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MHT-CET Maths · Applications of Derivative

Increasing and Decreasing Functions

The sign of the derivative decides where a function rises or falls: f prime greater than zero means increasing, f prime less than zero means decreasing. Find where f prime is zero, split the line, and sign-test each piece.

All Applications of Derivative notesMHT-CET Maths strategy

Why this matters

This is the workhorse subtopic of the chapter — 29 PYQs sit directly here (9 HARD, 14 MODERATE, 6 EASY). The moves recur exactly: factor a cubic's f prime and read intervals, use a discriminant to prove f prime keeps one sign, take a rational or rational-trig quotient down to a constant-sign ad minus bc condition, or run a chain-rule sign analysis on a product with exp or log. The recurring MHT-CET traps live here too: the decreasing case needs f prime LESS than zero (so a rational quotient decreasing forces ad minus bc less than zero, not greater), an interval option must be a SUBSET of the true monotonic set, and a strictly-increasing cubic needs its quadratic f prime to have negative discriminant.

Concept 1 of 6

The Sign of the Derivative Decides Monotonicity

Intuition

Where the tangent slopes up the curve rises; where it slopes down the curve falls. So the sign of f′(x)f'(x) on an interval — not the size or sign of ff itself — is what decides whether ff is increasing or decreasing there. The whole subtopic reduces to building the sign chart of f′f'.

Definition

On an interval II:

  • f′(x)>0f'(x) > 0 for all x∈I⇒fx \in I \Rightarrow f is strictly increasing on II.
  • f′(x)<0f'(x) < 0 for all x∈I⇒fx \in I \Rightarrow f is strictly decreasing on II.

Method (the sign chart): solve f′(x)=0f'(x) = 0 (and note where f′f' is undefined); these critical points split the number line into intervals. Test the sign of f′f' in each interval — a factored form like (x−a)(x−b)(x−c)(x-a)(x-b)(x-c) flips sign at each simple root. Where f′f' is ++, ff increases; where −-, it decreases.

Monotonicity from the sign of f prime

f′(x)>0  ⇒  f increasing,f′(x)<0  ⇒  f decreasingf'(x) > 0 \;\Rightarrow\; f \text{ increasing}, \qquad f'(x) < 0 \;\Rightarrow\; f \text{ decreasing}
  • f'(x)the slope of the tangent at xx — its SIGN is all that matters
f′ > 0 ↑f′ < 0 ↓f′ > 0 ↑f′=0

Worked example

On which intervals is f(x)=x3−12x+5f(x) = x^3 - 12x + 5 increasing, and on which decreasing?
  1. f′(x)=3x2−12=3(x−2)(x+2)f'(x) = 3x^2 - 12 = 3(x-2)(x+2).
  2. Critical points where f′=0f'=0: x=−2x = -2 and x=2x = 2; they split the line into three pieces.
  3. Sign-test f′f': on (−∞,−2)(-\infty,-2) it is ++; on (−2,2)(-2,2) it is −-; on (2,∞)(2,\infty) it is ++.
Answer:Increasing on (−∞,−2)(-\infty,-2) and (2,∞)(2,\infty); decreasing on (−2,2)(-2,2).
Practice this conceptself-check · 4 quick reps

Try it yourself

Find where f(x)=2x3−3x2−12x+1f(x) = 2x^3 - 3x^2 - 12x + 1 is decreasing.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    f′(x)<0f'(x) < 0 on an interval means ff is?
  2. 2.
    First step to find monotonic intervals?
  3. 3.
    f(x)=x2−4xf(x)=x^2-4x: where is f′=0f'=0?
  4. 4.
    Where is f(x)=x3−3xf(x)=x^3-3x decreasing?

Monotonicity is decided by the sign of f′f', not by ff

ff increasing   ⟺  f′(x)≥0\iff f'(x) \ge 0 on the interval; decreasing   ⟺  f′(x)≤0\iff f'(x) \le 0. A large or positive VALUE of ff tells you nothing — read the sign of the DERIVATIVE. Build the sign chart of f′f' piece by piece between its zeros.
Drill 3 more on the sign of the derivative decides monotonicity

Concept 2 of 6

Polynomial Monotonicity via a Factored Derivative

Intuition

For a polynomial, f′f' is a lower-degree polynomial that factors. Once f′f' is in factored form, its roots are the only places the sign can change, and a product of linear factors flips sign at each simple root. So factor f′f', mark its roots, and alternate signs across the line.

