MHT-CET Maths · Differentiation

Foundations, the Chain Rule, and Differentiability

Differentiation measures instantaneous rate of change. Master the standard-derivative table, the sum/product/quotient rules, and the chain rule for composite functions — then know exactly where a derivative can fail to exist.

All Differentiation notes MHT-CET Maths strategy

Why this matters

This subtopic is the on-ramp to the whole chapter: 15 PYQs sit directly here (4 HARD, 11 MODERATE). Every harder differentiation question — implicit, logarithmic, parametric, applications — reduces to applying the chain rule cleanly and recalling the table cold. The recurring MHT-CET traps live here too: forgetting the inner factor of a composite, treating any modulus as a corner, and slipping on the exponential derivative aˣ log a.

Concept 1 of 7

Standard Derivatives and the Rules of Differentiation

Intuition

About a dozen derivatives are the alphabet of the whole chapter — power, exponential, logarithmic, and the six trig functions. Combined with three rules (sum, product, quotient) they cover most of what the paper asks before any chain rule appears. Know them as reflexes.

Definition

The standard derivatives you must recall instantly:

$\dfrac{d}{dx}x^n = nx^{n-1}$ , $\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2\sqrt{x}}$ , $\dfrac{d}{dx}\dfrac{1}{x} = -\dfrac{1}{x^2}$
$\dfrac{d}{dx}e^x = e^x$ , and $\dfrac{d}{dx}a^x = a^x \log a$
$\dfrac{d}{dx}\log x = \dfrac{1}{x}$ , and $\dfrac{d}{dx}\log_a x = \dfrac{1}{x \log a}$
$\dfrac{d}{dx}\sin x = \cos x$ , $\dfrac{d}{dx}\cos x = -\sin x$ , $\dfrac{d}{dx}\tan x = \sec^2 x$
$\dfrac{d}{dx}\sec x = \sec x \tan x$ , $\dfrac{d}{dx}\csc x = -\csc x \cot x$ , $\dfrac{d}{dx}\cot x = -\csc^2 x$

The three combining rules:

Sum/difference: $\dfrac{d}{dx}[u \pm v] = u' \pm v'$
Product: $\dfrac{d}{dx}[u\,v] = u'v + uv'$
Quotient: $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}$

Product rule

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

u, vthe two factors being multiplied

Worked example

Differentiate

y = x^2 e^x

This is a product: take $u = x^2$ , $v = e^x$ .
Then $u' = 2x$ and $v' = e^x$ .
Apply the product rule: $y' = u'v + uv' = 2x\,e^x + x^2 e^x$ .
Factor: $y' = x e^x(2 + x)$ .

Answer:

y' = x e^x(x + 2)

Practice this conceptself-check · 4 quick reps

Try it yourself

Differentiate

y = 2^x

and evaluate

y'

x = 0

Practice — Level 1 (4 reps)

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

1.
$\dfrac{d}{dx}(5^x)$
2.
$\dfrac{d}{dx}(x^3 \sin x)$
3.
$\dfrac{d}{dx}\!\left(\dfrac{x}{\cos x}\right)$
4.
$\dfrac{d}{dx}(\log_3 x)$

From the bank · past-year question

Example 1DifferentiationMODERATE

f(x)=3^x

g(x)=4^x

, then

\frac{f'(0)-g'(0)}{1+f'(0)g'(0)}

[Q130 · 11th May Shift 2 · 2024]

$\dfrac{d}{dx}a^x$ 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'.

Concept 2 of 7

The Chain Rule and Composite Functions

Intuition

When a function sits inside another — like

\sin(x^2)

(3x+1)^5

— you differentiate the outer function first (treating the inside as a single block), then MULTIPLY by the derivative of the inside. Peel the layers from outside in, multiplying as you go.

Definition

If $y = f(g(x))$ , then $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$ . For multiple nested layers, multiply the derivative of every layer:

\dfrac{d}{dx}f(g(h(x))) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x).

