreasing to increasing → local minima. |
| No Sign Change | \displaystyle +\to+ or \displaystyle -\to- → neither a local maxima nor a local minima. |
| Point of Inflexion | If \displaystyle f'(x) does not change sign through a critical point \displaystyle c, then \displaystyle c is a point of inflexion. For example, \displaystyle f(x)=x^3 at \displaystyle x=0. |
| Sign-Chart Working | Find critical points → arrange them on the number line → choose a test point in each interval → determine the sign of \displaystyle f'(x) → apply the sign-change rule. |
| Quick Sign Pattern | \displaystyle +\ |\ c\ |\ - → maxima. \displaystyle -\ |\ c\ |\ + → minima. Same signs → neither. |
| Important Point | The first derivative test works by studying the change in monotonicity of the function around the critical point. |
Maths Better Tip… For the first derivative test, remember the two decisive patterns: \displaystyle +\to- gives a local maximum and \displaystyle -\to+ gives a local minimum. No sign change means neither.
View or Download the free Application of Derivatives QRT – Class 12 Maths PDF for quick revision anytime.
Table 6: Application of Derivatives QRT – Second Derivative Test
| Concept | Formula / Result |
|---|---|
| Second Derivative Test | Let \displaystyle x=c be a critical point where \displaystyle f'(c)=0 and \displaystyle f''(x) exists in a neighbourhood of \displaystyle c. Examine \displaystyle f''(c). |
| Local Maximum | If \displaystyle f'(c)=0 and \displaystyle f''(c)\lt0, then \displaystyle f(c) is a local maximum value. |
| Local Minimum | If \displaystyle f'(c)=0 and \displaystyle f''(c)\gt0, then \displaystyle f(c) is a local minimum value. |
| If \displaystyle f''(c)=0 | The second derivative test is inconclusive. Use the first derivative test to determine the nature of the critical point. |
| Quick Sign Rule | \displaystyle f''(c)\lt0 → local maxima. \displaystyle f''(c)\gt0 → local minima. |
| What Must Be Checked | The test requires a critical point satisfying \displaystyle f'(c)=0 and the existence of \displaystyle f'' in a neighbourhood of \displaystyle c. |
| First vs Second Test | First derivative test: checks the sign change of \displaystyle f'(x). Second derivative test: uses \displaystyle f''(c) when \displaystyle f'(c)=0. |
| Working Rule | 1. Find \displaystyle f'(x). 2. Solve \displaystyle f'(x)=0. 3. Find \displaystyle f''(x). 4. Evaluate \displaystyle f''(c). 5. Apply the sign rule. |
| Common Trap | Do not conclude “no extremum” (maxima/minima) when \displaystyle f''(c)=0. The test simply fails to decide; use the first derivative test. |
Maths Better Tip… Second derivative test shortcut: at a critical point with \displaystyle f'(c)=0, \displaystyle f''(c)\lt0 gives a local maximum and \displaystyle f''(c)\gt0 gives a local minimum. If \displaystyle f''(c)=0, the test is inconclusive.
Table 7: Application of Derivatives QRT – Absolute Maximum & Minimum on a Closed Interval
| Concept | Formula / Result |
|---|---|
| Absolute Maximum | \displaystyle f(c) is the absolute maximum value of \displaystyle f on \displaystyle [a,b] if \displaystyle f(c)\ge f(x) for every \displaystyle x\in[a,b]. |
| Absolute Minimum | \displaystyle f(c) is the absolute minimum value of \displaystyle f on \displaystyle [a,b] if \displaystyle f(c)\le f(x) for every \displaystyle x\in[a,b]. |
| Existence Theorem | If \displaystyle f is continuous on a closed interval \displaystyle [a,b], then \displaystyle f has both an absolute maximum and an absolute minimum on \displaystyle [a,b]. |
| Where Can Absolute Extrema Occur? | For a continuous function on \displaystyle [a,b], an absolute maximum or minimum can occur at an interior critical point or at an endpoint \displaystyle a or \displaystyle b. |
| Critical Points | Find points in the interval where \displaystyle f'(x)=0 or \displaystyle f is not differentiable, provided the points belong to the domain. |
| Working Rule | 1. Find all critical points of \displaystyle f in the interval. 2. Take the end points of the interval. 3. Evaluate \displaystyle f at all these points. 4. The greatest value is the absolute maximum; the least value is the absolute minimum. |
| Candidate Values | Compare \displaystyle f(a),\ f(b) and \displaystyle f(c) for every critical point \displaystyle c in the interval. |
| Point vs Value | The point is the value of \displaystyle x where the extremum occurs; the corresponding value is \displaystyle f(c). |
| Repeated Values | If the greatest or least value occurs at more than one point, all such points give the corresponding absolute extremum value. |
| Important Caution | Do not check only \displaystyle f'(x)=0. The end points \displaystyle a and \displaystyle b must also be checked. |
Maths Better Tip… For absolute maximum/minimum on \displaystyle [a,b], remember: Critical Points + Endpoints → Evaluate → Compare. Never skip the endpoints.