Definition

For ff a polynomial:

  • Compute f′(x)f'(x) and factor it fully into linear (and irreducible-quadratic) factors.
  • The simple real roots of f′f' are the sign-change points. A product like (x−a)(x−b)(x−c)(x-a)(x-b)(x-c) is ++ to the right of the largest root and alternates as you cross each root going left.
  • A squared factor (double root) does NOT change sign — it touches zero and keeps the same sign on both sides.

Read off the increasing (f′>0f'>0) and decreasing (f′<0f'<0) intervals directly from the chart.

Cubic derivative factors to a quadratic

f(x)=ax3+…  ⇒  f′(x)=3a (x−r1)(x−r2)f(x) = ax^3 + \dots \;\Rightarrow\; f'(x) = 3a\,(x - r_1)(x - r_2)
  • r_1, r_2roots of f′f'; the sign of f′f' flips at each simple root

Worked example

For what xx is f(x)=x3−3x2−9x+4f(x) = x^3 - 3x^2 - 9x + 4 increasing?
  1. f′(x)=3x2−6x−9=3(x−3)(x+1)f'(x) = 3x^2 - 6x - 9 = 3(x-3)(x+1).
  2. Roots x=−1,3x = -1, 3. Sign of the product: ++ on (−∞,−1)(-\infty,-1), −- on (−1,3)(-1,3), ++ on (3,∞)(3,\infty).
  3. Increasing where f′>0f' > 0.
Answer:Increasing on (−∞,−1)(-\infty,-1) and (3,∞)(3,\infty).
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the set where f(x)=[x(x−2)]2f(x) = [x(x-2)]^2 is increasing.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    f(x)=2x3−6x+5f(x)=2x^3-6x+5: increasing set?
  2. 2.
    f(x)=x3−6x2+9x+3f(x)=x^3-6x^2+9x+3: decreasing set?
  3. 3.
    f(x)=x3+6x2−36x+7f(x)=x^3+6x^2-36x+7: increasing set?
  4. 4.
    Does a squared factor of f′f' change its sign?

From the bank · past-year question

Example 2Applications of DerivativeMODERATE
The function f(x)=2x3−9x2+12x+2f(x) = 2x^3-9x^2+12x+2 is decreasing in

[Q118 · 15th May Shift 2 · 2023]

An option must be a SUBSET of the true monotonic set

For f(x)=x2+2xf(x)=\dfrac{x}{2}+\dfrac{2}{x} the true decreasing set is (−2,0)∪(0,2)(-2,0)\cup(0,2). The correct MHT-CET option is (1,2)(1,2) — not because that is the whole set, but because it is a valid SUBSET on which ff decreases. Test each option for 'is this interval inside the monotonic set?', not 'does this equal the full set?'.

Factor f′f' before reading signs

Trying to sign-test f′f' without factoring invites arithmetic slips. Always factor f′(x)f'(x) fully first; the roots are the only sign-change points, and a clean factored product makes the alternating-sign chart automatic.
Drill 8 more on polynomial monotonicity via a factored derivative

Concept 3 of 6

Discriminant Test for a Strictly Monotonic Cubic

Intuition

If f′f' is a quadratic that never crosses zero, it can never change sign — so ff is monotonic on the whole real line. A quadratic with positive leading coefficient and negative discriminant stays strictly positive everywhere, forcing ff to be strictly increasing.

Definition

For a cubic ff, f′(x)=Ax2+Bx+Cf'(x) = Ax^2 + Bx + C (with A>0A > 0). Then:

  • **Discriminant B2−4AC<0B^2 - 4AC < 0** ⇒f′\Rightarrow f' has no real roots ⇒f′(x)>0\Rightarrow f'(x) > 0 for all xx ⇒f\Rightarrow f is **strictly increasing on R\mathbb{R}** (no turning points).
  • Symmetrically, A<0A<0 with B2−4AC<0B^2-4AC<0 gives f′<0f'<0 everywhere (strictly decreasing).