The composite standard forms (outer derivative times inner derivative

g'(x)

$\dfrac{d}{dx}[g(x)]^n = n[g(x)]^{n-1}\,g'(x)$
$\dfrac{d}{dx}\sin[g(x)] = \cos[g(x)]\,g'(x)$ , $\dfrac{d}{dx}\cos[g(x)] = -\sin[g(x)]\,g'(x)$
$\dfrac{d}{dx}\log[g(x)] = \dfrac{g'(x)}{g(x)}$ , $\dfrac{d}{dx}e^{g(x)} = e^{g(x)}\,g'(x)$

Chain rule

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

fouter function
g(x)inner function — its derivative is the multiplying factor

Worked example

Differentiate

y = \sin(3x^2 + 1)

Outer function is $\sin$ , inner is $g(x) = 3x^2 + 1$ .
Derivative of the outer at the inner: $\cos(3x^2 + 1)$ .
Multiply by the inner derivative $g'(x) = 6x$ .
So $\dfrac{dy}{dx} = \cos(3x^2 + 1) \cdot 6x$ .

Answer:

\dfrac{dy}{dx} = 6x \cos(3x^2 + 1)

Practice this conceptself-check · 4 quick reps

Try it yourself

Differentiate

y = (2x^2 - 5)^4

Practice — Level 1 (4 reps)

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

1.
$\dfrac{d}{dx}\cos(5x)$
2.
$\dfrac{d}{dx}e^{x^2}$
3.
$\dfrac{d}{dx}\log(x^2 + 1)$
4.
$\dfrac{d}{dx}(\sin x)^3$

From the bank · past-year question

Example 2DifferentiationMODERATE

y = \cos(\sin x^2)

, then

\frac{dy}{dx}

x = \sqrt{\frac{\pi}{2}}

[Q135 · Shift 1 · 2022]

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.

Drill 1 more on the chain rule and composite functions

Concept 3 of 7

Differentiating Iterated Functions f(f(x))

Intuition

When the same function is applied repeatedly —

f(f(x))

f(f(f(x)))

— and you are only given

f

and

f'

at one point, the chain rule still works one layer at a time. You never need the formula for

f

; you just multiply the slope at each layer.

Definition

By the chain rule, $\dfrac{d}{dx}f(f(x)) = f'(f(x)) \cdot f'(x)$ , and for three layers $\dfrac{d}{dx}f(f(f(x))) = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x)$ . If a fixed point is given (e.g. $f(1) = 1$ ), each nested $f$ at that point is still $1$ , so every factor becomes $f'(1)$ . When an inner expression carries its own coefficient (such as $f(2f(x) + 2)$ ), the chain rule pulls out that extra factor too — do not drop it.

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)

Worked example

f(2) = 2

and

f'(2) = 5

, find the derivative of

f(f(x))

x = 2

Chain rule: $\dfrac{d}{dx}f(f(x)) = f'(f(x)) \cdot f'(x)$ .
At $x = 2$ : the inner $f(2) = 2$ , so $f'(f(2)) = f'(2) = 5$ .
And $f'(2) = 5$ . Multiply: $5 \times 5 = 25$ .

Answer:

25

Practice this conceptself-check · 4 quick reps

Try it yourself

f(0) = 0

and

f'(0) = 4

, find the derivative of

g(x) = f(3f(x))

x = 0

Practice — Level 1 (4 reps)

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

1.
$f(1)=1, f'(1)=2$ . Find $\dfrac{d}{dx}f(f(x))$ at $x=1$ .
2.
$f(3)=3, f'(3)=2$ . Find $\dfrac{d}{dx}f(f(f(x)))$ at $x=3$ .
3.
$f(0)=0, f'(0)=5$ . Find $\dfrac{d}{dx}(f(x))^2$ at $x=0$ .
4.
$f(0)=0, f'(0)=3$ . Find $\dfrac{d}{dx}f(2f(x))$ at $x=0$ .

