MHT-CET Maths · Differentiation

Implicit Differentiation and Special Forms

When y is tangled up with x in one equation, differentiate the whole equation as it stands — treating y as a hidden function of x — and then solve for dy/dx.

All Differentiation notes MHT-CET Maths strategy

Why this matters

This is the broadest subtopic in the chapter — 43 PYQs (25 HARD, 17 MODERATE, 1 EASY) — and the paper's reliable source of non-routine relations. Beyond the core implicit method, MHT-CET keeps recycling a handful of signature shapes: the log(x+y)=2xy family, exponential relations you must log first, tan y written as a rational in x, 'prove this relation' problems, self-referential infinite expressions, and functional equations. Recognise the shape and the method follows.

Concept 1 of 7: Implicit Differentiation — the Core Method

Sometimes you cannot solve an equation for y cleanly (it is buried inside the equation with x). You do not need to. Differentiate every term of the equation with respect to x as it stands — wherever y appears, its derivative carries a dy/dx by the chain rule — then collect the dy/dx terms and solve.

Definition

For a relation $F(x,y)=0$ where $y$ is an (implicit) function of $x$ :

Differentiate both sides with respect to $x$ .
Every $y$ -term picks up a factor $\dfrac{dy}{dx}$ (chain rule): $\dfrac{d}{dx}(y^n)=n y^{n-1}\dfrac{dy}{dx}$ , $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$ .
Gather all $\dfrac{dy}{dx}$ terms on one side and solve for $\dfrac{dy}{dx}$ .

The slope of the tangent at a point is $\dfrac{dy}{dx}$ evaluated there.

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 $\frac{dy}{dx}$

Worked example

Find

\dfrac{dy}{dx}

for

x^3 + y^3 = 3axy

(the folium of Descartes).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q115Moderate

Example 1 · Differentiation · Implicit Differentiation & Special Forms

x^{\frac{2}{5}}+y^{\frac{2}{5}}=a^{\frac{2}{5}}

then

\frac{dy}{\text{ }dx}=

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.

Drill 4 more on implicit differentiation — the core method

Concept 2 of 7: Implicit Relations like log(x + y) = 2xy

These ask for dy/dx at a single point (almost always x = 0), not a general formula. The trick is in two halves: first use the equation itself to find the y-value at that x, then differentiate implicitly and plug in BOTH coordinates to read off the slope.

Definition

For a relation tying $x+y$ to a product or transcendental expression, evaluated at a point:

Find the point. Substitute the given $x$ into the original equation to solve for $y$ . For $\log(x+y)=2xy$ at $x=0$ : $\log y = 0 \Rightarrow y = 1$ .
Differentiate implicitly, then substitute the full point $(x,y)$ and solve for $\dfrac{dy}{dx}$ .

For $\log(x+y)=2xy$ : $\dfrac{1+y'}{x+y} = 2(y + xy')$ ; at $(0,1)$ this gives $1+y' = 2y' \Rightarrow y' = 1$ .

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 $x+y$

Worked example

\log(x + y) = xy

, find

\dfrac{dy}{dx}

x = 0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q133Moderate

Example 2 · Differentiation · Implicit Differentiation & Special Forms

\log(x+y)=2xy

, then

\frac{dy}{dx}

x=0

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.

Drill 5 more on implicit relations like log(x + y) = 2xy

Concept 3 of 7: Exponential Relations — Take Logs, Then Differentiate

When the variable y sits in an EXPONENT (like (2x) raised to the power 2y), you cannot differentiate directly — the exponent is variable. Take natural logs of the whole equation first. That pulls the exponent down as a multiplier, turning a power into a product you can differentiate implicitly.

Definition

For a relation where $y$ appears in an exponent on one or both sides:

Take $\log$ of both sides to bring exponents down: $\log(A^{B}) = B\log A$ .
Differentiate implicitly (product rule on each $B\log A$ term).
Collect and solve for $\dfrac{dy}{dx}$ .

Example shape: $(2x)^{2y}=4e^{2x-2y}$ becomes $2y\log(2x) = \log 4 + 2x - 2y$ ; differentiating yields $(1+\log 2x)^2\dfrac{dy}{dx} = x\log 2x - \log 2x$ .

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

Worked example

x^{y} = e^{x-y}

, find

\dfrac{dy}{dx}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q128Hard

Example 3 · Differentiation · Implicit Differentiation & Special Forms

For

x > 1

, if

(2x)^{2y} = 4e^{2x-2y}

, then

\left(1+\log_e 2x\right)^2 \frac{dy}{dx}

is equal to

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.

