MHT-CET Maths · Formula sheet

Applications of Derivative formulas

46 formulas and 91 common traps for MHT-CET Maths Applications of Derivative, grouped by subtopic.

Tangents, Normals, and the Slope of a Curve

Slope of a Curve: Tangent Slope and Normal Slope

Tangent slope and normal slope

m_{\text{tangent}} = \left.\dfrac{dy}{dx}\right|_{(x_1,y_1)} \qquad m_{\text{normal}} = -\dfrac{1}{m_{\text{tangent}}}

$m$ slope of the tangent = value of the derivative at the point
$-1/m$ slope of the normal — negative reciprocal of the tangent slope

Equations of the Tangent and Normal Lines

Tangent and normal at a point

\text{tangent: } y - y_1 = m(x - x_1) \qquad \text{normal: } y - y_1 = -\dfrac{1}{m}(x - x_1)

Tangent Parallel to the X-axis or Y-axis

Horizontal vs vertical tangent

\text{horizontal: } \dfrac{dy}{dx} = 0 \qquad \text{vertical: } \dfrac{dx}{dy} = 0

Tangent or Normal at an Axis-Crossing or Special Point

Locate the special point, then the line

\text{Y-axis: } x = 0 \qquad \text{X-axis: } y = 0 \qquad \text{ordinate = abscissa: } y = x

Tangents and Normals to Parametric Curves

Parametric slope and second derivative

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \qquad \dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\cdot\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

Normal Parallel (or Perpendicular) to a Given Line: Solve for the Point

Normal parallel to a line ⇒ tangent slope condition

\text{normal} \parallel \text{line of slope } s \;\Rightarrow\; \dfrac{dy}{dx} = -\dfrac{1}{s}

Finding a Curve's Constants from Tangency Conditions

Touches the X-axis at (p, 0): two conditions

y(p) = 0 \quad \text{and} \quad y'(p) = 0

Tangent Line Given: Solve for the Curve's Parameters

Given tangent line at a point: two conditions

(x_0, y_0) \text{ on the curve} \qquad \left.\dfrac{dy}{dx}\right|_{(x_0,y_0)} = m_{\text{line}}

Lengths of Tangent/Normal, Intercepts, and Fixed Points

Length of normal and length of tangent

\ell_{\text{normal}} = \left|y\sqrt{1 + y'^2}\right| \qquad \ell_{\text{tangent}} = \left|\dfrac{y\sqrt{1 + y'^2}}{y'}\right|

$y$ ordinate at the point of contact
$y'$ slope at the point of contact

Common traps

Normal slope is the NEGATIVE reciprocal, not the reciprocal or the negative

If the tangent slope is

m

, the normal slope is

-\dfrac{1}{m}

— both the minus sign AND the reciprocal. Writing

1/m

-m

gives a wrong normal line. When the normal itself is given (e.g. its angle with the X-axis), remember the tangent slope is

-1/(\text{normal slope})

: a normal at

\tfrac{3\pi}{4}

has slope

-1

, so

f' = -1/(-1) = 1

, not

-1

"Parallel to a line" copies the slope; "perpendicular" flips it

A tangent/normal parallel to a line has that line's slope. Perpendicular means negative reciprocal. Read whether it is the TANGENT or the NORMAL that is parallel — that decides which of

m

-1/m

equals the line's slope.

Use the tangent slope for the tangent, the negative reciprocal for the normal

The single most common slip: writing the normal line with the tangent slope. If the question asks for the NORMAL, substitute

-1/m

into point-slope, not

m

Find the point first, then the slope AT that point

Many stems only describe the point ('where the curve crosses the Y-axis', 'where ordinate = abscissa'). Pin down

(x_1, y_1)

exactly before evaluating the derivative — the slope must be computed at that point, not at a generic

x

Vertical tangent means dx/dy = 0, not dy/dx = 0

For a tangent parallel to the Y-axis, the slope

\dfrac{dy}{dx}

blows up. Instead of setting the (infinite)

dy/dx = 0

, set the reciprocal

\dfrac{dx}{dy} = 0

. On

4y^2 - 4y + 2x - 1 = 0

, differentiating w.r.t.

y

gives

\dfrac{dx}{dy} = 2 - 4y = 0 \Rightarrow y = \tfrac12

— the clean route.

Don't stop at the slope condition — substitute back for the point

Solving

dy/dx = 0

gives you the x-value (or y-value); the answer is the POINT. Plug back into the curve to get the missing coordinate before matching options.

Read the axis correctly: Y-axis ⇒ x = 0, X-axis ⇒ y = 0

'Crosses the Y-axis' means the x-coordinate is

0

(set

x = 0

); 'crosses the X-axis' means

y = 0

. Swapping these puts you at the wrong point and every later step is wrong.

'Ordinate = abscissa' means y = x, not a numerical guess

Substitute

y = x

into the curve and solve — e.g. on

y = \sqrt{9 - 2x^2}

x = \sqrt{9 - 2x^2} \Rightarrow x = \sqrt{3}

. Then differentiate at that point.

The parametric second derivative has an extra 1/(dx/dt) factor

\dfrac{d^2y}{dx^2} \neq \dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

. You must divide that by

\dfrac{dx}{dt}

again:

\dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

. Forgetting this factor is the classic error on

x = 3\tan t,\ y = 3\sec t

type problems.

Get the point from the parameter, not from x-alone

Substitute the given

t

(or

\theta

) into BOTH

x(t)

and

y(t)

for the point of contact. Then the tangent/normal line uses that point with the parametric slope.