From the bank · past-year question

Example 3DifferentiationHARD

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

, then the derivative of

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

x=1

[Q142 · 15th May Shift 2 · 2023]

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

Drill 5 more on differentiating iterated functions f(f(x))

Concept 4 of 7

Simplify the Expression Before Differentiating

Intuition

Many integrands and derivands look intimidating only because they are written badly. Cancel common factors, combine fractions, or use an algebraic identity FIRST — a 'hard' derivative often collapses to a one-line quotient rule once the expression is clean. Always simplify before differentiating.

Definition

Before differentiating, look for cheap algebraic simplifications:

Common factors in a quotient that cancel.
Negative/fractional powers that combine — e.g. multiply top and bottom by the lower power to clear them.
Identities that reduce a product or ratio to a standard form.

Only after the expression is in its simplest form do you apply the rules. This converts an ugly derivative into a routine one and removes most of the error surface.

Quotient rule (used after simplifying)

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

Worked example

y = \dfrac{x^{1/2} + x^{-1/2}}{x^{1/2} - x^{-1/2}}

, simplify

y

and find

\dfrac{dy}{dx}

Multiply numerator and denominator by $x^{1/2}$ : $y = \dfrac{x + 1}{x - 1}$ .
Now apply the quotient rule with $u = x+1$ , $v = x-1$ : $u' = 1$ , $v' = 1$ .
$\dfrac{dy}{dx} = \dfrac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \dfrac{-2}{(x-1)^2}$ .

Answer:

y = \dfrac{x+1}{x-1}, \quad \dfrac{dy}{dx} = \dfrac{-2}{(x-1)^2}

Practice this conceptself-check · 4 quick reps

Try it yourself

y = \dfrac{1 - \tan^2 x}{1 + \tan^2 x}

, simplify and differentiate.

Practice — Level 1 (4 reps)

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

1.
Simplify and differentiate $y = \dfrac{x^2 - 1}{x - 1}$ .
2.
Simplify and differentiate $y = \dfrac{2\tan x}{1 + \tan^2 x}$ .
3.
Simplify $y = \dfrac{x^{3/2}}{x^{1/2}}$ before differentiating.
4.
Differentiate $y = \log(e^{2x})$ .

From the bank · past-year question

Example 4DifferentiationMODERATE

y = \frac{x^{2/3} - x^{-1/3}}{x^{2/3} + x^{-1/3}},\, x \neq 0

, then

(x+1)^2 y_1 =

[Q103 · 15th May Shift 2 · 2023]

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.

Concept 5 of 7

Linear Approximation Using the Derivative

Intuition

Near a known point, a smooth curve is almost a straight line — its tangent. So to estimate a value slightly away from an easy point, start at the easy value and add the tangent's rise,

h \times \text{slope}

. This turns a hard arithmetic value into a one-line estimate.

Definition

For a small change $h$ about a point $a$ : $f(a + h) \approx f(a) + h\,f'(a)$ . Choose $a$ so that $f(a)$ is easy to compute exactly; let $h$ be the small (possibly negative) gap to the target. The term $h\,f'(a)$ is the tangent-line correction. The closer $h$ is to zero, the better the estimate.

Linear approximation

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

anearby point with an easy exact value
hsmall gap to the target (may be negative)

Worked example

Use linear approximation to estimate

\sqrt{36.6}

Take $f(x) = \sqrt{x}$ , $a = 36$ (easy: $\sqrt{36} = 6$ ), $h = 0.6$ .
$f'(x) = \dfrac{1}{2\sqrt{x}}$ , so $f'(36) = \dfrac{1}{12}$ .
Apply the formula: $\sqrt{36.6} \approx 6 + 0.6 \times \dfrac{1}{12} = 6 + 0.05 = 6.05$ .

Answer:

\sqrt{36.6} \approx 6.05

Practice this conceptself-check · 4 quick reps

Try it yourself

Estimate

(1.02)^5

using linear approximation.

Practice — Level 1 (4 reps)

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

1.
Estimate $\sqrt{25.3}$ .
2.
Estimate $(8.1)^{1/3}$ .
3.
Estimate $\log_{10} 1002$ given $\log_{10} e = 0.4343$ .
4.
Estimate $\sin 31^\circ$ (use $1^\circ \approx 0.01745$ rad).