This is the standard way to prove 'increasing throughout the real line' or to impose 'no local extremum' as a parameter condition.

Strictly increasing everywhere

f′(x)=Ax2+Bx+C, A>0, B2−4AC<0  ⇒  f′(x)>0 ∀xf'(x) = Ax^2 + Bx + C,\ A > 0,\ B^2 - 4AC < 0 \;\Rightarrow\; f'(x) > 0 \ \forall x
  • B^2 - 4ACdiscriminant of f′f'; negative means f′f' never touches zero

Worked example

Show that f(x)=x3−10x2+200x−10f(x) = x^3 - 10x^2 + 200x - 10 is increasing throughout R\mathbb{R}.
  1. f′(x)=3x2−20x+200f'(x) = 3x^2 - 20x + 200.
  2. Discriminant =(−20)2−4(3)(200)=400−2400=−2000<0= (-20)^2 - 4(3)(200) = 400 - 2400 = -2000 < 0.
  3. Leading coefficient 3>03 > 0 and no real roots, so f′(x)>0f'(x) > 0 for every xx.
Answer:ff is strictly increasing on the whole real line.
Practice this conceptself-check · 4 quick reps

Try it yourself

If f(x)=x3+bx2+cx+df(x) = x^3 + bx^2 + cx + d with 0<b2<c0 < b^2 < c, what can you say about ff on (−∞,∞)(-\infty,\infty)?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    f′(x)=3x2−20x+200f'(x)=3x^2-20x+200: sign for all xx?
  2. 2.
    Cubic strictly increasing on R\mathbb{R} needs f′f' (a quadratic) to have?
  3. 3.
    3x2+2bx+c3x^2+2bx+c, 0<b2<c0<b^2<c: sign of discriminant?
  4. 4.
    A>0A>0, B2−4AC<0B^2-4AC<0: does the cubic have any local extremum?

From the bank · past-year question

Example 3Applications of DerivativeMODERATE
If f(x)=x3−10x2+200x−10f(x) = x^{3}-10x^{2}+200x-10, then

[Q118 · 15th May Shift 1 · 2023]

'Increasing throughout' is a discriminant statement, not an interval statement

When the options include 'increasing throughout the real line', check the discriminant of f′f' FIRST. If B2−4AC<0B^2-4AC<0 with A>0A>0, f′f' has no roots so there are no intervals to split — the answer is 'increasing everywhere', and any option offering split intervals is a distractor.

0<b2<c0 < b^2 < c is engineered to make the discriminant negative

The condition 0<b2<c0<b^2<c for f′=3x2+2bx+cf'=3x^2+2bx+c gives discriminant 4(b2−3c)4(b^2-3c), and c>b2c>b^2 forces b2−3c<0b^2-3c<0. Recognise this family: whenever the constant term dominates the middle coefficient squared, the quadratic f′f' stays one-signed.
Drill 1 more on discriminant test for a strictly monotonic cubic

Concept 4 of 6

Rational and Rational-Trig Quotients: the ad minus bc Condition

Intuition

For a quotient of two linear-in-(sin, cos) or linear expressions, the quotient rule collapses: the denominator becomes a squared (always positive) term, so the sign of f′f' is entirely decided by ONE constant, ad−bcad-bc. Monotonicity then reduces to the sign of that single constant.

Definition

For f(x)=asin⁡x+bcos⁡xcsin⁡x+dcos⁡xf(x) = \dfrac{a\sin x + b\cos x}{c\sin x + d\cos x}, the quotient rule gives

f′(x)=ad−bc(csin⁡x+dcos⁡x)2.f'(x) = \dfrac{ad - bc}{(c\sin x + d\cos x)^2}.
The denominator is a square, hence >0> 0 wherever defined, so:

  • **ff increasing for all xx**   ⟺  ad−bc>0\iff ad - bc > 0.
  • **ff decreasing for all xx**   ⟺  ad−bc<0\iff ad - bc < 0.

