MHT-CET Maths · Differentiation

Parametric Differentiation, Second Derivatives & Proving Relations

When x and y are each given through a parameter t (or theta), differentiate each with respect to the parameter and divide; for the second derivative, differentiate dy/dx again with respect to the parameter and divide once more.

All Differentiation notes MHT-CET Maths strategy

Why this matters

This subtopic carries 10 PYQs — 5 HARD, 3 MODERATE, 2 EASY — and is the part of Differentiation MHT-CET likes most. Three shapes recur: parametric forms (x and y through a parameter), second derivatives of those forms, and 'prove this relation' problems where you must show y satisfies an equation like y'' + n squared times y = 0. The single most-punished mistake is computing the parametric second derivative as a ratio of second derivatives — it is not — so that trap is drilled hard below.

Concept 1 of 5

Parametric Differentiation

Intuition

When you cannot (or do not want to) eliminate the parameter to write y as a function of x, differentiate x and y separately with respect to the parameter, then divide. The parameter cancels in spirit, leaving the genuine slope dy/dx.

Definition

If $x = x(t)$ and $y = y(t)$ are both differentiable and $\dfrac{dx}{dt} \neq 0$ , then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ . The result is usually still a function of the parameter $t$ (or $\theta$ ) — that is fine; you evaluate it at the required parameter value.

Step 1: differentiate $y$ with respect to the parameter.
Step 2: differentiate $x$ with respect to the parameter.
Step 3: divide $dy/dt$ by $dx/dt$ .

Parametric first derivative

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

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

Worked example

For the parabola in parametric form

x = at^2,\ y = 2at

, find

\dfrac{dy}{dx}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1DifferentiationHARD

x=\log_e\!\frac{\cos\frac{y}{2}-\sin\frac{y}{2}}{\cos\frac{y}{2}+\sin\frac{y}{2}},\ \tan\frac{y}{2}=\frac{1-t}{1+t}

. Then

y'\!\!\left|_{t=\frac{1}{2}}\right.

has the value

[Q122 · 13th May Shift 2 · 2024]

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.

Drill 1 more on parametric differentiation

Concept 2 of 5

Second Derivative of a Parametric Function

Intuition

The second derivative is NOT the ratio of the two second derivatives. You already have dy/dx as a function of the parameter; differentiate THAT with respect to the parameter, then divide by dx/dt one more time — exactly the same divide-by-dx/dt move as before.

Definition

Given $x = x(t),\ y = y(t)$ , first find $\dfrac{dy}{dx}$ (a function of $t$ ). Then $\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$ . The key point: apply the chain rule — $\dfrac{d}{dx} = \dfrac{1}{dx/dt}\cdot\dfrac{d}{dt}$ — to the quantity $\dfrac{dy}{dx}$ , not to $y$ . It is emphatically not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ .

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/dtdivide by it once more — the chain-rule leftover

Worked example

For the parabola

x = at^2,\ y = 2at

, find

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

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2DifferentiationHARD

x=2\cos\theta-\cos 2\theta

and

y=2\sin\theta-\sin 2\theta

, then

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

is equal to

[Q133 · 15th May Shift 1 · 2023]

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

Drill 2 more on second derivative of a parametric function

Concept 3 of 5

Proving Second-Order Relations

Intuition

Some questions give y in a form (a sum of sin/cos, a power combination, an exponential) and ask you to show it satisfies a differential relation. The recipe is mechanical: differentiate twice, then look for the original y (or x times y) staring back at you, and substitute.

Definition

To verify that $y$ satisfies a relation such as $y'' + n^2 y = 0$ or $x^2 y'' = n(n+1)y$ : differentiate $y$ once, then again, and rearrange $y''$ until the bracket that appears is exactly $y$ (or a known multiple of it). Two patterns dominate the bank:

Trigonometric: $y = A\cos(nx) + B\sin(nx)$ gives $y'' = -n^2 y$ , i.e. $y'' + n^2 y = 0$ .
Power combination: $y = a x^{n+1} + b x^{-n}$ gives $x^2 y'' = n(n+1)y$ .

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 ythe SHM-type relation from the sin/cos combination
n(n+1)ythe multiple that appears for the power combination

Worked example

y = e^{mx}

, show that

y'' - m^2 y = 0

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3DifferentiationEASY

y=A\cos(nx)+B\sin(nx)

, then

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

[Q139 · 4th May Shift 2 · 2023]

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.