Drill 1 more on exponential relations — take logs, then differentiate

Concept 4 of 7: Relations of the Form tan y = (rational in x)

When y is given by tan y equal to a rational function of x, you differentiate using sec-squared on the left and the quotient rule on the right. The magic is that sec-squared y rebuilds itself from 1 + tan-squared y, and the tan-squared cancels neatly against the rational expression — leaving a clean quadratic denominator in x.

Definition

Given $\tan y = \dfrac{x\sin\alpha}{1 - x\cos\alpha}$ :

Differentiate: $\sec^2 y\,\dfrac{dy}{dx} = \dfrac{d}{dx}\!\left[\dfrac{x\sin\alpha}{1-x\cos\alpha}\right]$ (quotient rule on the right).
Replace $\sec^2 y = 1 + \tan^2 y$ and substitute $\tan y$ from the given relation; the algebra collapses to

$\dfrac{dy}{dx} = \dfrac{\sin\alpha}{1 - 2x\cos\alpha + x^2}$ . Matching to $\dfrac{m}{x^2 + 2nx + 1}$ gives $m = \sin\alpha$ , $n = -\cos\alpha$ , so $m^2 + n^2 = 1$ .

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

Worked example

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

, find

\dfrac{dy}{dx}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q101Hard

Example 4 · Differentiation · Implicit Differentiation & Special Forms

\tan y = \frac{x\sin\alpha}{1 - x\cos\alpha}

and

\frac{dy}{dx} = \frac{m}{x^2 + 2nx + 1}

, then

m^2 + n^2

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.

Concept 5 of 7: Proving a Given Differential Relation

These show a relation that hides an explicit form of y (often through a substitution), then ask you to verify an identity in y and its derivative. The reliable route: unwrap the relation to get y explicitly, differentiate, and confirm the claimed identity falls out.

Definition

Typical shape: $y^{1/m} + y^{-1/m} = 2x$ (or similar), prove $(x^2-1)(y')^2 = m^2 y^2$ .

Unwrap to explicit $y$ . Set $t = y^{1/m}$ ; then $t + \dfrac{1}{t} = 2x$ is a quadratic $t^2 - 2xt + 1 = 0$ , giving $t = x + \sqrt{x^2-1}$ , so $y = \left(x+\sqrt{x^2-1}\right)^{m}$ .
Differentiate. $\dfrac{dy}{dx} = \dfrac{m y}{\sqrt{x^2-1}}$ .
Square and clear the root: $(x^2-1)(y')^2 = m^2 y^2$ . Done.

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 $t$

Worked example

y = \left(x + \sqrt{x^2+1}\right)^{n}

, prove that

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q140Hard

Example 5 · Differentiation · Implicit Differentiation & Special Forms

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

x \neq 1

, then

\left(x^2-1\right)\!\left(\frac{dy}{dx}\right)^2

is equal to

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.

Drill 2 more on proving a given differential relation

Concept 6 of 7: Self-Referential Infinite Expressions

An infinite nested radical or continued product repeats itself forever — so the whole thing equals one copy of the pattern wrapped around the whole thing again. Set y equal to the entire expression; the inner copy is also y. That self-reference turns an infinite tower into a simple finite equation you can differentiate implicitly.

Definition

For $y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x) + \cdots}}}$ :

The expression under the first root is $f(x)$ plus the SAME infinite expression, i.e. $f(x) + y$ .
So $y = \sqrt{f(x) + y}$ , giving the finite equation $y^2 = f(x) + y$ .
Differentiate implicitly: $2y\,y' = f'(x) + y'$ , so $\dfrac{dy}{dx} = \dfrac{f'(x)}{2y - 1}$ .

For $f(x) = x - \sin x$ this gives $\dfrac{dy}{dx} = \dfrac{1 - \cos x}{2y - 1}$ .

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}

ythe whole infinite expression — it reappears under the first root

Worked example

y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}

, find

\dfrac{dy}{dx}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q112Moderate

Example 6 · Differentiation · Implicit Differentiation & Special Forms

y = \sqrt{(x-\sin x) + \sqrt{(x-\sin x) + \sqrt{x - \sin x + \cdots}}}

, then

\frac{dy}{dx} =

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.

Drill 1 more on self-referential infinite expressions

Concept 7 of 7: Functional Equations — Find f, Then Differentiate

These give an equation that f must satisfy (often involving f(x) and f(1/x), or f together with its own derivatives) rather than f directly. First pin down f — by substituting x to 1/x and solving the pair, or by comparing coefficients — and only then differentiate to answer the question.