'Normal parallel to the line' means the NORMAL slope equals the line slope

It's the normal, not the tangent, that matches the given line's slope. So set the tangent slope to the NEGATIVE RECIPROCAL of the line slope. On

y = x\log x

with line slope

1

: tangent slope

= -1

, giving

1 + \log x = -1

. Matching the tangent slope to

1

instead is the standard wrong turn.

For a tangent PARALLEL to a line, match the tangent slope directly

Don't blanket-apply the negative reciprocal. If the TANGENT is parallel to the line, set

\dfrac{dy}{dx} =

line slope. The reciprocal flip is only for a NORMAL-parallel (or tangent-perpendicular) condition. Also watch

y = \cos(x + y)

: differentiate implicitly to

\dfrac{dy}{dx} = \dfrac{-\sin(x+y)}{1 + \sin(x+y)}

before applying the slope condition.

'Touches the axis' is TWO conditions, not one

A curve that touches (is tangent to) the X-axis at

(p, 0)

satisfies both

y(p) = 0

(point on the axis) and

y'(p) = 0

(slope zero there). Using only

y(p) = 0

loses an equation and you can't solve for all the constants.

'Gradient at the Y-axis' means evaluate y' at x = 0

The gradient where the curve cuts the Y-axis is

y'(0)

. For

y = ax^3 + bx^2 + cx + 5

y'(0) = c

, so 'gradient 3 at the Y-axis' immediately gives

c = 3

— the fastest first equation.

You need BOTH the point-on-curve equation and the slope equation

One condition alone under-determines the two unknowns. Use

(x_0, y_0)

on the curve for one equation, and the curve's slope at that point equal to the line's slope for the other. Skipping the slope match leaves a free parameter.

Differentiate the curve implicitly, not the line

The slope you match is the CURVE's derivative at the point (in terms of

p, q, a, b

), set equal to the line's known slope. Differentiate

y^2 = px^3 + q

implicitly as

2y\,y' = 3px^2

; don't confuse the line's slope with the curve's derivative expression.

Length of NORMAL and length of TANGENT are different formulas

Length of normal

= \left|y\sqrt{1 + y'^2}\right|

; length of tangent

= \left|\dfrac{y\sqrt{1 + y'^2}}{y'}\right|

— the tangent version has the extra

1/y'

. Using the tangent formula where the normal is asked (or vice versa) is a classic slip; on 'length of normal from a point' problems, pick the one WITHOUT the

1/y'

Distance from the origin uses only the constant term

For a normal written as

Ax + By + C = 0

, the perpendicular distance from the origin is

\dfrac{|C|}{\sqrt{A^2 + B^2}}

. After reducing a parametric normal to

x\cos t + y\sin t = 2

, the distance is

\dfrac{2}{\sqrt{\cos^2 t + \sin^2 t}} = 2

— the

\theta

-dependence cancels.

Angle Between Curves, Orthogonality, and Nearest Distance

Learn this subtopic in the notes

Tangent Slopes of Two Curves at Their Meeting Point

Slope at a point on a curve

m = \left.\dfrac{dy}{dx}\right|_{(x_0,\,y_0)}

$m$ tangent slope of one curve at the shared point
$(x_0, y_0)$ the intersection point, found by solving the two curves together

The Angle Between Two Curves

Angle between two curves

\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

$m_1, m_2$ the two tangent slopes at the intersection point
$\theta$ the acute angle between the curves

The Angle a Curve Makes With a Coordinate Axis

Angle a curve makes with the X-axis

\tan\theta = |m| \qquad \theta = \tan^{-1}|m|

$m$ tangent slope of the curve at the point on the axis

Orthogonal Curves and Solving for a Parameter

Orthogonality condition

m_1 \, m_2 = -1

$m_1, m_2$ the two tangent slopes at the point of intersection

Shortest Distance From a Line to a Curve (Parallel-Tangent Trick)

Point-to-line distance (used at the parallel-tangent point)

d = \dfrac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}

$(x_0, y_0)$ the point on the curve where its tangent is parallel to the line
$a, b, c$ coefficients of the line written as $ax + by + c = 0$

Common traps

You need slopes at the SHARED point, not at any point

The angle between two curves is defined only where they intersect. Always solve the two equations together for

P

FIRST, then substitute those coordinates into each dy/dx. Evaluating the slopes at convenient but different points gives a meaningless angle.

Differentiate the implicit curve fully

For a curve like

9x^2 + by^2 = 16

, every term differentiates:

18x + 2by\dfrac{dy}{dx} = 0

, giving

\dfrac{dy}{dx} = -\dfrac{9x}{by}

. Forgetting the

\dfrac{dy}{dx}

factor on the

y

-term drops the slope entirely.

Keep the modulus for the ACUTE angle

The formula

\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

uses a modulus so

\theta

comes out acute. Dropping it can hand you a negative tangent (an obtuse angle) — check the answer options: MHT-CET usually lists the acute value.

$\dfrac{d}{dx}a^x = a^x\log a$ , not $x\,a^{x-1}$

For

y = 3^x, y = 7^x

the base is constant and the exponent is the variable — the derivative is

a^x\log a

, so the slopes at

(0,1)

are

\log 3

and

\log 7

. Using the power rule here is a common wipe-out.

Angle with the X-axis is $\tan^{-1}|m|$ , not the angle formula

You do NOT need the full

\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|

here — the axis has slope

0

, so that formula collapses to

\tan\theta = |m|

. Trying to force the two-slope formula wastes time and invites arithmetic slips.