QRT 8: Application of Maxima & Minima
| Concept | Formula / Result |
|---|---|
| Optimization | Maxima/minima are used to find the largest or smallest possible value of a quantity such as distance, area, volume, cost or profit, subject to given conditions. |
| Objective Function | Express the quantity to be maximised or minimised as a function \displaystyle y=f(x) of one variable. |
| Use the Constraint | Use the given condition to eliminate other variables and express the required quantity in terms of a single variable. |
| Domain | Determine the physically meaningful values of the variable from the conditions of the problem before finding the extremum. |
| Critical Point | Find \displaystyle f'(x) and solve \displaystyle f'(x)=0. Also check points where \displaystyle f is not differentiable, if they lie in the relevant domain. |
| Test for Maximum / Minimum | Use the first derivative test or second derivative test to determine the nature of the critical point. |
| Distance Problems | Write the distance or its square as a function of one variable. Minimising \displaystyle d^2 also minimises the positive distance \displaystyle d. |
| Area Problems | Express the area in terms of one variable using the given geometric constraint, then differentiate to find its maximum or minimum. |
| Volume Problems | Express the volume in terms of one variable using the given constraint, then find the critical point and test it. |
| Surface Area Problems | Express the required surface area as a function of one variable using the given dimensions or constraint, then optimise it. |
| Cost & Revenue | Express cost or revenue as a function of the number of units, then use derivatives to find the required optimum. |
| Profit | If \displaystyle C(x) is cost and \displaystyle R(x) is revenue, then \displaystyle P(x)=R(x)-C(x). Maximise \displaystyle P(x) using derivatives. |
| Working Rule — Optimization |
1. Identify the quantity to maximise/minimise. 2. Define the variable. 3. Use the constraint to form a one-variable function. 4. Determine its domain. 5. Find critical points. 6. Test the critical point(s). 7. State the required value with appropriate units. |
| Final Check | Check that the answer satisfies the original conditions and represents the required maximum or minimum, not merely a critical value. |
Maths Better Tip… In optimisation problems, the most important step is often forming the correct one-variable function. Use the given constraint first, then differentiate. A critical point is only a candidate until it is tested.
