MHT-CET Maths · Formula sheet

Differentiation formulas

33 formulas and 78 common traps for MHT-CET Maths Differentiation, grouped by subtopic.

Foundations, the Chain Rule, and Differentiability

Standard Derivatives and the Rules of Differentiation

Product rule

\dfrac{d}{dx}\big[u(x)\,v(x)\big] = u'(x)\,v(x) + u(x)\,v'(x)

$u, v$ the two factors being multiplied

The Chain Rule and Composite Functions

Chain rule

\dfrac{dy}{dx} = f'\big(g(x)\big) \cdot g'(x) \qquad \text{for } y = f(g(x))

$f$ outer function
$g(x)$ inner function — its derivative is the multiplying factor

Differentiating Iterated Functions f(f(x))

Chain rule on an iterated function

\dfrac{d}{dx}f\big(f(f(x))\big) = f'\big(f(f(x))\big)\cdot f'\big(f(x)\big)\cdot f'(x)

Simplify the Expression Before Differentiating

Quotient rule (used after simplifying)

\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}

Linear Approximation Using the Derivative

Linear approximation

f(a + h) \approx f(a) + h\,f'(a)

$a$ nearby point with an easy exact value
$h$ small gap to the target (may be negative)

The Derivative as the Slope of the Tangent

Slope of the tangent

m_{\text{tangent}} = \left.\dfrac{dy}{dx}\right|_{x=a} = f'(a)

$f'(a)$ instantaneous slope at $x = a$

Differentiability and Where a Derivative Fails to Exist

Differentiability test

\text{LHD} = \lim_{h\to0^-}\dfrac{f(a+h)-f(a)}{h} = \lim_{h\to0^+}\dfrac{f(a+h)-f(a)}{h} = \text{RHD}

Trigonometric Simplification Toolkit

The collapses you reach for most

1+\cos x = 2\cos^2\tfrac{x}{2},\quad 1-\cos x = 2\sin^2\tfrac{x}{2},\quad \sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|

$\tfrac{x}{2}$ half-angle — appears whenever you collapse $1\pm\cos x$
$|\cdots|$ the root of a perfect square is a MODULUS; fix the sign on the given interval

Common traps

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

Do not apply the power rule to a constant base.

2^x

has a variable EXPONENT, so its derivative is

2^x \log 2

. The power rule

nx^{n-1}

applies only to

x^n

(variable base, constant exponent). At

x = 0

\dfrac{d}{dx}3^x = \log 3

— exactly the kind of value the bank tests.

Quotient rule sign: numerator is $u'v - uv'$

The order matters —

u'v - uv'

, not

uv' - u'v

. Writing it backwards flips the sign of the whole answer. Memorise it as 'low d-high minus high d-low, over low squared'.

Never forget the inner derivative factor

Differentiating

\sin(3x^2+1)

as just

\cos(3x^2+1)

drops the

\times 6x

— the single most common chain-rule error. Every layer contributes a multiplying factor.

Evaluate the inner argument, not the outer, when a factor is zero

For

y = \cos(\sin x^2)

\dfrac{dy}{dx} = -\sin(\sin x^2)\cdot \cos x^2 \cdot 2x

. At

x = \sqrt{\pi/2}

x^2 = \pi/2

\cos x^2 = 0

— the whole product is

0

. Spot the vanishing middle factor before grinding the arithmetic.

Drop the inner coefficient and you lose a factor

For

g(x) = [f(2f(x) + 2)]^2

the chain rule contributes

\times 2

from the inner

2f(x)

. Students who differentiate

f(2f(x)+2)

as if the inner were just

f(x)

miss this factor and get half the answer.

Don't try to find a formula for $f$

These questions give only

f

and

f'

at a point. You never need an explicit rule for

f

— evaluate each chain-rule factor at the appropriate point and multiply. For

f(f(f(x))) + (f(x))^2

x=1

with

f(1)=1, f'(1)=3

3\cdot3\cdot3 + 2\cdot1\cdot3 = 33

Simplify first, or the algebra buries you

Differentiating

\dfrac{x^{2/3} - x^{-1/3}}{x^{2/3} + x^{-1/3}}

directly with the quotient rule is error-prone. Multiply through by

x^{1/3}

to get

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

— then

y' = \dfrac{2}{(x+1)^2}

and

(x+1)^2 y' = 2

falls out instantly.