The same collapse happens for a simple rational ax+bcx+d\dfrac{ax+b}{cx+d}: f′=ad−bc(cx+d)2f'=\dfrac{ad-bc}{(cx+d)^2}. A parameter version (e.g. ksin⁡x+2cos⁡xsin⁡x+cos⁡x\dfrac{k\sin x+2\cos x}{\sin x+\cos x}) turns 'strictly increasing' into a linear inequality in the parameter.

Sign of the derivative of a bilinear-trig quotient

f(x)=asin⁡x+bcos⁡xcsin⁡x+dcos⁡x  ⇒  f′(x)=ad−bc(csin⁡x+dcos⁡x)2f(x) = \frac{a\sin x + b\cos x}{c\sin x + d\cos x} \;\Rightarrow\; f'(x) = \frac{ad - bc}{(c\sin x + d\cos x)^2}
  • ad - bcthe ONLY thing whose sign matters; >0>0 increasing, <0<0 decreasing

Worked example

If f(x)=asin⁡x+bcos⁡xcsin⁡x+dcos⁡xf(x) = \dfrac{a\sin x + b\cos x}{c\sin x + d\cos x} is decreasing for all xx, what condition must hold?
  1. Quotient rule: f′(x)=ad−bc(csin⁡x+dcos⁡x)2f'(x) = \dfrac{ad - bc}{(c\sin x + d\cos x)^2}.
  2. The denominator is a square, so it is >0> 0 wherever defined; the sign of f′f' equals the sign of ad−bcad - bc.
  3. Decreasing means f′(x)<0f'(x) < 0 for all xx, so we need ad−bc<0ad - bc < 0.
Answer:ad−bc<0ad - bc < 0.
Practice this conceptself-check · 4 quick reps

Try it yourself

For what kk is f(x)=ksin⁡x+2cos⁡xsin⁡x+cos⁡xf(x) = \dfrac{k\sin x + 2\cos x}{\sin x + \cos x} strictly increasing for all real xx?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    asin⁡x+bcos⁡xcsin⁡x+dcos⁡x\dfrac{a\sin x+b\cos x}{c\sin x+d\cos x} increasing for all xx needs?
  2. 2.
    Why does the denominator not affect the sign of f′f'?
  3. 3.
    f(x)=2x+1x−3f(x)=\dfrac{2x+1}{x-3}: sign of f′f'?
  4. 4.
    ksin⁡x+2cos⁡xsin⁡x+cos⁡x\dfrac{k\sin x+2\cos x}{\sin x+\cos x} strictly increasing needs?

From the bank · past-year question

Example 4Applications of DerivativeMODERATE
If f(x)=asin⁡x+bcos⁡xcsin⁡x+dcos⁡xf(x) = \frac{a\sin x + b\cos x}{c\sin x + d\cos x} is decreasing for all xx, then

[Q120 · Shift 1 · 2023]

Decreasing needs f′<0f' < 0: the sign FLIPS

The single most common error here is setting f′>0f'>0 for a 'decreasing' function. Decreasing means f′(x)<0f'(x) < 0, so a bilinear-trig quotient decreasing forces ad−bc<0ad - bc < 0 — NOT ad−bc>0ad-bc>0. Read 'decreasing' →\to 'negative derivative' →\to 'negative ad−bcad-bc'.

It is ad−bcad - bc, not ab−cdab - cd

The determinant of the coefficient pattern is ad−bcad - bc (main-diagonal minus off-diagonal of abcd\begin{smallmatrix}a&b\\c&d\end{smallmatrix}). Distractor options offer ab−cdab-cd or a swapped sign — write out the quotient-rule numerator once to lock in ad−bcad-bc.
Drill 2 more on rational and rational-trig quotients: the ad minus bc condition

Concept 5 of 6

Products and Composites with exp and log: Chain-Rule Sign Analysis

Intuition

Exponentials are always positive and logs of positive arguments are defined only on part of the line — so when ff is a product or composite involving e(⋅)e^{(\cdot)} or log⁡(⋅)\log(\cdot), the always-positive exponential factor drops out of the sign test, and the monotonicity is decided by the remaining polynomial or rational factor. Differentiate, pull out the guaranteed-positive part, and sign-test what is left.