From the bank · past-year question

Example 5DifferentiationMODERATE

The approximate value of

\log_{10}998

is (given that

\log_{10}e=0.4343

)

[Q115 · 9th May Shift 1 · 2024]

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.

Concept 6 of 7

The Derivative as the Slope of the Tangent

Intuition

Geometrically, the derivative at a point IS the slope of the tangent line there. As the second point of a secant slides toward the first, the secant's slope approaches the tangent's slope — that limiting slope is

f'(x)

. So 'find where the slope is maximum/minimum' means 'analyse

f'(x)

Definition

The slope of the tangent to $y = f(x)$ at $x = a$ is $f'(a)$ . To find where the slope itself is greatest or least, treat the slope function $s(x) = f'(x)$ as a new function and analyse IT: set $s'(x) = f''(x) = 0$ to locate the candidate points, then compare $s$ -values. Equation of the tangent at $(a, f(a))$ : $y - f(a) = f'(a)(x - a)$ .

Slope of the tangent

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

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

Worked example

Find the slope of the tangent to

y = x^3 - 3x

x = 2

The slope function is $\dfrac{dy}{dx} = 3x^2 - 3$ .
Substitute $x = 2$ : $3(4) - 3 = 12 - 3 = 9$ .

Answer:Slope

= 9

Practice this conceptself-check · 4 quick reps

Try it yourself

For

y = e^x \cos x

, find where the SLOPE of the tangent is minimum on

0 \leq x \leq 2\pi

Practice — Level 1 (4 reps)

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

1.
Slope of $y = x^2$ at $x = 3$ .
2.
Slope of $y = \sin x$ at $x = 0$ .
3.
Where is the slope of $y = x^2 - 4x$ zero?
4.
Slope of the tangent to $y = \log x$ at $x = e$ .

From the bank · past-year question

Example 6DifferentiationHARD

Slope of the tangent to the curve

y=2e^x\sin\!\left(\frac{\pi}{4}-\frac{x}{2}\right)\cos\!\left(\frac{\pi}{4}-\frac{x}{2}\right)

where

0\leq x\leq2\pi

is minimum at

x=

[Q107 · 13th May Shift 2 · 2024]

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.

Concept 7 of 7

Differentiability and Where a Derivative Fails to Exist

Intuition

A function is differentiable at a point only if it has ONE well-defined tangent slope there — the slope coming from the left must equal the slope from the right. Smooth curves are fine everywhere; sharp corners (like the tip of

|x|

) and breaks are where the derivative dies. But a vanishing factor can smooth a corner so the function IS differentiable after all.

Definition

$f$ is differentiable at $x = a$ if the left-hand derivative equals the right-hand derivative:

\text{LHD} = \lim_{h \to 0^-}\dfrac{f(a+h) - f(a)}{h} = \lim_{h \to 0^+}\dfrac{f(a+h) - f(a)}{h} = \text{RHD}.

Key facts:

**Differentiable $\Rightarrow$ continuous** (but NOT the converse — $|x|$ is continuous yet not differentiable at $0$ ).
A derivative typically fails at corners ( $|x|$ at $0$ ), cusps, breaks (jump discontinuities), and vertical tangents.
A modulus inside a product can be smoothed: if another factor vanishes at the corner, the product may be differentiable everywhere.

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}

Worked example

f(x) = x\,|x|

differentiable at

x = 0

Write piecewise: $f(x) = x^2$ for $x \geq 0$ and $f(x) = -x^2$ for $x < 0$ .
RHD: $\dfrac{d}{dx}x^2 = 2x \to 0$ as $x \to 0^+$ .
LHD: $\dfrac{d}{dx}(-x^2) = -2x \to 0$ as $x \to 0^-$ .
LHD $= 0 =$ RHD, so $f$ IS differentiable at $0$ — the corner of $|x|$ was smoothed by the extra factor $x$ .

Answer:Yes —

f(x) = x|x|

is differentiable at

x = 0

(and everywhere), with

f'(0) = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

At which point does the derivative of

f(x) = |x - 3|

fail to exist?