Pick $h$ small and signed correctly

To estimate

\log_{10} 998

, use

a = 1000

and

h = -2

(negative, since

998 < 1000

). Getting the sign of

h

wrong pushes the estimate the wrong way. With

f'(x) = \dfrac{0.4343}{x}

\log_{10} 998 \approx 3 - 2\cdot\dfrac{0.4343}{1000} = 2.99913

The slope is the DERIVATIVE at $a$ , not at the target

Evaluate

f'(a)

at the easy anchor point

a

, not at

a + h

. Using

f'(a+h)

defeats the purpose — you wanted an easy slope.

Minimum SLOPE means differentiate twice

When a question asks where the slope (not the function) is minimum, you must differentiate again. First derivative

f'(x)

IS the slope; set its derivative

f''(x) = 0

. Confusing 'minimum of

y

' with 'minimum of

y'

' is a classic MHT-CET trap.

Simplify the curve before differentiating

2\sin\theta\cos\theta = \sin 2\theta

: the curve

y = 2e^x \sin(\tfrac{\pi}{4} - \tfrac{x}{2})\cos(\tfrac{\pi}{4} - \tfrac{x}{2})

collapses to

y = e^x \cos x

before you ever differentiate. Spot the double-angle identity first.

Not every modulus is a non-differentiable point

Students reflexively answer 'fails at the corner' whenever they see

|\cdot|

. But

f(x) = \dfrac{x}{1 + |x|}

is differentiable on ALL of

\mathbb{R}

: the LHD and RHD at

x = 0

both equal

1

. Always test LHD vs RHD at the suspect point — a vanishing or matching factor can smooth the corner, sometimes making the 'set of failure points' EMPTY.

Continuous does not mean differentiable

|x|

is continuous everywhere but has no derivative at

0

. Differentiability is the stronger condition: differentiable

\Rightarrow

continuous, never the reverse.

$1\pm\cos x$ (half-angle) vs $1\pm\cos 2x$ (power-reduction)

Different collapses:

1-\cos x = 2\sin^2\tfrac{x}{2}

but

1-\cos 2x = 2\sin^2 x

. Read the angle inside the cosine before choosing the factor — the wrong one halves or doubles the argument.

The root of a perfect square is a MODULUS

\sqrt{(\sin x-\cos x)^2} = |\sin x-\cos x|

, not

\sin x-\cos x

. Resolve the sign on the given interval: on

(\tfrac\pi4,\tfrac\pi2)

it is

+(\sin x-\cos x)

; on

(0,\tfrac\pi4)

it is

-(\sin x-\cos x)

. Dropping the modulus flips the sign of the whole derivative.

$\sec\pm\tan$ — mind which way the half-angle shifts

\sec x+\tan x=\tan\!\left(\tfrac\pi4+\tfrac x2\right)

but

\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)

. Useful check:

(\sec x+\tan x)(\sec x-\tan x)=\sec^2x-\tan^2x=1

Logarithmic Differentiation — Logs, Powers, and Long Products

Learn this subtopic in the notes

Logarithmic Differentiation — the Method

Derivative of f(x) raised to g(x)

\frac{d}{dx}\left[f(x)^{g(x)}\right] = f(x)^{g(x)}\left[g'(x)\,\log f(x) + g(x)\,\frac{f'(x)}{f(x)}\right]

$g'(x)\,\log f(x)$ the term from differentiating the exponent (treat base as constant)
$g(x)\,f'(x)/f(x)$ the term from differentiating the base (treat exponent as constant)

Products, Quotients and Powers via Logs

Log of a power-product

\log\!\left(\frac{a^{p}\,b^{q}}{c^{r}}\right) = p\log a + q\log b - r\log c

The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0

Power sums (the leftover at x=0)

\sum_{k=1}^{n} k = \frac{n(n+1)}{2} \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

$p$ the outer power (e.g. $2$ , $4$ , $3/2$ , or $n$ ) — it just multiplies the sum
$k$ the coefficient of $x$ in the $k$ -th factor; squared factors give $\sum k^2$

Change of Base and log-of-a-log Forms

Change of base

\log_a b = \frac{\log b}{\log a}

$\log$ natural log (base $e$ ) throughout this chapter
$a$ the base — when it depends on $x$ , this is why you must convert

Square-Root Quotients with Inverse-Trig Arguments

Log of a square-root quotient

\log\sqrt{\frac{1-u}{1+u}} = \frac{1}{2}\Big[\log(1-u) - \log(1+u)\Big]

$u$ the inner function, e.g. $\sin^{-1}x$ , with $u' = 1/\sqrt{1-x^2}$
$1/2$ the coefficient produced by the outer square root

Common traps

Both terms appear — never use just one

For

f^g

, the "power rule" alone (

g f^{g-1}f'

) and the "exponential rule" alone (

f^g\log f\cdot g'

) are each HALF the answer. Logarithmic differentiation produces BOTH terms and adds them. Using only one is the single most common error here.

