NDA Maths · Differentiation

Parametric, Implicit & Higher-Order Derivatives

When y is tangled with x (an implicit equation), routed through a parameter t, or you need the second derivative, the chain rule is still the engine — applied a little differently.

All Differentiation notes NDA Maths strategy

Why this matters

These are the advanced-form questions: differentiate both sides of F(x,y)=0, divide parametric rates, or differentiate twice. They reward knowing the right setup — especially the parametric second-derivative formula and the d²x/dy² reciprocal identity, which are easy marks once memorised and easy to botch otherwise.

Concept 1 of 6

Implicit differentiation

Intuition

When

y

can't be isolated, differentiate both sides with respect to

x

, treating

y

as a function of

x

— so every

y

-term picks up a

\frac{dy}{dx}

factor by the chain rule. Then solve the linear equation for

\frac{dy}{dx}

Definition

Differentiate $F(x,y)=0$ term by term w.r.t. $x$ : a term in $y$ gives its derivative $\times \frac{dy}{dx}$ (chain rule), and products use the product rule. Collect the $\frac{dy}{dx}$ terms and solve. No need to express $y$ explicitly.

Implicit rule of thumb

\frac{d}{dx}\big[\,y\text{-term}\,\big] = (\text{its derivative})\cdot\frac{dy}{dx}

Worked example

Find

\dfrac{dy}{dx}

for

x^2 + y^2 = 25

Differentiate both sides: $2x + 2y\dfrac{dy}{dx} = 0$ .
Solve: $\dfrac{dy}{dx} = -\dfrac{x}{y}$ .

Answer:

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

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

\dfrac{dy}{dx}

for

xy = 1

Practice — Level 1 (4 reps)

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

1.
Differentiating $y^3$ w.r.t. $x$ gives?
2.
$\frac{dy}{dx}$ for $x^2+y^2=r^2$ ?
3.
Which rule for the term $xy$ ?
4.
Must you isolate $y$ first?

From the bank · past-year question

Example 1DifferentiationMODERATE

x^m y^n=a^{m+n}

, then what is

\frac{dy}{dx}

equal to?

[Q87 · Apr · 2020]

Drill 4 more on implicit differentiation

Concept 2 of 6

Logarithmic differentiation of implicit power relations

Intuition

When both

x

and

y

appear in exponents —

x^y

y^x

x^m y^n

— take

\ln

of both sides FIRST. Logs turn products into sums and exponents into coefficients, after which implicit differentiation is routine.

Definition

For relations like $x^y y^x = c$ : take $\ln$ to get $y\ln x + x\ln y = \ln c$ , then differentiate implicitly (product rule on each term, $\frac{dy}{dx}$ on $y$ -terms) and solve. For $x^m y^n = k$ : $\,m\ln x + n\ln y = \ln k \Rightarrow \frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = 0$ .

Worked example

Use logarithms to find

\dfrac{dy}{dx}

for

x^2 y^3 = 1

Take $\ln$ : $2\ln x + 3\ln y = 0$ .
Differentiate: $\dfrac{2}{x} + \dfrac{3}{y}\dfrac{dy}{dx} = 0$ .
Solve: $\dfrac{dy}{dx} = -\dfrac{2y}{3x}$ .

Answer:

\dfrac{dy}{dx} = -\dfrac{2y}{3x}

Practice this conceptself-check · 4 quick reps

Try it yourself

For

x^y = e

(constant), find

\dfrac{dy}{dx}

Practice — Level 1 (4 reps)

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

1.
First step for $x^y y^x = c$ ?
2.
$\ln(x^m y^n)$ ?
3.
$\frac{dy}{dx}$ for $x^m y^n = k$ ?
4.
Why take logs first?

From the bank · past-year question

Example 2DifferentiationMODERATE

x^y y^x=1

, then what is

\frac{dy}{dx}

(1,1)

equal to?

[Q69 · Apr · 2022]

Drill 2 more on logarithmic differentiation of implicit power relations

Concept 3 of 6

Parametric differentiation

Intuition

When

x

and

y

are both given in terms of a parameter

t

, you don't eliminate

t

. Differentiate each w.r.t.

t

and divide:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

. The parameter cancels out of the ratio.