At the origin, most terms die — keep only the linear ones

Substituting

(0,0)

into an implicit derivative kills every term carrying an

x

y

factor. Only the constant-coefficient linear terms (here

3x

and

-3y

) survive, so the slope reads off in one line:

3 - 3\dfrac{dy}{dx} = 0

Orthogonal means slope PRODUCT $= -1$ , not slope sum $= 0$

Perpendicular tangents satisfy

m_1 m_2 = -1

(negative reciprocals). Writing

m_1 + m_2 = 0

m_1 = m_2

is a different condition and gives the wrong parameter value.

Let the intersection relation cancel the coordinates

After imposing

m_1 m_2 = -1

you are left with

x, y

in the equation. Don't panic — substitute the simpler curve relation (

y^2 = 6x

) and the coordinates cancel, leaving a clean equation in the parameter alone.

Nearest point is the PARALLEL-tangent point, not the closest-looking one

The minimum gap happens exactly where the curve's tangent is parallel to the line. Guessing a point or plugging in the vertex usually overshoots — set

\dfrac{dy}{dx}

equal to the line's slope and solve.

Rationalise before matching the options

\dfrac{3/4}{\sqrt2} = \dfrac{3}{4\sqrt2} = \dfrac{3\sqrt2}{8}

. MHT-CET options are usually written with a rational denominator, so rationalise or you may not spot your answer in the list.

Approximations Using Differentials

Learn this subtopic in the notes

The Differential dy and the Linear-Approximation Formula

Linear approximation

f(a + h) \approx f(a) + h\,f'(a) \qquad \big(dy = f'(x)\,dx\big)

$a$ nearby point with an easy exact value
$h$ small gap to the target (may be negative)
$f'(a)$ slope at the anchor a — the multiplier of h

Approximating Roots and Powers

Power/root approximation

(a + h)^{p/q} \approx a^{p/q} + h\cdot\dfrac{p}{q}\,a^{\,p/q - 1}

$a$ nearest perfect power (perfect cube for a cube root, etc.)
$p/q$ the exponent — carries through to the derivative

Approximating Trigonometric Values

Trig approximation (h in radians)

\sin(a + h) \approx \sin a + h\cos a, \qquad \cos(a + h) \approx \cos a - h\sin a

$a$ nearby standard angle (30°, 45°, 60° …)
$h$ the small angular gap, CONVERTED TO RADIANS

Approximating Logarithms and Exponentials

Log & exponential approximation

\dfrac{d}{dx}\log_{10} x = \dfrac{0.4343}{x}, \qquad \dfrac{d}{dx}a^x = a^x \log a

$0.4343$ $\log_{10} e$ — the base-conversion factor for a base-10 log
$\log a$ natural log of the base, in the exponential derivative

Approximating Polynomial Values

Polynomial approximation / reconstruction

f(a+h) \approx f(a) + h\,f'(a); \quad P(x) = P(a) + P'(a)(x-a) + \tfrac12 P''(a)(x-a)^2

Common traps

The slope is $f'(a)$ — evaluate at the anchor, not the target

The correction term is

h\,f'(a)

, using the derivative at the EASY point

a

. Evaluating

f'(a+h)

(at the target) defeats the whole purpose — you chose

a

precisely so the slope there is easy. For

\sqrt{25.3}

use

f'(25)=1/10

, not

f'(25.3)

Get the sign of $h$ right

If the target is BELOW the anchor,

h

is negative. To estimate

\sqrt{24.7}

with

a=25

, take

h=-0.3

, giving

5 - 0.03 = 4.97

. A wrong sign pushes the estimate the wrong way by twice the correction.

Anchor at a perfect power, not just any round number

For

\sqrt[3]{0.026}

, do NOT anchor at

0

0.025

— neither has a clean cube root. Use

0.027 = 0.3^3

f(a)=0.3

is exact. Choosing an anchor whose value you cannot compute exactly wrecks the whole method.

Watch $x^{p/q - 1}$ in the derivative

For

x^{3/2}

the derivative is

\tfrac32 x^{1/2} = \tfrac32\sqrt{x}

, so

f'(4) = \tfrac32\cdot 2 = 3

— a clean integer, which is why

a=4

is the right anchor. Subtracting

1

from the exponent wrongly (e.g. leaving

x^{3/2}

) inflates the correction.

Convert the gap to RADIANS before multiplying

The derivatives

\cos x,\;-\sin x

are rates per radian. If you plug

h = 0.5

(the degree count) instead of

0.00875

rad, the correction is off by a factor of ~57. Always convert minutes/seconds → degrees → radians first.

Cosine's derivative carries a minus sign

\dfrac{d}{dx}\cos x = -\sin x

. For

\cos(30^\circ 30')

the correction is

h\cdot(-\sin 30^\circ)

, which DECREASES the value (cosine falls as the angle rises past 0). Dropping the minus pushes the estimate the wrong way.

$\dfrac{d}{dx}\log_{10} x$ carries the $0.4343$ factor

A base-10 log is NOT

\dfrac{1}{x}

— that is the natural log.

\dfrac{d}{dx}\log_{10} x = \dfrac{\log_{10} e}{x} = \dfrac{0.4343}{x}

. Forgetting the factor makes the correction ~2.3× too big.

$\dfrac{d}{dx}a^x = a^x\log a$ , not $x\,a^{x-1}$

The base is constant and the EXPONENT is the variable, so the power rule does not apply. For

3^{2.001}

use

f'(x) = 3^x \log 3

; here

\log 3 = 1.0986

is the natural log, supplied in the question.