A variable in the exponent kills the power rule

\dfrac{d}{dx}(x^x)

is NOT

x\cdot x^{x-1}

. The power rule

\dfrac{d}{dx}x^n = nx^{n-1}

requires

n

to be CONSTANT. When the exponent itself depends on

x

, take logs.

cos⁻¹(sin θ) collapses before you differentiate

In mixed stems like

\cos^{-1}(\sin\theta) + x^x

, use

\cos^{-1}(\sin\theta) = \tfrac{\pi}{2} - \theta

(for

\theta\in[0,\tfrac{\pi}{2}]

) to flatten the inverse-trig piece to a simple

-\theta'

; only the

x^x

part needs log differentiation. Mixing the two methods up wastes time.

If y is already a log, there is no 1/y

For

y = \log(\text{expression})

, expand the inside with log laws and differentiate the SUM directly. The

\dfrac{1}{y}\dfrac{dy}{dx}

form is only for

\log y = \cdots

(i.e. you took the log yourself).

Simplify before you differentiate — $\log\sqrt{\frac{1+\sin x}{1-\sin x}}$

Some quotient-log stems collapse to a tidy single function.

\log\sqrt{\tfrac{1+\sin x}{1-\sin x}} = \log\tan\!\left(\tfrac{\pi}{4}+\tfrac{x}{2}\right)

, whose derivative is the clean

\sec x

. Charging in with the quotient rule on the raw fraction works but is far slower and error-prone.

A fractional exponent becomes a fractional COEFFICIENT

\log[(x+5)^{4/3}] = \tfrac{4}{3}\log(x+5)

, not

\tfrac{4}{3}(x+5)

and not

(x+5)^{4/3}\log

. Drop the exponent out front; do not leave it on the argument.

Substitute x=0 only AFTER differentiating

If you plug

x=0

into

y

before differentiating, you get the constant

1

and derivative

0

— wrong. Differentiate the log-sum fully (keep

x

), THEN set

x=0

so the denominators collapse to

1

Squared factor $\Rightarrow$ $\sum k^2$ , not $\sum k$

If the

k

-th factor is

k^2x+1

(i.e.

1, 4x+1, 9x+1,\ldots

), differentiating

\log(k^2x+1)

gives

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

, so the leftover sum is

\sum k^2 = \tfrac{n(n+1)(2n+1)}{6}

. Don't reflexively write

\sum k

Product like (1-x)(2-x)⋯(n-x) at x=1 — factor, don't sum

When a factor itself VANISHES at the evaluation point (here

(1-x)=0

x=1

), the

\dfrac{1}{y}y'

form blows up. Instead write

y=(1-x)\,g(x)

; then

y'(1) = -1\cdot g(1)

, where

g(1)

is the product of the remaining factors at

x=1

The outer power just multiplies the sum

A power

3/2

n

on the whole product becomes a coefficient

p

\log y = p\sum\log(\cdots)

. So the answer is always

p\sum k

(or

p\sum k^2

) — e.g. power

3/2

gives

\tfrac{3}{2}\cdot\tfrac{n(n+1)}{2} = \tfrac{3n(n+1)}{4}

Convert the variable base BEFORE differentiating

\dfrac{d}{dx}\log_{\sin x}(\tan x)

cannot be done with the

\dfrac{1}{u}

rule directly — the base

\sin x

is not constant. Always rewrite as

\dfrac{\log\tan x}{\log\sin x}

first, then use the quotient rule.

A vanishing log term kills half the quotient rule

At nice points (

x=\pi/4

gives

\log\tan x = 0

;

x=e

gives

\log\log x = 0

), the quotient-rule term multiplying that zero disappears. Spot the vanishing term FIRST — it saves most of the algebra. The bank answer for

\log_{\sin x}\tan x

\pi/4

-4\log 2

log of a log is NOT (log)²

\log(\log x)

is a composition (log applied to

\log x

), with derivative

\dfrac{1}{\log x}\cdot\dfrac{1}{x}

. It is not

(\log x)^2

, whose derivative would be

2\log x\cdot\tfrac{1}{x}

. Keep the two straight.

Don't forget the chain factor u' on the inverse-trig inner

When

u = \sin^{-1}x

, each

\log(1\pm u)

differentiates to

\dfrac{\pm u'}{1\pm u}

with

u' = \dfrac{1}{\sqrt{1-x^2}}

. Dropping the

u'

(treating

u

x

) is a frequent error; at

x=0

it happens to equal

1

, but you must include it in general.

