MHT-CET Maths · Applications of Derivative

Approximations Using Differentials

Near an easy point, a smooth curve is almost its tangent line — so f(a + h) is roughly f(a) plus the tangent's rise h·f'(a). This one formula estimates roots, powers, trig values, logs, exponentials, and polynomial values.

All Applications of Derivative notes MHT-CET Maths strategy

Why this matters

This subtopic is a reliable easy-to-moderate scorer on MHT-CET: 11 PYQs sit here (10 MODERATE, 1 EASY), and every one is the SAME single-line move — pick a nearby exact point, add the tangent correction. The recurring traps are all mechanical: choosing an anchor whose value you cannot compute exactly, getting the sign of h wrong, and — the biggest one — using degrees instead of radians for a trig derivative. Master the formula once and the whole subtopic collapses into arithmetic.

Concept 1 of 5

The Differential dy and the Linear-Approximation Formula

Intuition

For a tiny input change

dx

, the actual change in

y

is almost exactly the tangent's rise:

dy = f'(x)\,dx

. Geometrically the tangent line hugs the curve near the point of contact, so replacing the curve by its tangent gives a fast, accurate estimate. To evaluate

f

at a point slightly off an easy one, start at the easy value and add the tangent's rise.

Definition

For a differentiable function, the differential is $dy = f'(x)\,dx$ — the change predicted by the tangent line. Writing the target as $a + h$ where $a$ is a nearby point with an easy exact value and $h$ is a small (possibly negative) gap:

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

Two disciplines make this work every time:

Choose $a$ so $f(a)$ is exact and clean — a perfect square/cube, a standard angle, a round power of 10.
Get the sign of $h$ right — if the target is below the anchor, $h$ is negative.

The correction term $h\,f'(a)$ uses the slope AT the anchor $a$ , never at the target.

Linear approximation

f(a + h) \approx f(a) + h\,f'(a) \qquad \big(dy = f'(x)\,dx\big)

anearby point with an easy exact value
hsmall gap to the target (may be negative)
f'(a)slope at the anchor a — the multiplier of h

Worked example

Estimate

\sqrt{25.3}

using differentials.

Practice this conceptself-check · 4 quick reps

The slope is $f'(a)$ — evaluate at the anchor, not the target

The correction term is

h\,f'(a)

, using the derivative at the EASY point

a

. Evaluating

f'(a+h)

(at the target) defeats the whole purpose — you chose

a

precisely so the slope there is easy. For

\sqrt{25.3}

use

f'(25)=1/10

, not

f'(25.3)

Get the sign of $h$ right

If the target is BELOW the anchor,

h

is negative. To estimate

\sqrt{24.7}

with

a=25

, take

h=-0.3

, giving

5 - 0.03 = 4.97

. A wrong sign pushes the estimate the wrong way by twice the correction.

Concept 2 of 5

Approximating Roots and Powers

Intuition

The most common target is a root or a fractional power near a perfect value —

\sqrt[3]{64.04}

(3.978)^{3/2}

\sqrt[3]{0.026}

. Anchor at the nearest perfect power so

f(a)

is a whole number, then add one tangent correction. The whole difficulty is picking the right anchor and computing

f'(a)

cleanly.

Definition

For $f(x) = x^{p/q}$ : $f'(x) = \dfrac{p}{q}\,x^{p/q - 1}$ , and $f(a+h) \approx a^{p/q} + h\cdot\dfrac{p}{q}\,a^{p/q-1}$ .

Cube root $f(x)=x^{1/3}$ : $f'(x)=\dfrac{1}{3x^{2/3}}$ . Anchor at a perfect cube ( $64,\;0.027,\;8$ ).
Three-halves power $f(x)=x^{3/2}$ : $f'(x)=\dfrac{3}{2}\sqrt{x}$ . Anchor at a perfect square ( $4,\;9,\;16$ ).

Small decimals like $0.026$ still work — anchor at the nearby perfect cube $0.027 = 0.3^3$ .

Power/root approximation

(a + h)^{p/q} \approx a^{p/q} + h\cdot\dfrac{p}{q}\,a^{\,p/q - 1}

anearest perfect power (perfect cube for a cube root, etc.)
p/qthe exponent — carries through to the derivative

Worked example

Estimate

\sqrt[3]{64.04}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Applications of DerivativeMODERATE

The approximate value of

(3.978)^{3/2}

[Q125 · 11th May Shift 1 · 2023]

Anchor at a perfect power, not just any round number

For

\sqrt[3]{0.026}

, do NOT anchor at

0

0.025

— neither has a clean cube root. Use

0.027 = 0.3^3

f(a)=0.3

is exact. Choosing an anchor whose value you cannot compute exactly wrecks the whole method.