Anchor at the integer nearest the TARGET

In the reconstruction question the data is at

x = 2

, but

P(1.001)

is asked — anchor at

a = 1

, not

2

. Blindly linearising at the data point

a = 2

uses

P(2) = -1

and

P'(2) = 0

and gives the wrong answer. Reconstruct

P

first, then anchor near the target.

The $\tfrac12$ in the reconstruction is essential

P(x) = P(a) + P'(a)(x-a) + \tfrac12 P''(a)(x-a)^2

— the second-order term carries a

\tfrac12

. Dropping it doubles the quadratic coefficient. Here

\tfrac12 P''(2) = \tfrac12(2) = 1

, so the leading coefficient is

1

, not

2

Rate of Change and Related Rates

Learn this subtopic in the notes

Rate of Change as a Chain of Derivatives

The two rate relations

\dfrac{dQ}{dt} = \dfrac{dQ}{dx}\cdot\dfrac{dx}{dt} \qquad \dfrac{dQ}{dP} = \dfrac{dQ/dt}{dP/dt}

$Q, P$ the two quantities being compared
$dx/dt$ the given rate of the driving variable

Related Rates: Circle, Sphere, and Square

Sphere volume and surface area

V = \tfrac{4}{3}\pi r^3,\quad S = 4\pi r^2,\qquad A_{\text{circle}} = \pi r^2

$r$ radius (the driving variable)

Related Rates: Cone, Hemispherical Bowl, and Cylinder

Cone and hemispherical-bowl volumes

V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h,\qquad V_{\text{bowl}} = \pi\!\left(Rx^2 - \tfrac{x^3}{3}\right),\qquad V_{\text{cyl}} = \pi R^2 h

$R$ fixed radius (bowl / cylinder)
$x, h$ the changing depth / height

Ladder and Sliding-Rod Problems (Pythagorean Rates)

Pythagorean length constraint

x^2 + y^2 = L^2 \;\Rightarrow\; \dfrac{dy}{dt} = -\dfrac{x}{y}\,\dfrac{dx}{dt}

$L$ fixed ladder/rod length
$x, y$ horizontal and vertical distances of the ends

A Point Moving Along a Curve

Coordinate rate and distance rate on a curve

\dfrac{dy}{dt} = f'(x)\,\dfrac{dx}{dt} \qquad \dfrac{d}{dt}\sqrt{x^2 + y^2} = \dfrac{x\dot x + y\dot y}{\sqrt{x^2 + y^2}}

$\dot x, \dot y$ $dx/dt$ and $dy/dt$

Rectilinear Motion: Displacement, Velocity, Acceleration

Velocity, acceleration, resultant acceleration

v = \dfrac{ds}{dt},\quad a = \dfrac{d^2s}{dt^2},\qquad a_{\text{res}} = \sqrt{\ddot x^2 + \ddot y^2}

Recovering a Quantity from Its Rate (Integrate Back)

Recover a quantity by integrating its rate

P = P_0 + \int_0^{n}\dfrac{dP}{dx}\,dx \qquad v(t) = \int_0^{t} a(\tau)\,d\tau

$P_0$ the base value that must be added back

Common traps

'Rate of $A$ w.r.t. $B$ ' is a RATIO of derivatives, not $A/B$

For volume w.r.t. surface area, do NOT compute

V/S

. Use

\dfrac{dV}{dS} = \dfrac{dV/dr}{dS/dr} = \dfrac{r}{2}

. The single most common slip here is dividing the quantities instead of their derivatives.

Everything moves in time — differentiate w.r.t. $t$

A relation like

A = \pi r^2

is static. The moment a rate

\tfrac{dr}{dt}

is given, differentiate w.r.t.

t

\tfrac{dA}{dt} = 2\pi r\,\tfrac{dr}{dt}

. Forgetting the

\tfrac{dr}{dt}

factor leaves you with

2\pi r

, which is not a rate.

Sign: a decreasing rate is negative — report the magnitude

A contracting plate has

\dfrac{dA}{dt} = -4

(negative because it shrinks). Carry the minus sign through the algebra, then report the rate of decrease as the magnitude. Dropping the sign mid-way flips the answer.

Volume rate vs. surface-area rate — different factors

For a sphere

\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}

but

\dfrac{dS}{dt} = 8\pi r\dfrac{dr}{dt}

. When a surface-area rate is GIVEN and a volume rate wanted, first back out

\dfrac{dr}{dt}

from the

8\pi r

relation, then feed it into the

4\pi r^2

one.

Substitute $r = h/2$ BEFORE differentiating a cone

If you keep both

r

and

h

V = \tfrac13\pi r^2 h

and differentiate, you get a two-rate mess. The semi-vertical angle fixes

r = kh

; substitute first so

V

has one variable — the algebra collapses.

Melting shell: differentiate the OUTER radius, keep the inner fixed

For an iron ball (radius

R

) coated with ice of thickness

r

, the ice volume differentiates via the outer radius

R + r

\dfrac{dV}{dt} = 4\pi(R + r)^2\dfrac{dr}{dt}

. Using the total radius wrongly (or the inner

R

) is the classic error — plug the FULL outer radius

R + r

into

4\pi(\cdot)^2

Convert units before substituting

A ladder is

5

m but the top slides at

10

cm/s. Work in ONE unit:

10

cm/s

= 0.1

m/s, or the length

5

= 500

cm. Mixing metres and centimetres is the single most common wrong answer in these problems.