Compute y at the point — usually y=1 at x=0

You multiply

\dfrac{1}{y}y'

y

to finish. At

x=0

the quotient under the root is

\tfrac{1}{1}=1

, so

y=1

and the multiply-back is trivial. Forgetting to put

y

back (leaving only

\tfrac{1}{y}y'

) gives the wrong magnitude when

y\neq 1

Watch which factor is on top — it sets the sign

\sqrt{\tfrac{1-\sin^{-1}x}{1+\sin^{-1}x}}

gives

-1

x=0

; flipping to

\sqrt{\tfrac{1+\sin^{-1}x}{1-\sin^{-1}x}}

gives

+1

. The numerator/denominator order flips every sign — read the stem carefully before reaching for a memorised answer.

Implicit Differentiation and Special Forms

Learn this subtopic in the notes

Implicit Differentiation — the Core Method

Implicit chain rule

\frac{d}{dx}\big[g(y)\big] = g'(y)\,\frac{dy}{dx}, \qquad \frac{d}{dx}(xy)=y+x\frac{dy}{dx}

$\frac{dy}{dx}$ the unknown you collect and solve for
$g(y)$ any function of $y$ ; its $x$ -derivative carries $\frac{dy}{dx}$

Implicit Relations like log(x + y) = 2xy

Differentiating log(x + y)

\frac{d}{dx}\log(x+y) = \frac{1}{x+y}\left(1 + \frac{dy}{dx}\right)

$1 + \frac{dy}{dx}$ the chain-rule derivative of the inner $x+y$

Exponential Relations — Take Logs, Then Differentiate

Log first, then differentiate

\frac{d}{dx}\big[u(x)\,\log v(x)\big] = u'\log v + u\cdot\frac{v'}{v}

$u(x)$ the exponent (often containing $y$ )
$\log v(x)$ log of the base, after taking logs of both sides

Relations of the Form tan y = (rational in x)

Standard result

\tan y = \frac{x\sin\alpha}{1 - x\cos\alpha} \;\Rightarrow\; \frac{dy}{dx} = \frac{\sin\alpha}{1 - 2x\cos\alpha + x^2}

$\sec^2 y$ rewritten as $1 + \tan^2 y$ to substitute the given expression

Proving a Given Differential Relation

Key explicit form

y^{1/m} + y^{-1/m} = 2x \;\Rightarrow\; y = \left(x + \sqrt{x^2-1}\right)^{m}

$t = y^{1/m}$ substitution that turns the relation into a quadratic in $t$

Self-Referential Infinite Expressions

Self-reference for a nested radical

y = \sqrt{f(x) + y} \;\Rightarrow\; y^2 = f(x) + y \;\Rightarrow\; \frac{dy}{dx} = \frac{f'(x)}{2y - 1}

$y$ the whole infinite expression — it reappears under the first root

Functional Equations — Find f, Then Differentiate

Reciprocal-substitution setup

a f(x) + b f\!\left(\tfrac{1}{x}\right) = g(x), \quad\text{then } x\to\tfrac1x:\; a f\!\left(\tfrac1x\right) + b f(x) = g\!\left(\tfrac1x\right)

$f'(1), f''(2)$ treat as unknown CONSTANTS, solve via coefficient comparison

Common traps

Differentiating a y-term without the dy/dx factor

\dfrac{d}{dx}(y^2)

2y\dfrac{dy}{dx}

, NOT

2y

. The whole method rests on attaching

\dfrac{dy}{dx}

to every

y

-derivative by the chain rule. Drop it and every answer is wrong.

Forgetting the product rule on the xy term

\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}

— it has TWO terms because both factors carry an

x

-dependence. Writing just

x\dfrac{dy}{dx}

or just

y

loses half the term.

Find the y-value before substituting into the derivative

The derivative formula contains both

x

and

y

. At

x=0

you still need

y

; get it from the ORIGINAL equation (e.g.

\log y = 0 \Rightarrow y=1

) before plugging into

y'

. Substituting only

x=0

leaves the answer undetermined.

log(x + y) = sin(x + y) collapses to slope -1

When both sides are functions of the single quantity

x+y

, differentiating factors out

(1+y')

. Setting it to zero gives

y' = -1

, independent of the functions — recognise this shortcut for

\log(x+y)=\sin(x+y)

and its cousins.

You cannot use the power rule when the exponent contains y

\dfrac{d}{dx}(x^{y})

is NOT

yx^{y-1}

— that rule needs a CONSTANT power. Because

y

varies, take logs first:

\log(x^{y}) = y\log x

, then differentiate the product.