Definition

If $x=f(t)$ and $y=g(t)$ , then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{g'(t)}{f'(t)}$ (provided $f'(t)\neq 0$ ). The result is usually left in terms of $t$ .

Parametric first derivative

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Worked example

x = t^2

and

y = t^3

, find

\dfrac{dy}{dx}

$\dfrac{dx}{dt} = 2t$ , $\dfrac{dy}{dt} = 3t^2$ .
Divide: $\dfrac{dy}{dx} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$ .

Answer:

\dfrac{dy}{dx} = \dfrac{3t}{2}

Practice this conceptself-check · 4 quick reps

Try it yourself

x = a\cos\theta

y = a\sin\theta

, find

\dfrac{dy}{dx}

Practice — Level 1 (4 reps)

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

1.
Parametric $\frac{dy}{dx}$ formula?
2.
$x=t^2, y=t^3$ : $\frac{dy}{dx}$ ?
3.
Do you eliminate $t$ first?
4.
Condition for the formula?

From the bank · past-year question

Example 3DifferentiationEASY

x=e^t\cos t

and

y=e^t\sin t

, then what is

\frac{dx}{dy}

t=0

equal to?

[Q89 · Apr · 2021]

Drill 5 more on parametric differentiation

Concept 4 of 6

Higher-order derivatives

Intuition

The second derivative is just the derivative of the first; the

n

-th is the derivative applied

n

times. For named functions there are clean patterns; for a specific point, differentiate the required number of times and substitute last.

Definition

$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)$ , and so on. Useful standard results: $(e^{ax})^{(n)} = a^n e^{ax}$ ; $(\sin x)^{(n)} = \sin\!\left(x + \tfrac{n\pi}{2}\right)$ . To evaluate at a point, differentiate first and substitute the value at the end.

Second derivative

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

Worked example

y = x^4

, find

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

First derivative: $\dfrac{dy}{dx} = 4x^3$ .
Differentiate again: $\dfrac{d^2y}{dx^2} = 12x^2$ .

Answer:

\dfrac{d^2y}{dx^2} = 12x^2

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the second derivative of

y = \sin x

x = 0

Practice — Level 1 (4 reps)

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

1.
$\frac{d^2y}{dx^2}$ means?
2.
$n$ -th derivative of $e^{ax}$ ?
3.
$y=x^4$ : $y''$ ?
4.
Evaluate at a point — before or after differentiating?

From the bank · past-year question

Example 4DifferentiationHARD

Directions for the following two (02) items : Read the following information and answer the two items that follow : Consider the equation

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

What is

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

x = 1

equal to ?

[Q85 · Sep · 2019]

Drill 2 more on higher-order derivatives

Concept 5 of 6

Second derivative of an inverse — the d²x/dy² identity

Intuition

The first derivatives are reciprocals:

\frac{dx}{dy} = 1/\frac{dy}{dx}

. But the SECOND derivatives are NOT reciprocals —

\frac{d^2x}{dy^2} \neq 1/\frac{d^2y}{dx^2}

. The correct relation carries a cube and a sign.

Definition

$\dfrac{dx}{dy} = \left(\dfrac{dy}{dx}\right)^{-1}$ , and differentiating this w.r.t. $y$ (chain rule) gives $\dfrac{d^2x}{dy^2} = -\dfrac{d^2y/dx^2}{\left(dy/dx\right)^{3}}$ . Memorise the cube in the denominator and the minus sign — the classic trap.

Second derivative of the inverse

\frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{\left(dy/dx\right)^{3}}

Worked example

\dfrac{dy}{dx} = 2

and

\dfrac{d^2y}{dx^2} = 3

at a point, find

\dfrac{d^2x}{dy^2}

there.

Apply $\dfrac{d^2x}{dy^2} = -\dfrac{d^2y/dx^2}{(dy/dx)^3}$ .
Substitute: $-\dfrac{3}{2^3} = -\dfrac{3}{8}$ .

Answer:

\dfrac{d^2x}{dy^2} = -\dfrac{3}{8}

Practice this conceptself-check · 4 quick reps

Try it yourself

Why is

\dfrac{d^2x}{dy^2}

not simply

1\big/\dfrac{d^2y}{dx^2}

Practice — Level 1 (4 reps)

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

1.
$\frac{dx}{dy}$ in terms of $\frac{dy}{dx}$ ?
2.
$\frac{d^2x}{dy^2} = ?$
3.
Common wrong answer?
4.
$dy/dx=1, d^2y/dx^2=5$ : $d^2x/dy^2$ ?