Definition

Differentiate with the product/chain rule, then isolate the factor whose sign is fixed:

  • eg(x)>0e^{g(x)} > 0 always, so in f′(x)=eg(x)⋅(stuff)f'(x) = e^{g(x)} \cdot (\text{stuff}) the sign is the sign of stuff.
  • ddxlog⁡(u)=u′u\dfrac{d}{dx}\log(u) = \dfrac{u'}{u}; on the domain u>0u>0, the sign is the sign of u′u'.
  • For a composite (f∘g)′(x)=f′(g(x)) g′(x)(f\circ g)'(x) = f'(g(x))\,g'(x), each factor's sign multiplies. Reduce to the product of the non-trivial factors and build their combined sign chart.

The exponential factor drops out of the sign test

f′(x)=eg(x) h(x)  ⇒  sign⁡f′(x)=sign⁡h(x)(since eg(x)>0)f'(x) = e^{g(x)}\,h(x) \;\Rightarrow\; \operatorname{sign} f'(x) = \operatorname{sign} h(x) \quad (\text{since } e^{g(x)} > 0)
  • e^{g(x)}strictly positive — never changes the sign of f′f'
  • h(x)the remaining factor whose sign chart you must build

Worked example

Find the interval on which f(x)=x2e−xf(x) = x^2 e^{-x} strictly increases.
  1. Product rule: f′(x)=2x e−x+x2(−e−x)=xe−x(2−x)f'(x) = 2x\,e^{-x} + x^2(-e^{-x}) = x e^{-x}(2 - x).
  2. e−x>0e^{-x} > 0 always, so sign⁡f′=sign⁡(x(2−x))\operatorname{sign} f' = \operatorname{sign}\big(x(2-x)\big).
  3. x(2−x)>0⇒0<x<2x(2-x) > 0 \Rightarrow 0 < x < 2.
Answer:Strictly increasing on (0,2)(0,2).
Practice this conceptself-check · 4 quick reps

Try it yourself

On what interval is f(x)=(x+2)e−xf(x) = (x+2)e^{-x} increasing?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    f(x)=log⁡xxf(x)=\dfrac{\log x}{x} (x>0)(x>0): increasing set?
  2. 2.
    In f′(x)=eg(x)h(x)f'(x)=e^{g(x)}h(x), what decides the sign?
  3. 3.
    f(x)=log⁡(1+x)−2x2+xf(x)=\log(1+x)-\dfrac{2x}{2+x}: increasing set?
  4. 4.
    f(x)=xex(1−x)f(x)=xe^{x(1-x)}: increasing set?

From the bank · past-year question

Example 5Applications of DerivativeEASY
The set of all points, for which f(x)=x2e−xf(x) = x^2 e^{-x} strictly increases, is

[Q125 · 16th May Shift 1 · 2023]

Don't sign-test the exponential — it is always positive

In f′(x)=xe−x(2−x)f'(x) = xe^{-x}(2-x), the e−xe^{-x} is strictly positive and contributes nothing to the sign. Only the polynomial factor x(2−x)x(2-x) matters. Wasting effort on e−xe^{-x} (or, worse, treating it as sometimes negative) derails the whole sign chart.

For a log, respect the domain before reading the sign

ddxlog⁡(u)=u′u\dfrac{d}{dx}\log(u) = \dfrac{u'}{u} only where u>0u > 0. For f(x)=log⁡e(π+x)log⁡e(e+x)f(x)=\dfrac{\log_e(\pi+x)}{\log_e(e+x)}, the whole analysis lives on the domain where both logs are defined and positive; there π>e\pi>e forces the numerator of f′f' negative, so ff is DECREASING on (0,∞)(0,\infty) — a case where the 'obvious' increasing answer is wrong.
Drill 8 more on products and composites with exp and log: chain-rule sign analysis

Concept 6 of 6

Trigonometric Monotonicity: Reduce to a Single Sinusoid

Intuition

A trig expression's derivative is far easier once the expression is collapsed to a single sin⁡\sin or cos⁡\cos of one angle. Use identities (triple-angle, power-reduction) to rewrite ff, differentiate to a lone ±sin⁡(kx)\pm\sin(kx) or ±cos⁡(kx)\pm\cos(kx), and solve the elementary inequality sin⁡<0\sin < 0 or cos⁡>0\cos > 0 for the monotonic intervals.

