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CBSE Class 12 · Mathematics

Application of Derivatives

Official NCERT chapter from Mathematics Part I–II (book code lemh1). ExamMaster notes are original teaching at CBSE Class 12 depth.

This lesson follows the official chapter “Application of Derivatives” in Mathematics Part I–II. The words below are ExamMaster’s teaching, not a paste from the book. Use the NCERT chapter for the classroom sequence; use these notes to hold the idea without copying exercises or figures.

  • CBSE Class 12
  • Medium level
  • 8 concepts

1Rate of Change of Quantities

Rate of change: dy/dx as a rate when x is time or another given. If y=x^2 and dx/dt=2 at x=3, dy/dt=2x dx/dt=12. Rate is a linked-change, not a pretty slope-sticker.

Forgetting to multiply by dx/dt is a chain-skip.

Figure. The rate at an instant is the slope of the tangent at that point, not the chord across a minute.

How it works

  1. Name y and the given rate of xThe given.
  2. Form dy/dt = (dy/dx)(dx/dt)The rate.
  3. Keep units if the lesson used themHonest.

Linked rate

y=x^2, x=3, dx/dt=2. Find dy/dt.

  • dy/dx2x=6
  • dy/dt6×2=12
  • Ready changes 12 units per unit time

Pro tip. Chain the taught rates.

2Increasing and Decreasing Functions

Increasing / decreasing: f′>0 on an interval means increasing as taught; f′<0 decreasing. A sign-chart of f′ is the test. f(x)=x^2 is decreasing on (−∞,0) and increasing on (0,∞) as the sign of 2x says.

A graph that “looks up” with no f′ sign is a weak claim.

Figure. f is increasing while the graph climbs, decreasing while it falls. The peak is where the slope changes sign.

How it works

  1. Find f′ and its sign on the intervalThe test.
  2. Read + as increasing, − as decreasingThe word.
  3. Keep endpoints as the lesson framed themClosed or open.

3Maxima and Minima

Maxima and minima: f′=0 (or undefined) as candidates; first or second derivative test as taught. f(x)=x^2 has a min at 0. Max/min is a candidate-plus-test, not “the tallest-looking point”.

Calling every f′=0 a max is a miss — it may be a min or a flat.

Figure. A local max is a hill. A local min is a valley. Neither has to be the highest or lowest on a closed interval.

How it works

  1. Solve f′=0 (and ends if a closed interval)Candidates.
  2. Use first- or second-derivative test as taughtThe sort.
  3. Read the y-value as the max/min valueThe height.

4A definition is a test you can run

Application-word is a test: a linked rate, a sign of f′, or a max/min candidate. If you only say “slope”, you have a heading.

A hill-sticker is not the test.

Figure. The first-derivative test: plus then minus at c means a local max. The sign change is the test, not the name.

How it works

  1. Name rate, increase, or extremumThe object.
  2. Give the school sentenceThe test.
  3. Then the word has contentThe definition ran.

5Name the given before the unknown

The given is y(x) and a number. The unknown is the rate or the extremum. Copy x=3, dx/dt=2 before you output 6 as dy/dt.

Using dy/dx as dy/dt is a silent skip of the chain.

Figure. Name walk and ladder before asking for wall. A 6-8-10 right triangle keeps the rate pair honest. The square is the right angle, not to scale with the forces.

How it works

  1. Copy the rule and the given rate or intervalThe given.
  2. Name dy/dt, increase, or maxThe unknown.
  3. Then computeGiven first.

6One worked case is enough at this class

One worked case is enough: y=x^2, dy/dt=12 at that instant. Do not stack four related-rates. The case teaches the chain.

A poster of ten hills is not more science.

Figure. One worked slope: at P, f'=2x. The curve starts at the origin and rises faster as x grows.

How it works

  1. Take one linked rate or one minThe case.
  2. Run the testThe teach.
  3. Stop — one caseThis class.

7A miss: swapping the school name for the picture

A miss: swapping the school name for a picture. “Local min” is a test on f′ — a valley-drawing is a setting.

A pretty valley without f′ is a mute picture.

Figure. On a closed interval the largest value can sit at an endpoint. A turning point is local; A is the interval max here.

How it works

  1. Keep the taught name (rate, increase, min)The science.
  2. Use a picture only as a settingHonest.
  3. Refuse a swap of name for glowThe miss named.

8Check by the opposite action or the opposite test

Check by the opposite: if you claimed increasing, f′ should not be negative; if you claimed a min, a nearby left-down right-up (or f″>0) should hold. The opposite sign kills the claim.

A single plotted point as “proved max” is a miss.

Figure. The opposite test at a claimed max: slope must flip from plus to minus. Plus to plus is not a max.

How it works

  1. Name the claimIncrease or min.
  2. Look for the opposite sign of f′The check.
  3. If the opposite sits, redoHonest.
If y=x^2, x=3, dx/dt=2, then dy/dt is
  1. 12
  2. 6
  3. 2

2x times dx/dt.

Notes

  • The official chapter title is “Application of Derivatives”. Teach the school test for that title, not a contest shortcut.
  • If a step needs a later class, stop. The next official chapter will pick it up.

Formulas

  • dy/dt=(dy/dx)(dx/dt)
  • f′>0 increasing; f′<0 decreasing

Recap

Hold these pegs from the official chapter “Application of Derivatives”. The wording is ExamMaster’s teaching, not a textbook recap.

Rate of Change of Quantities
Rate of change: dy/dx as a rate when x is time or another given.
Increasing and Decreasing Functions
Increasing / decreasing: f′>0 on an interval means increasing as taught; f′<0 decreasing.
Maxima and Minima
Maxima and minima: f′=0 (or undefined) as candidates; first or second derivative test as taught.
A definition is a test you can run
Application-word is a test: a linked rate, a sign of f′, or a max/min candidate.
Name the given before the unknown
The given is y(x) and a number. The unknown is the rate or the extremum. Copy x=3, dx/dt=2 before you output 6 as dy/dt.
One worked case is enough at this class
One worked case is enough: y=x^2, dy/dt=12 at that instant.

Practise Application of Derivatives

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