Use the original (logged) relation to simplify the final answer

These answers are meant to come out clean. After collecting

y'

, substitute

y

from the logged equation (e.g.

y = x/(1+\log x)

) — that is what turns a messy fraction into the tidy form the options expect.

Differentiate tan y as sec-squared y times dy/dx

\dfrac{d}{dx}(\tan y) = \sec^2 y\,\dfrac{dy}{dx}

— the

\dfrac{dy}{dx}

is essential (chain rule on the implicit

y

). Then convert

\sec^2 y

1+\tan^2 y

so the given relation can be substituted.

Spotting a hidden inverse-tangent shortcut

If the rational in

x

is exactly

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

(or

\dfrac{x+a}{1-ax}

), it is a

\tan

addition/double-angle in disguise. Recognising it lets you write

y

explicitly and skip the heavy quotient differentiation.

Use the substitution to get y explicitly first

Trying to differentiate

y^{1/m}+y^{-1/m}=2x

directly is painful. Set

t=y^{1/m}

, solve the resulting quadratic for

y

, THEN differentiate — the proof falls out in two lines.

Square only after isolating the root

The target identities carry a

(x^2-1)

(x^2+1)

factor because

y'

has a

\sqrt{x^2\pm1}

in its denominator. Square

y'

to clear that root — squaring is what produces the polynomial coefficient.

The inner expression equals the WHOLE y, not part of it

Because the nesting is infinite, what sits under the first root is

f(x) + (\text{the same infinite expression}) = f(x) + y

. Treating it as a finite tower (or as just

f(x)

) breaks the self-reference that makes the problem solvable.

Square before differentiating, not after

Convert

y = \sqrt{f(x)+y}

into

y^2 = f(x)+y

FIRST, then differentiate. Differentiating the square root directly forces a chain-rule mess; the squared form differentiates in one clean line.

f'(1), f''(2) are CONSTANTS — name them and solve

f(x)=x^3+x^2f'(1)+xf''(2)+6

, the terms

f'(1)

and

f''(2)

are fixed numbers, not functions. Let them be

a, b

, build

f'

and

f''

, evaluate at the stated points, and solve the resulting linear system.

For f(x) and f(1/x), substitute x to 1/x to get a second equation

One equation in two unknowns

f(x)

and

f(1/x)

is not enough. Replacing

x

1/x

gives an independent equation; solve the pair simultaneously to isolate

f(x)

before differentiating.

f'(x) = f(x) means exponential

The only functions satisfying

f'=f

are

f(x)=Ce^{x}

. Use the given value (e.g.

f(1)=2

) to fix

C

, then differentiate compositions like

h(x)=f(f(x))

by the chain rule.

Inverse Functions and Inverse Trigonometric Differentiation

Learn this subtopic in the notes

Derivative of an inverse function

g'(x) = \dfrac{1}{f'\big(g(x)\big)} \qquad\text{where } g = f^{-1}

$g(x)$ the inverse $f^{-1}(x)$ — the input that maps to $x$ under $f$
$f'(g(x))$ slope of $f$ at the matching point, NOT $f'(x)$

The Inverse Trigonometric Derivative Table

Chain rule on an inverse-trig function

\dfrac{d}{dx}\tan^{-1}(u) = \dfrac{1}{1+u^2}\cdot\dfrac{du}{dx}

$u$ the inner function (e.g. $x^2$ , $3x$ , $\log x$ )
$du/dx$ derivative of the inner function — never forget it

Collapsing Inverse-Trig with a Substitution

The two workhorse collapses

\sin^{-1}\!\big(3x - 4x^3\big) = 3\sin^{-1}x, \qquad \tan^{-1}\!\dfrac{2x}{1-x^2} = 2\tan^{-1}x

$x = \sin\theta$ use when the argument is a sine multiple-angle ( $3x-4x^3$ , $2x\sqrt{1-x^2}$ )
$x = \tan\theta$ use when the argument is a tangent/double-angle ratio

tan inverse Addition and Complementary Identities

Arctan addition + complementary pair

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\dfrac{x+y}{1-xy}, \qquad \sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}

$1 - xy$ denominator of the combined argument; sign flips for the subtraction form
$\pi/2$ the constant a complementary pair collapses to — derivative $0$

Differentiating One Inverse-Trig with Respect to Another

Ratio of angle-multiples

\text{If } u = a\,\theta \text{ and } v = b\,\theta, \text{ then } \dfrac{du}{dv} = \dfrac{a}{b}

$a, b$ the constant multiples after each function collapses to a multiple of $\theta$
$d\theta$ cancels in the ratio — never appears in the final answer