The sign tells you sliding up vs. down — then take the magnitude

\dfrac{dy}{dt} = -\dfrac{x}{y}\dfrac{dx}{dt}

is negative when the top descends. The magnitude is the 'rate of decrease' the option lists (e.g.

\tfrac85

ft/s downwards). Report direction from the sign, value from the magnitude.

Find $\dot y$ from the curve before using it

In a distance-rate problem you are usually given only

\dot x

. Get

\dot y = f'(x)\dot x

from the curve first, THEN substitute into

\dfrac{x\dot x + y\dot y}{\sqrt{x^2+y^2}}

. Using

\dot y = \dot x

by accident is a common slip.

' $y$ changes $k$ times $x$ ' means $f'(x) = k$

The condition

\dfrac{dy}{dt} = k\dfrac{dx}{dt}

cancels the common

\dfrac{dx}{dt}

to give

f'(x) = k

. Solve that for

x

, then read off

y

from the curve for each root — usually a

\pm

pair.

'At rest' is $v = 0$ ; 'acceleration zero' is $a = 0$ — don't swap them

Read the trigger carefully. 'When the bullet comes to rest' sets

v = 0

(solve for

t

, then find distance). 'When the acceleration is zero' sets

a = 0

(then find velocity). Using the wrong condition finds the wrong

t

Resultant acceleration uses SECOND derivatives of both coordinates

For parametric

x(t), y(t)

, differentiate each TWICE, then combine:

\sqrt{\ddot x^2 + \ddot y^2}

. Using first derivatives gives speed, not acceleration — a factor-of-

t

error.

Add the base value back — the integral is only the CHANGE

\int_0^{25}(100 - 12\sqrt x)\,dx = 1500

is the ADDED production, not the total. The new level is

2000 + 1500 = 3500

. Forgetting the initial

2000

gives

1500

, a listed wrong option.

Integrate to go from rate up to quantity

Given acceleration, integrate ONCE for velocity and TWICE for displacement (from rest, the constants vanish). Differentiating instead — because 'rate' primes you to differentiate — is the reflex to resist here.

Increasing and Decreasing Functions

Learn this subtopic in the notes

The Sign of the Derivative Decides Monotonicity

Monotonicity from the sign of f prime

f'(x) > 0 \;\Rightarrow\; f \text{ increasing}, \qquad f'(x) < 0 \;\Rightarrow\; f \text{ decreasing}

$f'(x)$ the slope of the tangent at $x$ — its SIGN is all that matters

Polynomial Monotonicity via a Factored Derivative

Cubic derivative factors to a quadratic

f(x) = ax^3 + \dots \;\Rightarrow\; f'(x) = 3a\,(x - r_1)(x - r_2)

$r_1, r_2$ roots of $f'$ ; the sign of $f'$ flips at each simple root

Discriminant Test for a Strictly Monotonic Cubic

Strictly increasing everywhere

f'(x) = Ax^2 + Bx + C,\ A > 0,\ B^2 - 4AC < 0 \;\Rightarrow\; f'(x) > 0 \ \forall x

$B^2 - 4AC$ discriminant of $f'$ ; negative means $f'$ never touches zero

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

Sign of the derivative of a bilinear-trig quotient

f(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}

$ad - bc$ the ONLY thing whose sign matters; $>0$ increasing, $<0$ decreasing

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

The exponential factor drops out of the sign test

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 $f'$
$h(x)$ the remaining factor whose sign chart you must build

Trigonometric Monotonicity: Reduce to a Single Sinusoid

Collapse to one angle, then read the sinusoid

3\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

Common traps

Monotonicity is decided by the sign of $f'$ , not by $f$

f

increasing

\iff f'(x) \ge 0

on the interval; decreasing

\iff f'(x) \le 0

. A large or positive VALUE of

f

tells you nothing — read the sign of the DERIVATIVE. Build the sign chart of

f'

piece by piece between its zeros.

An option must be a SUBSET of the true monotonic set

For

f(x)=\dfrac{x}{2}+\dfrac{2}{x}

the true decreasing set is

(-2,0)\cup(0,2)

. The correct MHT-CET option is

(1,2)

— not because that is the whole set, but because it is a valid SUBSET on which

f

decreases. Test each option for 'is this interval inside the monotonic set?', not 'does this equal the full set?'.

Factor $f'$ before reading signs

Trying to sign-test

f'

without factoring invites arithmetic slips. Always factor

f'(x)

fully first; the roots are the only sign-change points, and a clean factored product makes the alternating-sign chart automatic.

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

When the options include 'increasing throughout the real line', check the discriminant of

f'

FIRST. If

B^2-4AC<0

with

A>0

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.

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

The condition

0<b^2<c

for

f'=3x^2+2bx+c

gives discriminant

4(b^2-3c)

, and

c>b^2

forces

b^2-3c<0

. Recognise this family: whenever the constant term dominates the middle coefficient squared, the quadratic

f'

stays one-signed.

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

The single most common error here is setting

f'>0

for a 'decreasing' function. Decreasing means

f'(x) < 0

, so a bilinear-trig quotient decreasing forces

ad - bc < 0

— NOT

ad-bc>0

. Read 'decreasing'

\to

'negative derivative'

\to

'negative

ad-bc

It is $ad - bc$ , not $ab - cd$

The determinant of the coefficient pattern is

ad - bc

(main-diagonal minus off-diagonal of

\begin{smallmatrix}a&b\\c&d\end{smallmatrix}

). Distractor options offer

ab-cd

or a swapped sign — write out the quotient-rule numerator once to lock in

ad-bc

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

f'(x) = xe^{-x}(2-x)

, the

e^{-x}

is strictly positive and contributes nothing to the sign. Only the polynomial factor

x(2-x)

matters. Wasting effort on

e^{-x}

(or, worse, treating it as sometimes negative) derails the whole sign chart.