Watch $x^{p/q - 1}$ in the derivative

For

x^{3/2}

the derivative is

\tfrac32 x^{1/2} = \tfrac32\sqrt{x}

, so

f'(4) = \tfrac32\cdot 2 = 3

— a clean integer, which is why

a=4

is the right anchor. Subtracting

1

from the exponent wrongly (e.g. leaving

x^{3/2}

) inflates the correction.

Drill 3 more on approximating roots and powers

Concept 3 of 5

Approximating Trigonometric Values

Intuition

To estimate a trig value a few minutes/seconds away from a standard angle, anchor at the standard angle (30°, 45°, 60°) and add the tangent correction. The one rule that trips everyone: the derivative of a trig function is in RADIANS, so the gap

h

must be converted from degrees/minutes/seconds to radians first.

Definition

Use $f(a+h) \approx f(a) + h\,f'(a)$ with $f = \sin$ or $\cos$ :

$\dfrac{d}{dx}\sin x = \cos x$ , $\dfrac{d}{dx}\cos x = -\sin x$ (note the sign for cosine).
$h$ must be in radians: $1^\circ = 0.0175$ rad, $1' = \tfrac{1}{60}^\circ$ , $1'' = \tfrac{1}{3600}^\circ$ . So $30' = 0.5^\circ = 0.00875$ rad, $10'' \approx 0.0000485$ rad.
Anchor $a$ at the standard angle so $\sin a,\cos a$ are exact; if the target is below the anchor, $h < 0$ .

Trig approximation (h in radians)

\sin(a + h) \approx \sin a + h\cos a, \qquad \cos(a + h) \approx \cos a - h\sin a

anearby standard angle (30°, 45°, 60° …)
hthe small angular gap, CONVERTED TO RADIANS

Worked example

Estimate

\sin(46^\circ)

, given

1^\circ = 0.0175

rad and

\sin 45^\circ = \cos 45^\circ = 0.7071

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Applications of DerivativeMODERATE

The approximate value of

\cos30^\circ 30'

is, given that

1^\circ=0.0175

rad and

\cos30^\circ=0.8660

[Q143 · 3rd May 2nd Shift · 2023]

Convert the gap to RADIANS before multiplying

The derivatives

\cos x,\;-\sin x

are rates per radian. If you plug

h = 0.5

(the degree count) instead of

0.00875

rad, the correction is off by a factor of ~57. Always convert minutes/seconds → degrees → radians first.

Cosine's derivative carries a minus sign

\dfrac{d}{dx}\cos x = -\sin x

. For

\cos(30^\circ 30')

the correction is

h\cdot(-\sin 30^\circ)

, which DECREASES the value (cosine falls as the angle rises past 0). Dropping the minus pushes the estimate the wrong way.

Drill 2 more on approximating trigonometric values

Concept 4 of 5

Approximating Logarithms and Exponentials

Intuition

Logs and exponentials near a clean anchor estimate the same way — anchor at a round power (

1000

, an integer exponent) and add the tangent correction. The two facts to keep straight are the derivatives:

\dfrac{d}{dx}\log_{10} x = \dfrac{\log_{10} e}{x}

and

\dfrac{d}{dx}a^x = a^x \log a

Definition

For a base-10 log, $f(x) = \log_{10} x$ gives $f'(x) = \dfrac{\log_{10} e}{x} = \dfrac{0.4343}{x}$ (since $\log_{10} x = \dfrac{\log_e x}{\log_e 10}$ ). Anchor at a power of 10 so $f(a)$ is a whole number. For an exponential $f(x) = a^x$ , $f'(x) = a^x \log a$ (natural log). Anchor at an integer exponent so $f(a)$ is exact, then $f(a+h) \approx a^n + h\,a^n\log a$ . Throughout, an unqualified $\log$ means the natural logarithm; a base-10 log is written $\log_{10}$ .

Log & exponential approximation

\dfrac{d}{dx}\log_{10} x = \dfrac{0.4343}{x}, \qquad \dfrac{d}{dx}a^x = a^x \log a

0.4343 $\log_{10} e$ — the base-conversion factor for a base-10 log
$\log a$ natural log of the base, in the exponential derivative

Worked example

Estimate

\log_{10} 1002

, given

\log_{10} e = 0.4343

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Applications of DerivativeMODERATE

The approximate value of

\log_{10}1002

is (Given

\log_{10}e=0.4343

)

[Q121 · 4th May Shift 1 · 2023]

$\dfrac{d}{dx}\log_{10} x$ carries the $0.4343$ factor

A base-10 log is NOT

\dfrac{1}{x}

— that is the natural log.

\dfrac{d}{dx}\log_{10} x = \dfrac{\log_{10} e}{x} = \dfrac{0.4343}{x}

. Forgetting the factor makes the correction ~2.3× too big.