QRT 9: Important Mensuration & Optimisation Results
| Figure / Problem | Formula / Important Result |
|---|---|
| Circle | Area \displaystyle A=\pi r^2; circumference \displaystyle C=2\pi r |
| Sphere | Volume \displaystyle V=\frac{4}{3}\pi r^3; surface area \displaystyle S=4\pi r^2 |
| Cylinder | Volume \displaystyle V=\pi r^2h; curved surface area \displaystyle 2\pi rh; total surface area \displaystyle 2\pi r(r+h) |
| Cone | Volume \displaystyle V=\frac13\pi r^2h; curved surface area \displaystyle \pi rl; \displaystyle l^2=r^2+h^2 |
| Rectangle | Area \displaystyle A=xy; perimeter \displaystyle P=2(x+y) |
| Trapezium | Area \displaystyle A=\frac12(a+b)h, where \displaystyle a,b are the parallel sides. |
| Open Box | After cutting squares of side \displaystyle x from a rectangular sheet \displaystyle l\times b, \displaystyle V=x(l-2x)(b-2x) |
| Distance | To minimise a positive distance \displaystyle d, it is often simpler to minimise \displaystyle d^2. |
| Fixed Sum — Maximum Product | For two positive numbers with a fixed sum, their product is maximum when the two numbers are equal. |
| Rectangle in a Circle | Among rectangles inscribed in a given fixed circle, the square has maximum area. |
| Cylinder in a Sphere | For the cylinder of maximum volume inscribed in a sphere of radius \displaystyle R, its height is \displaystyle \frac{2R}{\sqrt3}. |
| Cylinder in a Cone | For the cylinder of maximum volume inscribed in a right circular cone, the height of the cylinder is one-third of the height of the cone. |
| Cylinder — Given Surface | For a right circular cylinder of given surface and maximum volume, the height equals the diameter of the base. |
| Cylinder — Given Volume | Among closed right circular cylinders of given volume, the one with minimum surface area has height equal to the diameter of the base. |
| Cone — Given Volume | The right circular cone of least curved surface and given volume has altitude equal to twice the radius of the base. |
| Cone in a Sphere | The largest cone that can be inscribed in a sphere of radius \displaystyle R has volume \displaystyle \frac{8}{27} of the volume of the sphere. |
| Profit | \displaystyle P(x)=R(x)-C(x). Maximum profit is found by optimising \displaystyle P(x). |
| Unit Check | Length → units; area → square units; volume → cubic units; rate of area → units²/time; rate of volume → units³/time. |
Maths Better Tip… In AOD optimisation problems, the geometry gives the constraint and the derivative finds the optimum. First build the correct mathematical relation; then differentiate.
Table 10: Application of Derivatives QRT – Quick Problem-Solving Results & Common Traps
| Quick Check | Result / What to Do |
|---|---|
| Find Increasing Intervals | Find \displaystyle f'(x) → find points where \displaystyle f'(x)=0 or \displaystyle f is not differentiable → divide the domain into intervals → check the sign of \displaystyle f'(x). |
| Increasing / Decreasing | \displaystyle f'(x)\gt0 → increasing; \displaystyle f'(x)\lt0 → decreasing. |
| Critical Point | Check points where \displaystyle f'(x)=0 or \displaystyle f is not differentiable, provided they belong to the domain. |
| Critical Point ≠ Extremum | Finding \displaystyle f'(c)=0 only gives a candidate. The point must be tested. |
| First Derivative Test | \displaystyle +\to- → local maximum; \displaystyle -\to+ → local minimum; no sign change → neither. |
| Second Derivative Test | At \displaystyle f'(c)=0: \displaystyle f''(c)\lt0 → maximum; \displaystyle f''(c)\gt0 → minimum; \displaystyle f''(c)=0 → inconclusive. |
| Absolute Maximum / Minimum | On a closed interval, evaluate \displaystyle f at all critical points and both endpoints, then compare the values. |
| Local vs Absolute | Local extremum → compare nearby values. Absolute extremum → compare values over the entire given domain or interval. |
| Optimisation Workflow | Objective → constraint → one-variable function → domain → critical points → test → required value. |
| Related Rates | Write the relation → differentiate with respect to \displaystyle t → substitute the given values/rates → state the answer with units. |
| Distance Problems | To minimise a positive distance \displaystyle d, it is often easier to minimise \displaystyle d^2. |
| Profit Problems | Use \displaystyle P(x)=R(x)-C(x), then find the production level giving maximum profit. |
| Do Not Ignore Domain | Critical points outside the allowed domain are not candidates for the required problem. |
| Do Not Skip Endpoints | For absolute extrema on \displaystyle [a,b], always check the endpoints along with the critical points. |
| Do Not Stop at \displaystyle f'(x)=0 | \displaystyle f'(c)=0 does not by itself prove a maximum or minimum. Apply an appropriate test. |
| Final Answer Check | Check the sign, domain, units and original condition. Make sure the result answers exactly what the question asks. |
Maths Better Tip… For AOD problems, the derivative is only part of the solution. Always check the domain, test the critical point when required, include endpoints for absolute extrema, and give the final answer with correct units.
Quick Questions — Test Yourself
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📘 Application of Derivatives QRT – Class 12 Maths
10 Quick Revision Tables + Quick Q&A