For a log, respect the domain before reading the sign

\dfrac{d}{dx}\log(u) = \dfrac{u'}{u}

only where

u > 0

. For

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

\pi>e

forces the numerator of

f'

negative, so

f

is DECREASING on

(0,\infty)

— a case where the 'obvious' increasing answer is wrong.

Collapse to one angle BEFORE differentiating

Differentiating

3\sin x - 4\sin^3 x

term by term is messy; recognising it as

\sin 3x

makes

f' = 3\cos 3x

a one-liner. Likewise

\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 $k$ when scaling the interval

For

f'=-\sin 4x

, you solve

\sin 4x<0

for the argument

4x

(a run of width

\pi

4x

), then divide by 4 to get the

x

-interval (width

\pi/4

). Forgetting to divide the argument's bounds by

k

inflates the interval by a factor of

k

Maxima, Minima & Optimisation

Learn this subtopic in the notes

Critical Points — Where the Slope Vanishes

Critical-point condition

f'(c) = 0 \quad \text{or} \quad f'(c) \text{ does not exist}

$c$ a candidate for a local maximum or minimum

The First-Derivative Test

First-derivative test

f' : + \to - \ \Rightarrow\ \text{max}, \qquad f' : - \to + \ \Rightarrow\ \text{min}

The Second-Derivative Test

Second-derivative test

f''(c) < 0 \Rightarrow \text{local max}, \qquad f''(c) > 0 \Rightarrow \text{local min}

$f''(c)$ concavity at the critical point c

Extreme Value at a Given Point ⇒ Solve for Parameters

Extremum condition at a given point

f'(p) = 0 \quad \text{for each stated extreme point } p

Absolute Max/Min on a Constrained Set S

Absolute extremum on a closed interval

\max_{[a,b]} f = \max\big\{ f(a),\ f(b),\ f(c_i) \big\}

$a, b$ endpoints of the interval from solving the inequality
$c_i$ critical points of f lying inside (a, b)

Applied Optimisation — Geometry & the AM-GM Shortcut

AM-GM optimisation shortcut

\min(ax + by)\ \text{s.t.}\ xy = c^2 \ =\ 2c\sqrt{ab} \quad (\text{equality at } ax = by)

Applied Optimisation — Tanks, Boxes & Cost

Open square-based tank, least surface

A(x) = x^2 + \dfrac{4V}{x}, \qquad A'(x) = 0 \Rightarrow x^3 = 2V,\ \ x = 2h

$x$ side of the square base
$h$ height; at the optimum x = 2h

Applied Optimisation — Profit, Revenue & Cost

Profit maximisation

P(x) = R(x) - C(x), \qquad P'(x) = 0,\ \ P''(x) < 0

Extrema of Trig and Rational Expressions

Key extremum formulas

\min(a\sec\theta - b\tan\theta) = \sqrt{a^2 - b^2}\ (a>b), \qquad \max/\min(a\sin\theta + b\cos\theta) = \pm\sqrt{a^2 + b^2}

$a, b$ coefficients; for the sec–tan form require a > b > 0

Common traps

$f'(c) = 0$ is NECESSARY, not sufficient

A critical point is a CANDIDATE, not a guarantee.

f(x) = x^3

has

f'(0) = 0

yet no extremum at

0

— the slope touches zero and keeps the same sign (a point of inflection). Always confirm with a genuine sign change of

f'

or the sign of

f''

Don't forget points where $f'$ is UNDEFINED

For

f(x) = x^{2/3} + (x-2)^{2/3}

f'(x) = \tfrac{2}{3}\big(x^{-1/3} + (x-2)^{-1/3}\big)

is undefined at

x = 0

and

x = 2

— those are critical points too. Restricting to

f' = 0

alone misses them.

A repeated root of $f'$ is NOT an extremum

For

f'(x) = 5x^2(x-1)(x-3)

, the factor

x^2

means

f'

touches zero at

x = 0

without changing sign — so

x = 0

is neither a max nor a min. Only

x = 1

and

x = 3

(simple roots, genuine sign flips) are extrema.

$f'' < 0$ is a MAXIMUM (the sign trips everyone)

Concave DOWN (

f'' < 0

) is a local MAX; concave UP (

f'' > 0

) is a local MIN. Students routinely guess the opposite. Picture the shape: a dome (

\cap

f'' < 0

) peaks; a bowl (

\cup

f'' > 0

) bottoms out.

When $f''(c) = 0$ , the test says NOTHING

f''(c) = 0

is inconclusive —

c

could be a max, a min, or an inflection. Do not conclude 'inflection' automatically. Switch to the first-derivative test and read the actual sign change of

f'

Read exactly which combination is asked

Every one of these questions has

a = 2, b = -\tfrac12

at its core, but they ask for different outputs:

a + b = \tfrac32

\tfrac{a}{b} + \tfrac{b}{a} = -\tfrac{17}{4}

a^2 + 2b = 3

, or just the pair itself. Solve the system once, then compute the specific expression requested — don't stop at

a, b

$\log x$ is natural log, and the $x = -1$ extremum is formal

Throughout MHT-CET

\log

means

\log_e

. The template uses

x = -1

as an extreme point by writing

\log|x|

(or reading

\log x

in the extended sense) — treat

\dfrac{d}{dx}\log|x| = \dfrac{1}{x}

and substitute

x = -1

directly.