Practice — Level 1 (4 reps)

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

1.
Is $|x|$ differentiable at $x = 0$ ?
2.
Where does $f(x) = |x + 2|$ fail to be differentiable?
3.
Does differentiable at $a$ imply continuous at $a$ ?
4.
Is $x^2|x|$ differentiable at $0$ ?

From the bank · past-year question

Example 7DifferentiationMODERATE

The set of all points, where the derivative of the function

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

exists, is

[Q115 · 11th May Shift 2 · 2024]

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.

Drill 2 more on differentiability and where a derivative fails to exist

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 (7)

Standard Derivatives and the Rules of Differentiation
Product rule
$\dfrac{d}{dx}\big[u(x)\,v(x)\big] = u'(x)\,v(x) + u(x)\,v'(x)$
The Chain Rule and Composite Functions
Chain rule
$\dfrac{dy}{dx} = f'\big(g(x)\big) \cdot g'(x) \qquad \text{for } y = f(g(x))$
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
$f(a + h) \approx f(a) + h\,f'(a)$
The Derivative as the Slope of the Tangent
Slope of the tangent
$m_{\text{tangent}} = \left.\dfrac{dy}{dx}\right|_{x=a} = f'(a)$
Differentiability and Where a Derivative Fails to Exist
Differentiability test
$\text{LHD} = \lim_{h\to0^-}\dfrac{f(a+h)-f(a)}{h} = \lim_{h\to0^+}\dfrac{f(a+h)-f(a)}{h} = \text{RHD}$

Watch out for (13)

$\dfrac{d}{dx}a^x$ is $a^x \log a$ , not $x\,a^{x-1}$
→ Standard Derivatives and the Rules of Differentiation
Quotient rule sign: numerator is $u'v - uv'$
→ Standard Derivatives and the Rules of Differentiation
Never forget the inner derivative factor
→ The Chain Rule and Composite Functions
Evaluate the inner argument, not the outer, when a factor is zero
→ The Chain Rule and Composite Functions
Drop the inner coefficient and you lose a factor
→ Differentiating Iterated Functions f(f(x))
Don't try to find a formula for $f$
→ Differentiating Iterated Functions f(f(x))
Simplify first, or the algebra buries you
→ Simplify the Expression Before Differentiating
Pick $h$ small and signed correctly
→ Linear Approximation Using the Derivative
The slope is the DERIVATIVE at $a$ , not at the target
→ Linear Approximation Using the Derivative
Minimum SLOPE means differentiate twice
→ The Derivative as the Slope of the Tangent
Simplify the curve before differentiating
→ The Derivative as the Slope of the Tangent
Not every modulus is a non-differentiable point
→ Differentiability and Where a Derivative Fails to Exist
Continuous does not mean differentiable
→ Differentiability and Where a Derivative Fails to Exist

Mastery check — 5 interleaved questions

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

Example 1DifferentiationMODERATE

f'(x) = \sin(\log x)

and

y = f\!\left(\frac{2x+3}{3-2x}\right)

, then

\frac{dy}{dx}

x=1

[Q109 · 14th May Shift 2 · 2024]

Example 2DifferentiationMODERATE

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

and

f(0) = -1

f'(0) = 1

, then

g'(0)

[Q101 · 16th May Shift 2 · 2023]

Example 3DifferentiationHARD

Let

K

be the set of all real values of

x

, where the function

f(x) = \sin|x| - |x| + 2(x-\pi)\cos|x|

is not differentiable. Then the set

K

[Q120 · 2nd May Shift 1 · 2023]

Example 4DifferentiationMODERATE

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

, then the derivative of

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

x=1

[Q125 · 9th May Shift 1 · 2024]

Example 5DifferentiationHARD

Let

S=\{t\in\mathbb{R}: f(x)=|x-\pi|(e^{|x|}-1)\sin|x|

is not differentiable at

t\}

, then

S

[Q123 · 10th May Shift 2 · 2024]

Drill every past-year question on this subtopic

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