Definition

Two recurring routes:

Reciprocal substitution. Given $a f(x) + b f(1/x) = g(x)$ , replace $x \to 1/x$ to get a second equation, then solve the two simultaneously for $f(x)$ .
Coefficient comparison. Given $f(x) = x^3 + x^2 f'(1) + x f''(2) + 6$ , let $f'(1)=a$ , $f''(2)=b$ (constants); differentiate to get $f'(x), f''(x)$ , evaluate at the stated points, and solve for $a, b$ . Here $a=-5, b=2$ , so $f(x)=x^3-5x^2+2x+6$ and $f(2)=-2$ .

Once $f$ is explicit, differentiate normally.

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

Worked example

2f(x) + f\!\left(\dfrac{1}{x}\right) = 3x

for

x \neq 0

, find

f'(2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q132Hard

Example 7 · Differentiation · Implicit Differentiation & Special Forms

Let

f:R\to R

be a function such that

f(x)=x^3+x^2f'(1)+xf''(2)+6,\ x\in R

, then

f(2)

equals

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.

Drill 3 more on functional equations — find f, then differentiate

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)

Implicit Differentiation — the Core Method
Implicit chain rule
$\frac{d}{dx}\big[g(y)\big] = g'(y)\,\frac{dy}{dx}, \qquad \frac{d}{dx}(xy)=y+x\frac{dy}{dx}$
Implicit Relations like log(x + y) = 2xy
Differentiating log(x + y)
$\frac{d}{dx}\log(x+y) = \frac{1}{x+y}\left(1 + \frac{dy}{dx}\right)$
Exponential Relations — Take Logs, Then Differentiate
Log first, then differentiate
$\frac{d}{dx}\big[u(x)\,\log v(x)\big] = u'\log v + u\cdot\frac{v'}{v}$
Relations of the Form tan y = (rational in x)
Standard result
$\tan y = \frac{x\sin\alpha}{1 - x\cos\alpha} \;\Rightarrow\; \frac{dy}{dx} = \frac{\sin\alpha}{1 - 2x\cos\alpha + x^2}$
Proving a Given Differential Relation
Key explicit form
$y^{1/m} + y^{-1/m} = 2x \;\Rightarrow\; y = \left(x + \sqrt{x^2-1}\right)^{m}$
Self-Referential Infinite Expressions
Self-reference for a nested radical
$y = \sqrt{f(x) + y} \;\Rightarrow\; y^2 = f(x) + y \;\Rightarrow\; \frac{dy}{dx} = \frac{f'(x)}{2y - 1}$
Functional Equations — Find f, Then Differentiate
Reciprocal-substitution setup
$a f(x) + b f\!\left(\tfrac{1}{x}\right) = g(x), \quad\text{then } x\to\tfrac1x:\; a f\!\left(\tfrac1x\right) + b f(x) = g\!\left(\tfrac1x\right)$

Watch out for (15)

Differentiating a y-term without the dy/dx factor
→ Implicit Differentiation — the Core Method
Forgetting the product rule on the xy term
→ Implicit Differentiation — the Core Method
Find the y-value before substituting into the derivative
→ Implicit Relations like log(x + y) = 2xy
log(x + y) = sin(x + y) collapses to slope -1
→ Implicit Relations like log(x + y) = 2xy
You cannot use the power rule when the exponent contains y
→ Exponential Relations — Take Logs, Then Differentiate
Use the original (logged) relation to simplify the final answer
→ Exponential Relations — Take Logs, Then Differentiate
Differentiate tan y as sec-squared y times dy/dx
→ Relations of the Form tan y = (rational in x)
Spotting a hidden inverse-tangent shortcut
→ Relations of the Form tan y = (rational in x)
Use the substitution to get y explicitly first
→ Proving a Given Differential Relation
Square only after isolating the root
→ Proving a Given Differential Relation
The inner expression equals the WHOLE y, not part of it
→ Self-Referential Infinite Expressions
Square before differentiating, not after
→ Self-Referential Infinite Expressions
f'(1), f''(2) are CONSTANTS — name them and solve
→ Functional Equations — Find f, Then Differentiate
For f(x) and f(1/x), substitute x to 1/x to get a second equation
→ Functional Equations — Find f, Then Differentiate
f'(x) = f(x) means exponential
→ Functional Equations — Find f, Then Differentiate

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