SOLVE the inequality first — S is not all of $\mathbb{R}$

The constraint

x^2 + 30 \le 11x

restricts

x

to a short interval

[5, 6]

. Optimising over all reals (finding

f' = 0

globally) gives the wrong answer — the global critical points

x = 1, 3

are not even in the set. Always convert the inequality into the interval before doing anything else.

On a closed interval, always compare the ENDPOINTS

f

is monotonic on the interval (no interior critical point), the extreme value sits at an endpoint. Even when there IS an interior critical point, you must still evaluate

f

at both endpoints and pick the winner — the endpoint value often beats the turning-point value.

AM-GM only maximises a PRODUCT (fixed sum) or minimises a SUM (fixed product)

The shortcut applies to a sum-with-fixed-product (minimise) or a product-with-fixed-sum (maximise), with all terms POSITIVE. It gives

2c\sqrt{ab}

for

\min(ax + by)

— not

2ab\sqrt{c}

-2c\sqrt{ab}

. Match the constraint shape before quoting the result.

Number-splitting: split in the ratio of the EXPONENTS

To maximise

x^m(k - x)^n

, the maximiser is

x = \dfrac{m}{m+n}\,k

. For 'cube of one × square of the other' of 20: ratio

3 : 2

gives

12

and

8

. Guessing

10, 10

(equal split) is wrong unless the exponents are equal.

OPEN tank has no top — count the faces carefully

An open square-based tank has surface area

x^2 + 4xh

(one base + four sides), NOT

2x^2 + 4xh

. Including a non-existent top changes the optimum. For a cost problem, weight only the faces that actually exist.

Eliminate the second variable via the volume constraint FIRST

You cannot differentiate

A = x^2 + 4xh

directly — it has two variables. Use

V = x^2 h

to write

h

in terms of

x

, reducing

A

to one variable before setting

A'(x) = 0

Build REVENUE as price × quantity, not just price

Revenue from selling

x

items at price

p

each is

x \cdot p

, not

p

. Forgetting the factor

x

gives a linear profit with no interior maximum. Multiply the per-item price by the number of items before subtracting cost.

Return the profit VALUE, not the quantity

Solving

P'(x) = 0

gives the optimal number of items

x^*

— but the question usually asks for the maximum PROFIT. Substitute

x^*

back into

P(x)

to get the rupee value.

$\sec$ – $\tan$ minimum is $\sqrt{a^2 - b^2}$ , not $\sqrt{a^2 + b^2}$

The sec–tan form gives

\sqrt{a^2 - b^2}

(a difference under the root), whereas the harmonic form

a\sin\theta + b\cos\theta

gives

\sqrt{a^2 + b^2}

(a sum). Mixing the two is the classic error — check whether you have

\sec/\tan

\sin/\cos

For a symmetric rational, both $x = \pm 1$ matter

\dfrac{x^2 - x + 1}{x^2 + x + 1}

has critical points at BOTH

x = 1

and

x = -1

, giving

\tfrac13

(min) and

3

(max). Evaluating only one loses either the greatest or the least value — you need both for a 'difference' or 'range' question.

Rolle's Theorem and the Mean Value Theorem

Learn this subtopic in the notes

Rolle's Theorem — the Three Hypotheses and the Conclusion

Rolle's theorem

f \text{ cont. on }[a,b],\ \text{diff. on }(a,b),\ f(a)=f(b)\ \Rightarrow\ \exists\, c\in(a,b):\ f'(c)=0

$c$ point inside (a, b) where the tangent is horizontal
$f(a)=f(b)$ the equal-endpoint hypothesis unique to Rolle

Finding c and Rejecting Roots Outside the Interval

Rolle point from f'(x) = 0

f'(c) = 0,\quad c \in (a,b)\ \ \text{(reject any root outside the interval)}

Counting the Number of Valid c

Number of Rolle points

\#\{\,c\in(a,b): f'(c)=0\,\} = \text{number of roots of } f'(x)=0 \text{ inside } (a,b)

Lagrange's Mean Value Theorem — Statement and Finding c

Lagrange's Mean Value Theorem

f'(c) = \dfrac{f(b) - f(a)}{b - a},\qquad c \in (a,b)

$\frac{f(b)-f(a)}{b-a}$ slope of the chord joining the endpoints
$f'(c)$ slope of the tangent at the guaranteed point c

Solving Unknown Parameters and the Tangent-Parallel-to-Chord View

Two equations from 'the theorem holds at c'

\underbrace{f(a)=f(b)}_{\text{endpoint}} \quad \text{and} \quad \underbrace{f'(c)=0}_{\text{Rolle at given }c}

Common traps

All THREE hypotheses are required, not just f(a) = f(b)

Rolle needs continuity on

[a,b]

AND differentiability on

(a,b)

AND

f(a)=f(b)

. A function like

f(x)=|x|

[-1,1]

has

f(-1)=f(1)=1

but is NOT differentiable at

0

, so Rolle does not apply — there is no

c

with

f'(c)=0

. Never skip the differentiability check.

Rolle gives f'(c) = 0, not the chord slope

Rolle's conclusion is specifically

f'(c)=0

(a horizontal tangent), because the endpoints are equal. If the endpoints differ, use the Mean Value Theorem instead — its conclusion is

f'(c)=\frac{f(b)-f(a)}{b-a}

, which reduces to

0

only when

f(a)=f(b)

Reject any root of f'(c) = 0 that lies OUTSIDE (a, b)

Solving

f'(x)=0

can give roots that Rolle never promised. For

f(x)=x(x+3)e^{-x/2}

[-3,0]

f'

vanishes at

x=3

and

x=-2

, but only

-2

lies in

(-3,0)

. Picking

x=3

is the single most common mistake — always filter candidates by the interval.