$\dfrac{d}{dx}a^x = a^x\log a$ , not $x\,a^{x-1}$

The base is constant and the EXPONENT is the variable, so the power rule does not apply. For

3^{2.001}

use

f'(x) = 3^x \log 3

; here

\log 3 = 1.0986

is the natural log, supplied in the question.

Drill 1 more on approximating logarithms and exponentials

Concept 5 of 5

Approximating Polynomial Values

Intuition

For a polynomial you could just substitute, but near a whole-number anchor the linear approximation is faster and is exactly what the paper tests. Anchor at the nearest integer, compute

f(a)

and

f'(a)

, and add the correction. When the question hands you

P(a)

P'(a)

P''(a)

instead of the polynomial, you often reconstruct

P

first, then approximate.

Definition

For $f(x) = a_n x^n + \dots + a_0$ : $f(a+h) \approx f(a) + h\,f'(a)$ , with $a$ the nearest integer to the target. When only derivative DATA is given (a Taylor-style setup): a degree-2 polynomial is fully determined by $P(a)$ , $P'(a)$ , $P''(a)$ via

P(x) = P(a) + P'(a)(x-a) + \tfrac12 P''(a)(x-a)^2.

Reconstruct

P

, then evaluate (or linearly approximate) at the required point. The target may be near a DIFFERENT integer than the one where the data is given — anchor at whatever integer is nearest the target.

Polynomial approximation / reconstruction

f(a+h) \approx f(a) + h\,f'(a); \quad P(x) = P(a) + P'(a)(x-a) + \tfrac12 P''(a)(x-a)^2

Worked example

Estimate

x^3 - 2x^2 + 3x + 2

x = 2.01

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Applications of DerivativeMODERATE

Let

P(x)

be a polynomial of degree 2 with

P(2)=-1, P'(2)=0, P''(2)=2

, then

P(1.001)

[Q140 · 14th May Shift 2 · 2024]

Anchor at the integer nearest the TARGET

In the reconstruction question the data is at

x = 2

, but

P(1.001)

is asked — anchor at

a = 1

, not

2

. Blindly linearising at the data point

a = 2

uses

P(2) = -1

and

P'(2) = 0

and gives the wrong answer. Reconstruct

P

first, then anchor near the target.

The $\tfrac12$ in the reconstruction is essential

P(x) = P(a) + P'(a)(x-a) + \tfrac12 P''(a)(x-a)^2

— the second-order term carries a

\tfrac12

. Dropping it doubles the quadratic coefficient. Here

\tfrac12 P''(2) = \tfrac12(2) = 1

, so the leading coefficient is

1

, not

2

Drill 1 more on approximating polynomial values

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)

The Differential dy and the Linear-Approximation Formula
Linear approximation
$f(a + h) \approx f(a) + h\,f'(a) \qquad \big(dy = f'(x)\,dx\big)$
Approximating Roots and Powers
Power/root approximation
$(a + h)^{p/q} \approx a^{p/q} + h\cdot\dfrac{p}{q}\,a^{\,p/q - 1}$
Approximating Trigonometric Values
Trig approximation (h in radians)
$\sin(a + h) \approx \sin a + h\cos a, \qquad \cos(a + h) \approx \cos a - h\sin a$
Approximating Logarithms and Exponentials
Log & exponential approximation
$\dfrac{d}{dx}\log_{10} x = \dfrac{0.4343}{x}, \qquad \dfrac{d}{dx}a^x = a^x \log a$
Approximating Polynomial Values
Polynomial approximation / reconstruction
$f(a+h) \approx f(a) + h\,f'(a); \quad P(x) = P(a) + P'(a)(x-a) + \tfrac12 P''(a)(x-a)^2$

Watch out for (10)

The slope is $f'(a)$ — evaluate at the anchor, not the target
→ The Differential dy and the Linear-Approximation Formula
Get the sign of $h$ right
→ The Differential dy and the Linear-Approximation Formula
Anchor at a perfect power, not just any round number
→ Approximating Roots and Powers
Watch $x^{p/q - 1}$ in the derivative
→ Approximating Roots and Powers
Convert the gap to RADIANS before multiplying
→ Approximating Trigonometric Values
Cosine's derivative carries a minus sign
→ Approximating Trigonometric Values
$\dfrac{d}{dx}\log_{10} x$ carries the $0.4343$ factor
→ Approximating Logarithms and Exponentials
$\dfrac{d}{dx}a^x = a^x\log a$ , not $x\,a^{x-1}$
→ Approximating Logarithms and Exponentials
Anchor at the integer nearest the TARGET
→ Approximating Polynomial Values
The $\tfrac12$ in the reconstruction is essential
→ Approximating Polynomial Values

Drill every past-year question on this subtopic

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

Drill the 11 questions

Drill every past-year question on this subtopic

Related notes