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

Practice this conceptself-check · 4 quick reps

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]

Every $y$ -term needs a $\frac{dy}{dx}$ factor

Differentiating

y^2

w.r.t.

x

gives

2y\dfrac{dy}{dx}

, NOT

2y

. Because

y

is a function of

x

, the chain rule attaches a

\frac{dy}{dx}

to every

y

-term. Forgetting it is the defining error of implicit differentiation — you'd never recover

\frac{dy}{dx}

to solve for.

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

Practice this conceptself-check · 4 quick reps

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}

Practice this conceptself-check · 4 quick reps

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]

Don't invert the parametric ratio

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

— the

y

-rate is on TOP, the

x

-rate on the bottom (it cancels like a fraction:

\frac{dy}{dt}\div\frac{dx}{dt}

). Writing

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

gives the reciprocal

\frac{dx}{dy}

instead. For

x=t^2, y=t^3

the slope is

\frac{3t^2}{2t}=\frac{3t}{2}

, not

\frac{2}{3t}

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}

Practice this conceptself-check · 4 quick reps

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]

$\frac{d^2y}{dx^2}$ is not $\left(\frac{dy}{dx}\right)^2$

The second derivative means 'differentiate the first derivative AGAIN', not 'square the first derivative'. For

y=x^3

\frac{dy}{dx}=3x^2

, so

\frac{d^2y}{dx^2}=6x

— whereas

\left(\frac{dy}{dx}\right)^2 = 9x^4

, a completely different (and wrong) object.

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.

Practice this conceptself-check · 4 quick reps

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

Practice this conceptself-check · 4 quick reps

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 (4)

Every $y$ -term needs a $\frac{dy}{dx}$ factor
→ Implicit differentiation
Don't invert the parametric ratio
→ Parametric differentiation
$\frac{d^2y}{dx^2}$ is not $\left(\frac{dy}{dx}\right)^2$
→ Higher-order derivatives
Second derivatives don't invert like first derivatives
→ Second derivative of an inverse — the d²x/dy² identity

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