Exponentials of Inverse-Trig Functions

Logarithmic-derivative ratio

h(x) = e^{g(x)} \;\Rightarrow\; \dfrac{h'(x)}{h(x)} = g'(x)

$g(x)$ the inner inverse-trig exponent (e.g. $\sin^{-1}x$ )
$h'/h$ the exponential cancels, leaving just $g'(x)$

Common traps

Evaluate $f'$ at $g(x)$ , never at $x$

The single most common error: writing

g'(x) = 1/f'(x)

. It is

g'(x) = 1/f'(g(x))

. At a numeric point you must first find

g(a)

(the input mapping to

a

), then plug THAT into

f'

You rarely need the formula for $f^{-1}$

For a point value, don't invert

f

algebraically — just find the matching input

b

with

f(b) = a

and take

1/f'(b)

. Inverting an awkward cubic-plus-exponential is impossible anyway; the reciprocal rule sidesteps it.

Don't forget the inner derivative $du/dx$

\dfrac{d}{dx}\sin^{-1}(3x)

is NOT

\dfrac{1}{\sqrt{1-9x^2}}

— you must multiply by the inner derivative

3

. The chain factor is what most option-traps omit.

The minus sign rides on the 'co' functions

\cos^{-1}

\cot^{-1}

\operatorname{cosec}^{-1}

carry the negative sign; their partners

\sin^{-1}

\tan^{-1}

\sec^{-1}

are positive. Mixing the sign flips the answer onto a distractor.

$\sec$ and $\operatorname{cosec}$ derivatives carry $|x|$

\dfrac{d}{dx}\sec^{-1}x = \dfrac{1}{|x|\sqrt{x^2-1}}

— the absolute value on

x

is part of the formula. Dropping it is a quiet error the bank tests.

Match the substitution to the argument's shape

3x-4x^3

screams

x=\sin\theta

(it is

\sin 3\theta

);

4x^3-3x

screams

x=\cos\theta

(it is

\cos 3\theta

); ratios with

1+x^2

1-x^2

scream

x=\tan\theta

. Picking the wrong one buries the simplification.

Watch the principal-value branch

\sin^{-1}(\sin\alpha) = \alpha

only inside

[-\pi/2,\pi/2]

. When the substituted angle leaves that range — e.g. at

x=\tfrac12

\sin^{-1}\!\frac{2\cdot 3^x}{1+9^x}

, where

2\theta = 2\pi/3 > \pi/2

— the collapse becomes

\pi -

(angle), flipping the sign of the derivative.

Exponential/log inner functions hide the same shapes

\dfrac{2\log x}{1+(\log x)^2}

\sin 2\phi

with

\log x = \tan\phi

, collapsing to

2\tan^{-1}(\log x)

. Substitute on the INNER expression (

\log x

3^x

), then chain the extra inner derivative when differentiating.

The constant differentiates to zero — but only if you SEE it

A sum of arctans that collapses to a constant has derivative

0

. Students grind out two quotient-rule derivatives and miss that the whole thing was

\pi/4 +

constant. Always test for the addition/complementary pattern first.

Mind the $1 \mp xy$ sign and the validity range

Addition uses

1-xy

in the denominator, subtraction uses

1+xy

. The split

\tan^{-1}\frac{x+y}{1-xy} = \tan^{-1}x + \tan^{-1}y

is exact only when

xy < 1

; outside that a

\pm\pi

correction appears (a constant, so the derivative is unchanged — but the function value differs).

Don't differentiate w.r.t. $x$ separately and then divide blindly

You CAN compute

\dfrac{du/dx}{dv/dx}

, but the elegant route is to collapse both to multiples of one angle and take the ratio. The shortcut avoids messy

\sqrt{1-x^2}

factors that cancel anyway.

Both functions must share ONE angle

If one collapses with

x=\sin\theta

and the other with

x=\cos\phi

, convert via

\cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x

so both are in the same

\theta

. Mixing two different angle variables corrupts the ratio (and can flip the sign).

$h'/h$ strips the exponential — don't carry it

Because

h'(x) = h(x)\,g'(x)

, the ratio

h'/h

is just

g'(x)

with no

e^{(\cdots)}

left. Distractors keep the exponential in the answer; the clean ratio doesn't.