From the bank · past-year question

Example 5DifferentiationHARD

What is

\dfrac{d^2x}{dy^2}

equal to?

[Q80 · Apr · 2017]

Second derivatives don't invert like first derivatives

\frac{dx}{dy}=1/\frac{dy}{dx}

is fine, but

\frac{d^2x}{dy^2}\neq 1/\frac{d^2y}{dx^2}

. The right formula has

(dy/dx)^3

in the denominator and a minus sign.

Drill 1 more on second derivative of an inverse — the d²x/dy² identity

Concept 6 of 6

Showing y satisfies a differential equation

Intuition

A common question gives

y

(often

\sin(\ln x)

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

, or similar) and asks you to verify it satisfies a relation among

y, y', y''

. Strategy: compute

y'

and

y''

, then combine them to cancel the transcendental parts back to

y

Definition

Differentiate $y$ once and twice, then substitute into the target relation and simplify — the goal is to eliminate $\sin, \ln,$ etc. and land on $0$ . Example: $y=\sin(\ln x)$ gives $xy' = \cos(\ln x)$ , and differentiating again yields $x^2 y'' + x y' + y = 0$ .

Worked example

Show that

y = e^{2x}

satisfies

y'' - 4y = 0

$y' = 2e^{2x}$ , $y'' = 4e^{2x}$ .
Substitute: $y'' - 4y = 4e^{2x} - 4e^{2x} = 0$ .

Answer:Verified —

y'' - 4y = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Verify that

y = \sin(\ln x)

satisfies

x^2 y'' + x y' + y = 0

Practice — Level 1 (4 reps)

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

1.
Strategy to prove a relation in $y, y', y''$ ?
2.
$y=e^{2x}$ : $y''$ ?
3.
For $y=\sin(\ln x)$ , $xy' = ?$
4.
Goal after substituting?

From the bank · past-year question

Example 6DifferentiationMODERATE

y=\sin(\ln x)

, then which one of the following is correct?

[Q92 · Sep · 2018]

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 (4)

Implicit differentiation
Implicit rule of thumb
$\frac{d}{dx}\big[\,y\text{-term}\,\big] = (\text{its derivative})\cdot\frac{dy}{dx}$
Parametric differentiation
Parametric first derivative
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
Higher-order derivatives
Second derivative
$\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right)$
Second derivative of an inverse — the d²x/dy² identity
Second derivative of the inverse
$\frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{\left(dy/dx\right)^{3}}$

Watch out for (1)

Second derivatives don't invert like first derivatives
→ Second derivative of an inverse — the d²x/dy² identity

Mastery check — 5 interleaved questions

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

Example 1DifferentiationHARD

\sqrt[4]{y}=x+\sqrt{x^2+4}

, then what is

\sqrt{x^2+4}\dfrac{dy}{dx}

equal to?

[Q76 · Apr · 2026]

Example 2DifferentiationMODERATE

Let

(x+y)^{p+q} = x^p y^q

, where

p, q

are positive integers.

The derivative of

y

with respect to

x

[Q75 · Apr · 2025]

Example 3DifferentiationHARD

For the following three (03) items: Let

y=f(x)=\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}+\ln\sqrt{1-x^2}

x=\sin\theta

, then what is

\dfrac{dy}{dx}

equal to?

[Q80 · Sep · 2025]

Example 4DifferentiationEASY

Let

f(x)

and

g(x)

be two functions such that

g(x)=x-\dfrac{1}{x}

and

f\!\circ\! g(x)=x^3-\dfrac{1}{x^3}

What is

f''(x)

equal to?

[Q86 · Apr · 2024]

Example 5DifferentiationHARD

for the items that follow: Consider the differential equation

e^{x+y}\dfrac{dy}{dx}=e^{x-y}

What is

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

equal to?

[Q97 · Apr · 2026]

Drill every past-year question on this subtopic

20 questions from the bank — paginated, with cart and Word-export support.

Drill the 20 questions

Drill every past-year question on this subtopic

Related notes