Drill 1 more on proving second-order relations

Concept 4 of 5

Showing an Expression Is Constant

Intuition

If you can show the derivative of an expression is identically zero, the expression cannot change — it is a constant. So its value at any one point equals its value everywhere. These questions hand you a value at one point and ask for it at another; the answer is just the same number.

Definition

If $\dfrac{d}{dx}\big[E(x)\big] = 0$ for all $x$ , then $E(x)$ is constant, so $E(b) = E(a)$ for any $a, b$ . The work is to differentiate the given expression and watch the terms cancel to zero, using the supplied relations (for example $f'' = -f$ and $g = f'$ ). Once $E' = 0$ , simply read off: whatever value is given at one point is the value at every point.

Zero derivative implies constant

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

Worked example

Verify that

E(x) = \sin^2 x + \cos^2 x

is constant by differentiation, and hence find

E(7)

given

E(0) = 1

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4DifferentiationMODERATE

Let f be twice differentiable function such that

f''(x)=-f(x)

f'(x)=g(x)

and

h(x)=(f(x))^2+(g(x))^2

. If

h(5)=1

, then the value of

h(10)

[Q144 · 11th May Shift 2 · 2023]

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.

Drill 2 more on showing an expression is constant

Concept 5 of 5

nth-Order Derivatives — Standard Results

Intuition

A handful of functions have a clean pattern when you keep differentiating: powers shed exponents factorially, the sine/cosine of a linear argument simply add quarter-turns to the phase, and the exponential reproduces a power of its coefficient. Knowing the pattern lets you jump straight to the nth derivative without grinding through every step. MHT-CET usually stops at the second order, so treat these as a completeness reference.

Definition

Standard nth-order derivatives (for a linear argument $ax + b$ ):

$\dfrac{d^n}{dx^n}(x^m) = \dfrac{m!}{(m-n)!}\,x^{m-n}$ (for $m \geq n$ )
$\dfrac{d^n}{dx^n}\!\left(\dfrac{1}{ax+b}\right) = \dfrac{(-1)^n\,n!\,a^n}{(ax+b)^{n+1}}$
$\dfrac{d^n}{dx^n}\big(\log(ax+b)\big) = \dfrac{(-1)^{n-1}(n-1)!\,a^n}{(ax+b)^n}$
$\dfrac{d^n}{dx^n}\big(\sin(ax+b)\big) = a^n \sin\!\left(ax+b+\dfrac{n\pi}{2}\right)$
$\dfrac{d^n}{dx^n}\big(\cos(ax+b)\big) = a^n \cos\!\left(ax+b+\dfrac{n\pi}{2}\right)$
$\dfrac{d^n}{dx^n}\big(e^{ax}\big) = a^n e^{ax}$

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

Worked example

Find the nth derivative of

\dfrac{1}{2x+1}

, and then write the 4th derivative of

\sin 2x

Practice this concept4 quick reps

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.

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 (5)

Parametric Differentiation
Parametric first derivative
$\dfrac{dy}{dx} = \dfrac{\,dy/dt\,}{\,dx/dt\,}, \qquad \dfrac{dx}{dt} \neq 0$
Second Derivative of a Parametric Function
Parametric second derivative
$\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$
Proving Second-Order Relations
Two standard second-order relations
$y = A\cos nx + B\sin nx \Rightarrow y'' = -n^2 y; \qquad y = ax^{n+1} + bx^{-n} \Rightarrow x^2 y'' = n(n+1)y$
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
$\dfrac{d^n}{dx^n}\big(\sin(ax+b)\big) = a^n \sin\!\left(ax+b+\dfrac{n\pi}{2}\right)$

Watch out for (10)

Do not flip the ratio
→ Parametric Differentiation
The slope can stay in terms of the parameter
→ Parametric Differentiation
NEVER divide the two second derivatives
→ Second Derivative of a Parametric Function
Differentiate dy/dx with respect to t, not x
→ Second Derivative of a Parametric Function
Carry the constants — they cancel cleanly
→ Proving Second-Order Relations
Match the power-combination exponents
→ Proving Second-Order Relations
Zero derivative means constant — the second point is a decoy
→ Showing an Expression Is Constant
Use the supplied relations during differentiation
→ Showing an Expression Is Constant
Sine and cosine cycle with period 4 in the order n
→ nth-Order Derivatives — Standard Results
The power-rule nth derivative stops at zero
→ nth-Order Derivatives — Standard Results

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