Definition

Standard collapses that make the derivative a single sinusoid:

  • Triple angle: 3sin⁡x−4sin⁡3x=sin⁡3x3\sin x - 4\sin^3 x = \sin 3x, so f′=3cos⁡3xf' = 3\cos 3x.
  • Power reduction: sin⁡4x+cos⁡4x=1−12sin⁡22x\sin^4 x + \cos^4 x = 1 - \tfrac12\sin^2 2x, giving f′=−sin⁡4xf' = -\sin 4x.

Then read monotonicity from the sinusoid: f′=3cos⁡3x>0  ⟺  cos⁡3x>0f' = 3\cos 3x > 0 \iff \cos 3x > 0; f′=−sin⁡4x>0  ⟺  sin⁡4x<0f' = -\sin 4x > 0 \iff \sin 4x < 0. The longest increasing interval of sin⁡(kx)\sin(kx)-type functions is the length of one rising quarter/half of the sinusoid — e.g. f=sin⁡3xf = \sin 3x rises on (−π6,π6)\left(-\tfrac{\pi}{6},\tfrac{\pi}{6}\right), a run of length π3\tfrac{\pi}{3}.

Collapse to one angle, then read the sinusoid

3sin⁡x−4sin⁡3x=sin⁡3x,sin⁡4x+cos⁡4x=1−12sin⁡22x  ⇒  f′=−sin⁡4x3\sin x - 4\sin^3 x = \sin 3x, \qquad \sin^4 x + \cos^4 x = 1 - \tfrac{1}{2}\sin^2 2x \;\Rightarrow\; f' = -\sin 4x

Worked example

Find the length of the longest interval on which f(x)=3sin⁡x−4sin⁡3xf(x) = 3\sin x - 4\sin^3 x is increasing.
  1. Collapse: 3sin⁡x−4sin⁡3x=sin⁡3x3\sin x - 4\sin^3 x = \sin 3x.
  2. f′(x)=3cos⁡3x≥0  ⟺  cos⁡3x≥0  ⟺  −π2≤3x≤π2f'(x) = 3\cos 3x \ge 0 \iff \cos 3x \ge 0 \iff -\tfrac{\pi}{2} \le 3x \le \tfrac{\pi}{2}.
  3. So −π6≤x≤π6-\tfrac{\pi}{6} \le x \le \tfrac{\pi}{6}; the interval has length π6−(−π6)=π3\tfrac{\pi}{6} - (-\tfrac{\pi}{6}) = \tfrac{\pi}{3}.
Answer:π3\dfrac{\pi}{3}.
Practice this conceptself-check · 4 quick reps

Try it yourself

On which interval does f(x)=sin⁡4x+cos⁡4xf(x) = \sin^4 x + \cos^4 x increase?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Simplify 3sin⁡x−4sin⁡3x3\sin x - 4\sin^3 x.
  2. 2.
    f=sin⁡4x+cos⁡4xf=\sin^4 x+\cos^4 x: what is f′f'?
  3. 3.
    sin⁡3x\sin 3x longest increasing interval length?
  4. 4.
    f′=−sin⁡4x>0f'=-\sin 4x>0 needs?

From the bank · past-year question

Example 6Applications of DerivativeMODERATE
The length of the longest interval, in which the function 3sin⁡x−4sin⁡3x3\sin x - 4\sin^3 x is increasing, is

[Q111 · 2nd May Shift 1 · 2023]

Collapse to one angle BEFORE differentiating

Differentiating 3sin⁡x−4sin⁡3x3\sin x - 4\sin^3 x term by term is messy; recognising it as sin⁡3x\sin 3x makes f′=3cos⁡3xf' = 3\cos 3x a one-liner. Likewise sin⁡4x+cos⁡4x\sin^4 x + \cos^4 x is best reduced with the double-angle identity first — spotting the standard form is the whole shortcut.