The sign comes from the inner inverse-trig

e^{\sin^{-1}x}

gives

+\frac{1}{\sqrt{1-x^2}}

;

e^{\cos^{-1}x}

gives

-\frac{1}{\sqrt{1-x^2}}

. The exponential is always positive, so the sign is decided entirely by

g'(x)

For monotonicity, check the SIGN of $g'$ , not its messiness

g'(u) = \frac{2e^u}{1+e^{2u}}

looks complicated, but

e^u>0

and the denominator

>0

, so

g'>0

everywhere — strictly increasing. You don't need to simplify further to conclude monotonicity.

Parametric Differentiation, Second Derivatives & Proving Relations

Learn this subtopic in the notes

Parametric Differentiation

Parametric first derivative

\dfrac{dy}{dx} = \dfrac{\,dy/dt\,}{\,dx/dt\,}, \qquad \dfrac{dx}{dt} \neq 0

$t$ the parameter (often $\theta$ ) linking $x$ and $y$
$dx/dt \neq 0$ needed so the slope is defined

Second Derivative of a Parametric Function

Parametric second derivative

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

$d/dt(dy/dx)$ differentiate the first slope (a function of $t$ ) again w.r.t. $t$
$dx/dt$ divide by it once more — the chain-rule leftover

Proving Second-Order Relations

Two standard second-order relations

y = A\cos nx + B\sin nx \Rightarrow y'' = -n^2 y; \qquad y = ax^{n+1} + bx^{-n} \Rightarrow x^2 y'' = n(n+1)y

$y'' = -n^2 y$ the SHM-type relation from the sin/cos combination
$n(n+1)y$ the multiple that appears for the power combination

Showing an Expression Is Constant

Zero derivative implies constant

\dfrac{d}{dx}E(x) = 0 \ \text{for all } x \ \Longrightarrow\ E(x) = \text{const}, \quad E(b) = E(a)

nth-Order Derivatives — Standard Results

nth derivative of a sine with linear argument

\dfrac{d^n}{dx^n}\big(\sin(ax+b)\big) = a^n \sin\!\left(ax+b+\dfrac{n\pi}{2}\right)

$a^n$ the coefficient $a$ factors out once per differentiation
$n\pi/2$ each derivative advances the phase by a quarter-turn

Common traps

Do not flip the ratio

The slope is

\dfrac{dy/dt}{dx/dt}

— the parameter-derivative of

y

on top, of

x

on the bottom. Writing

\dfrac{dx/dt}{dy/dt}

gives the reciprocal slope and a wrong answer.

The slope can stay in terms of the parameter

There is no rule that says

\dfrac{dy}{dx}

must be a function of

x

. Leaving it as

\dfrac{1}{t}

-\dfrac{b}{a}\cot\theta

is the final form; only substitute a parameter value when the question asks for the slope at a point.

NEVER divide the two second derivatives

\dfrac{d^2y}{dx^2} \neq \dfrac{d^2y/dt^2}{d^2x/dt^2}

. This is the single most common parametric error in MHT-CET. The correct route is: find

dy/dx

, differentiate it with respect to the parameter, then divide by

dx/dt

Differentiate dy/dx with respect to t, not x

After you have

\dfrac{dy}{dx}

as a function of

t

, you cannot differentiate it directly with respect to

x

. Differentiate it with respect to

t

and then divide by

\dfrac{dx}{dt}

to convert back to a derivative in

x

Carry the constants — they cancel cleanly

y = A\cos nx + B\sin nx

, differentiating twice brings down

-n^2

from BOTH terms identically, so the whole bracket reforms into

y

. Do not drop

A

B

— the relation only emerges because both terms behave the same way.

Match the power-combination exponents

For

y = ax^{n+1} + bx^{-n}

the relation

x^2 y'' = n(n+1)y

holds only because both exponents are tuned to give the SAME factor

n(n+1)

after two derivatives. If the exponents are arbitrary, no single relation appears — read them carefully.

Zero derivative means constant — the second point is a decoy

Once

E'(x) = 0

, the specific points (5 and 10, say) carry no information beyond the given value.

E(10) = E(5)

exactly. Students waste time trying to compute

E

at the new point from scratch.

Use the supplied relations during differentiation

Expressions like

(f)^2 + (g)^2

only collapse to zero because of the given conditions

f'' = -f

and

g = f'

. Substitute them as soon as

g'

f''

appears — that substitution is exactly what produces the cancellation.

Sine and cosine cycle with period 4 in the order n

Differentiating

\sin x

four times returns to

\sin x

; the

\tfrac{n\pi}{2}

phase term encodes exactly this 4-step cycle. Reduce

n

modulo 4 if you prefer to evaluate the phase directly.

The power-rule nth derivative stops at zero

\dfrac{d^n}{dx^n}(x^m) = \dfrac{m!}{(m-n)!}x^{m-n}

is valid only for

m \geq n

. Once

n > m

every further derivative of a polynomial term is

0

— the factorial pattern would otherwise give a meaningless negative factorial.