The exponential factor is never zero

In a form like

f'(x)=e^{-x/2}\,(\text{polynomial})

, the factor

e^{-x/2}>0

always. So

f'(c)=0

forces the POLYNOMIAL factor to be zero — never set the exponential to zero. It only affects sign/scaling, not the location of the Rolle point.

Count roots INSIDE the open interval only — mind the endpoints

When counting

c

, include only roots strictly between

a

and

b

. For

\sin 2\pi x

[-1,1]

the roots of

f'

are

\pm\tfrac14, \pm\tfrac34

— four values, all interior. A root landing exactly on an endpoint would NOT count. Sketch or list them; don't guess the count.

Solve the full trig equation — don't stop at the first solution

\cos(k\pi x)=0

has infinitely many solutions; over a finite interval several survive. Writing

k\pi x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \dots

systematically and filtering by the interval avoids under-counting (a common way to pick '02' when the answer is '04').

MVT needs DIFFERENTIABILITY, not just continuity

Both hypotheses must hold: continuous on

[a,b]

AND differentiable on

(a,b)

. A function continuous but with a corner (non-differentiable) inside the interval can fail to have a valid

c

. Continuity alone is not enough — always confirm differentiability before applying LMVT.

Chord slope uses f(b) − f(a), not f'(a) or f'(b)

The right-hand side of LMVT is the average slope

\frac{f(b)-f(a)}{b-a}

— a difference of FUNCTION values over the interval width, not a derivative at an endpoint. A frequent slip is to compute

f'(a)

f'(b)

instead of the chord slope.

Bounding trick: f(b) − f(a) ≤ (b − a)·max f'

When

f'(x) \le M

everywhere, LMVT gives

f(b) - f(a) = f'(c)(b-a) \le M(b-a)

. So if

f(1)=1

and

f'\le 5

[1,5]

, then

f(5) \le 1 + 5\cdot 4 = 21

. Use the theorem as an INEQUALITY to bound a value.

You need BOTH equations — endpoint and interior

Using only

f(a)=f(b)

fixes just one relation between

a

and

b

; some options may share that value. You must ALSO use

f'(c)=0

(or the chord condition) at the given

c

to pin down both. Solving one equation and matching a partial answer is how students land on the wrong option.

Watch coefficient order — 'a and b respectively'

Options like

(11, -6)

(-6, 11)

differ only by which value is

a

and which is

b

. Track exactly which unknown you solved for; re-substitute into the original

f

to confirm the pair actually satisfies

f(a)=f(b)

and

f'(c)=0

before selecting.

Applications of Derivative formulas

Tangents, Normals, and the Slope of a Curve

Slope of a Curve: Tangent Slope and Normal Slope

Equations of the Tangent and Normal Lines

Tangent Parallel to the X-axis or Y-axis

Tangent or Normal at an Axis-Crossing or Special Point

Tangents and Normals to Parametric Curves

Normal Parallel (or Perpendicular) to a Given Line: Solve for the Point

Finding a Curve's Constants from Tangency Conditions

Tangent Line Given: Solve for the Curve's Parameters

Lengths of Tangent/Normal, Intercepts, and Fixed Points

Common traps

Angle Between Curves, Orthogonality, and Nearest Distance

Tangent Slopes of Two Curves at Their Meeting Point

The Angle Between Two Curves

The Angle a Curve Makes With a Coordinate Axis

Orthogonal Curves and Solving for a Parameter

Shortest Distance From a Line to a Curve (Parallel-Tangent Trick)

Common traps

Approximations Using Differentials

The Differential dy and the Linear-Approximation Formula

Approximating Roots and Powers

Approximating Trigonometric Values

Approximating Logarithms and Exponentials

Approximating Polynomial Values

Common traps

Rate of Change and Related Rates

Rate of Change as a Chain of Derivatives

Related Rates: Circle, Sphere, and Square

Related Rates: Cone, Hemispherical Bowl, and Cylinder

Ladder and Sliding-Rod Problems (Pythagorean Rates)

A Point Moving Along a Curve

Rectilinear Motion: Displacement, Velocity, Acceleration

Recovering a Quantity from Its Rate (Integrate Back)

Common traps

Increasing and Decreasing Functions

The Sign of the Derivative Decides Monotonicity

Polynomial Monotonicity via a Factored Derivative

Discriminant Test for a Strictly Monotonic Cubic

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

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

Trigonometric Monotonicity: Reduce to a Single Sinusoid

Common traps

Maxima, Minima & Optimisation

Critical Points — Where the Slope Vanishes

The First-Derivative Test

The Second-Derivative Test

Extreme Value at a Given Point ⇒ Solve for Parameters

Absolute Max/Min on a Constrained Set S

Applied Optimisation — Geometry & the AM-GM Shortcut

Applied Optimisation — Tanks, Boxes & Cost

Applied Optimisation — Profit, Revenue & Cost

Extrema of Trig and Rational Expressions

Common traps

Rolle's Theorem and the Mean Value Theorem

Rolle's Theorem — the Three Hypotheses and the Conclusion

Finding c and Rejecting Roots Outside the Interval

Counting the Number of Valid c

Lagrange's Mean Value Theorem — Statement and Finding c

Solving Unknown Parameters and the Tangent-Parallel-to-Chord View

Common traps

More MHT-CET Maths formula sheets