Mind the kk when scaling the interval

For f′=−sin⁡4xf'=-\sin 4x, you solve sin⁡4x<0\sin 4x<0 for the argument 4x4x (a run of width π\pi in 4x4x), then divide by 4 to get the xx-interval (width π/4\pi/4). Forgetting to divide the argument's bounds by kk inflates the interval by a factor of kk.
Drill 2 more on trigonometric monotonicity: reduce to a single sinusoid

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

  • The Sign of the Derivative Decides Monotonicity

    Monotonicity from the sign of f prime

    f′(x)>0  ⇒  f increasing,f′(x)<0  ⇒  f decreasingf'(x) > 0 \;\Rightarrow\; f \text{ increasing}, \qquad f'(x) < 0 \;\Rightarrow\; f \text{ decreasing}
  • Polynomial Monotonicity via a Factored Derivative

    Cubic derivative factors to a quadratic

    f(x)=ax3+…  ⇒  f′(x)=3a (x−r1)(x−r2)f(x) = ax^3 + \dots \;\Rightarrow\; f'(x) = 3a\,(x - r_1)(x - r_2)
  • Discriminant Test for a Strictly Monotonic Cubic

    Strictly increasing everywhere

    f′(x)=Ax2+Bx+C, A>0, B2−4AC<0  ⇒  f′(x)>0 ∀xf'(x) = Ax^2 + Bx + C,\ A > 0,\ B^2 - 4AC < 0 \;\Rightarrow\; f'(x) > 0 \ \forall x
  • Rational and Rational-Trig Quotients: the ad minus bc Condition

    Sign of the derivative of a bilinear-trig quotient

    f(x)=asin⁡x+bcos⁡xcsin⁡x+dcos⁡x  ⇒  f′(x)=ad−bc(csin⁡x+dcos⁡x)2f(x) = \frac{a\sin x + b\cos x}{c\sin x + d\cos x} \;\Rightarrow\; f'(x) = \frac{ad - bc}{(c\sin x + d\cos x)^2}
  • Products and Composites with exp and log: Chain-Rule Sign Analysis

    The exponential factor drops out of the sign test

    f′(x)=eg(x) h(x)  ⇒  sign⁡f′(x)=sign⁡h(x)(since eg(x)>0)f'(x) = e^{g(x)}\,h(x) \;\Rightarrow\; \operatorname{sign} f'(x) = \operatorname{sign} h(x) \quad (\text{since } e^{g(x)} > 0)
  • Trigonometric Monotonicity: Reduce to a Single Sinusoid

    Collapse to one angle, then read the sinusoid

    3sin⁡x−4sin⁡3x=sin⁡3x,sin⁡4x+cos⁡4x=1−12sin⁡22x  ⇒  f′=−sin⁡4x3\sin x - 4\sin^3 x = \sin 3x, \qquad \sin^4 x + \cos^4 x = 1 - \tfrac{1}{2}\sin^2 2x \;\Rightarrow\; f' = -\sin 4x

Watch out for (11)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Applications of DerivativeMODERATE
Let f be a twice differentiable function such that f′′(x)=−f(x)f''(x)=-f(x), f′(x)=g(x)f'(x)=g(x) and h(x)=[f(x)]2+[g(x)]2h(x)=[f(x)]^2+[g(x)]^2. If h(5)=1h(5)=1, then h(10)h(10) is

[Q111 · 3rd May 2nd Shift · 2023]

Example 2Applications of DerivativeMODERATE
The range of values of xx for which f(x)=x3+6x2−36x+7f(x) = x^3 + 6x^2 - 36x + 7 is increasing in

[Q114 · 9th May Shift 2 · 2024]

Example 3Applications of DerivativeHARD
If f(x)=x3+bx2+cx+df(x)=x^{3}+bx^{2}+cx+d and 0<b2<c0<b^{2}<c, then in (−∞,∞)(-\infty,\infty)

[Q123 · 3rd May Shift 2 · 2023]

Example 4Applications of DerivativeMODERATE
If f(x)=ksinx+2cos⁡xsin⁡x+cos⁡xf(x) =\frac{ksinx+ 2\cos x}{\sin x+ \cos x} is strictly increasing for all real values of xx, then

[Q116 · 21 April Shift II · 2025]

Example 5Applications of DerivativeEASY
If f(x)=log⁡xxf(x) = \frac{\log x}{x} (x>0)(x>0), then it is increasing in

[Q120 · 9th May Shift 1 · 2023]

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