Differentiating One Function With Respect to Another

Learn this subtopic in the notes

Differentiating One Function With Respect to Another

Derivative of u with respect to v

\dfrac{du}{dv} = \dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}, \qquad \dfrac{dv}{dx} \neq 0

$u = f(x)$ the function being differentiated (the 'top')
$v = g(x)$ the function we differentiate with respect to (the 'bottom')
$dv/dx \neq 0$ ratio is undefined where the bottom's derivative vanishes

Composite Functions Using Given Derivatives f' and g'

Composite-over-composite ratio

\dfrac{d\,[f(p(x))]}{d\,[g(q(x))]} = \dfrac{f'(p(x))\,p'(x)}{g'(q(x))\,q'(x)}

$f'(p(x))$ outer derivative of the top, read from the given f' value
$p'(x)$ inner derivative of the top function
$g'(q(x))$ outer derivative of the bottom, read from the given g' value
$q'(x)$ inner derivative of the bottom function

Common traps

Do NOT differentiate one function directly by the other

\dfrac{du}{dv}

is not 'differentiate

u

and substitute

v

'. You must form both

\dfrac{du}{dx}

and

\dfrac{dv}{dx}

and divide. There is no shortcut that skips

x

Substitute the point only after dividing

If a value like

x = 5

is given, keep

x

symbolic while you form

\dfrac{du/dx}{dv/dx}

, then substitute. Plugging the point into

\dfrac{du}{dx}

and

\dfrac{dv}{dx}

before simplifying invites arithmetic slips.

The bottom's derivative must be non-zero

\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}

is only valid where

\dfrac{dv}{dx} \neq 0

. If

v = g(x)

has a stationary point at the value asked, the rate of change with respect to

v

is undefined there.

Each inner derivative must be carried through

When the top is

f(\sec x)

, its derivative is

f'(\sec x)\cdot\sec x\tan x

— the inner

\sec x\tan x

is part of it. Forgetting the inner derivative is the most common slip and silently drops a factor.

Match each supplied value to the right inner argument

A given

f'(\sqrt{2})

is meant for where the inner function equals

\sqrt{2}

(e.g.

\sec\frac{\pi}{4} = \sqrt{2}

) — not for

x = \sqrt{2}

. Evaluate the inner function at the given

x

first, then read off the matching

f'

value.

Keep the negative sign on falling inner functions

If the bottom is

g(\cos x)

, its inner derivative is

-\sin x

; the minus sign stays in the denominator and sets the sign of the final answer. Drop it and you get the right magnitude with the wrong sign.

Differentiation formulas

Foundations, the Chain Rule, and Differentiability

Standard Derivatives and the Rules of Differentiation

The Chain Rule and Composite Functions

Differentiating Iterated Functions f(f(x))

Simplify the Expression Before Differentiating

Linear Approximation Using the Derivative

The Derivative as the Slope of the Tangent

Differentiability and Where a Derivative Fails to Exist

Trigonometric Simplification Toolkit

Common traps

Logarithmic Differentiation — Logs, Powers, and Long Products

Logarithmic Differentiation — the Method

Products, Quotients and Powers via Logs

The Product Chain [(x+1)(2x+1)⋯(nx+1)] Evaluated at x=0

Change of Base and log-of-a-log Forms

Square-Root Quotients with Inverse-Trig Arguments

Common traps

Implicit Differentiation and Special Forms

Implicit Differentiation — the Core Method

Implicit Relations like log(x + y) = 2xy

Exponential Relations — Take Logs, Then Differentiate

Relations of the Form tan y = (rational in x)

Proving a Given Differential Relation

Self-Referential Infinite Expressions

Functional Equations — Find f, Then Differentiate

Common traps

Inverse Functions and Inverse Trigonometric Differentiation

The Inverse Trigonometric Derivative Table

Collapsing Inverse-Trig with a Substitution

tan inverse Addition and Complementary Identities

Differentiating One Inverse-Trig with Respect to Another

Exponentials of Inverse-Trig Functions

Common traps

Parametric Differentiation, Second Derivatives & Proving Relations

Parametric Differentiation

Second Derivative of a Parametric Function

Proving Second-Order Relations

Showing an Expression Is Constant

nth-Order Derivatives — Standard Results

Common traps

Differentiating One Function With Respect to Another

Differentiating One Function With Respect to Another

Composite Functions Using Given Derivatives f' and g'

Common traps

More MHT-CET Maths formula sheets