Some Basic Applications Of Derivatives

Q: What does the derivative mean as a rate of change?

The derivative f'(a) is the instantaneous rate at which f(x) changes as x passes through a. Its unit is the unit of f per unit of x, and its sign shows whether f is rising or falling. Velocity, marginal cost and the growth of an area with radius are all derivatives.

Q: What is the difference between average and instantaneous rate of change?

The average rate over [a, a + h] is the change in f divided by h, the slope of a secant. The instantaneous rate is its limit as h tends to zero, the slope of the tangent, which is f'(a). Average depends on the interval; instantaneous belongs to a single point.

Q: How do you solve related rates problems step by step?

Draw a figure and name the changing quantities, write one equation that links them, remove any extra variable using a fixed relation, differentiate both sides with respect to time, and only then substitute the values at the given instant. Finally state the unit and interpret the sign.

Q: What is the difference between delta y and dy?

Δy is the actual change f(x + Δx) - f(x) along the curve. The differential dy = f'(x)Δx is the change along the tangent line. For small Δx the two are nearly equal, which is the basis of approximation by differentials.

Q: How do you approximate a square root using derivatives?

Write the number as a + h with a a nearby perfect square, and use f(a + h) ≈ f(a) + hf'(a) with f(x) = √x. For example √25.3 ≈ 5 + 0.3/10 = 5.03. The closer a is to the number, the better the estimate.

Q: How is percentage error calculated using differentials?

Find dy = f'(x) dx and divide by y, then multiply by 100. For a power law y = kx^n, the relative error is n times the relative error in x, so a 1% error in a radius gives about 3% error in the volume of a sphere.

Q: How are rate of change questions asked in JEE Main?

JEE Main asks short related-rates problems with spheres, cubes, cones, ladders and shadows, often as a single numerical answer, and occasional error-estimate questions. Writing one relation, differentiating with respect to time and substituting values at the end solves almost all of them.

Q: Is the derivative as a rate of change useful for JEE Advanced?

Yes. JEE Advanced rarely asks a plain related-rates question, but rates appear inside multi-step problems: a changing angle or length in geometry, a rate step in a curve-sketching problem, and approximation ideas in limits and inequality proofs. A clear grip on dy and the tangent-line estimate helps there.

MathsApplication Of DerivativesFor JEE aspirants

The most basic applications of derivatives read $f^{'} (x)$ as a rate of change: how fast the area of a circle grows with its radius, how fast a ladder slides down a wall, how fast a sand heap rises. The same derivative also gives quick approximations and error estimates through the differential $d y = f^{'} (x) Δ x$ . These basic applications of derivatives come up in JEE Main as related-rates and small-error questions, and they are the base for tangents, monotonicity and maxima-minima.

On this page1Rate of change2Other variables3Related rates4Differentials5Approximation6Errors7Examples

Key Formulas - Quick Reference

★ Must learnAverage rate of $f$ over $[a, a + h]$ : $\frac{f ( a + h ) - f ( a )}{h}$ ; instantaneous rate at $a$ : $f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$
★ Must learnRelated rates (chain rule): $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ ; with a parameter, $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$
Circle: $\frac{d A}{d r} = 2 π r$ ; sphere: $\frac{d V}{d r} = 4 π r^{2}$ , $\frac{d S}{d r} = 8 π r$ ; cube: $\frac{d V}{d x} = 3 x^{2}$ , $\frac{d S}{d x} = 12 x$
Marginal cost $= \frac{d C}{d x}$ ; marginal revenue $= \frac{d R}{d x}$ ( $x$ = number of units)
★ Must learnDifferential: $d y = f^{'} (x) Δ x$ (with $d x = Δ x$ ); for small $Δ x$ , $Δ y \approx d y$
★ Must learnLinear approximation: $f (a + h) \approx f (a) + h f^{'} (a)$
Errors in $x$ : absolute $Δ x$ , relative $\frac{Δ x}{x}$ , percentage $\frac{Δ x}{x} \times 100$
★ Must learnIf $y = k x^{n}$ then $\frac{d y}{y} = n \frac{d x}{x}$ : the percentage error is multiplied by $n$

1. Derivative as a Rate of Change

In applied mathematics we often want to know how fast one quantity changes when another changes. The second quantity is usually time, but it need not be. An economist studies how investment changes as interest rates vary. A physician studies how the body's response changes with small changes in the dosage of a drug. A physicist studies how distance changes with time. All such questions are answered by the derivative.

1.1 Average and instantaneous rate

Average rate of change of $f (x)$ with respect to $x$ over the interval $[a, a + h]$ :

average rate = \frac{f ( a + h ) - f ( a )}{h}

Instantaneous rate of change of $f (x)$ with respect to $x$ at $x = a$ is the limit of the average rate as the interval shrinks to zero:

f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}, provided the limit exists.

So the derivative is the instantaneous rate of change. The word "instantaneous" is used even when $x$ is not time, and "rate of change" on its own always means the instantaneous rate.

Geometrically, the average rate is the slope of the secant joining $(a, f (a))$ and $(a + h, f (a + h))$ ; as $h \to 0$ the secant turns into the tangent at $x = a$ , and its slope is $f^{'} (a)$ .

Figure 1: Secants on

f (x) = \frac{x ^{2}}{4} + 1

from

a = 1.5

. As

h

shrinks from 3 to 2 to 1, the average rate (secant slope) falls from 1.5 to 1.25 to 1.0 and approaches the instantaneous rate

f^{'} (1.5) = 0.75

, the slope of the tangent at

P

Quantity	Rate of change	Meaning	Unit
Distance $s (t)$	$\frac{d s}{d t}$	velocity	$m s^{- 1}$
Velocity $v (t)$	$\frac{d v}{d t}$	acceleration	$m s^{- 2}$
Area $A (r)$ of a circle	$\frac{d A}{d r}$	area gained per unit increase in radius	$cm^{2}$ per cm
Volume $V (t)$ of water in a tank	$\frac{d V}{d t}$	flow rate (negative while draining)	$cm^{3} s^{- 1}$
Cost $C (x)$ of $x$ units	$\frac{d C}{d x}$	marginal cost: approximate cost of one more unit	rupees per unit
Revenue $R (x)$	$\frac{d R}{d x}$	marginal revenue	rupees per unit

The unit of a rate is always "unit of $y$ per unit of $x$ ", and its sign tells the direction: $\frac{d y}{d x} > 0$ means $y$ increases as $x$ increases; $\frac{d y}{d x} < 0$ means $y$ decreases.

1.2 Marginal cost and marginal revenue

If $C (x)$ is the total cost of producing $x$ units, the marginal cost $M C = \frac{d C}{d x}$ is the instantaneous rate at which cost rises with output. Since $C (x + 1) - C (x) \approx C^{'} (x) \times 1$ , it is approximately the cost of producing one more unit. In the same way, if $R (x)$ is the revenue from selling $x$ units, the marginal revenue is $M R = \frac{d R}{d x}$ .

Average rate

Over an interval $[a, a + h]$ : $\frac{f ( a + h ) - f ( a )}{h}$ .

Slope of a secant. Depends on $h$ .

Instantaneous rate

At a point $x = a$ : $f^{'} (a)$ .

Slope of the tangent. The limit of the average rate as $h \to 0$ .

Key idea

A rate of change is a derivative. Its unit is unit of

y

per unit of

x

, and its sign says whether

y

grows or shrinks.

2. Rate of Change with Respect to Another Variable

The variable need not be time. The area of a circle $A = π r^{2}$ changes with its radius at the rate $\frac{d A}{d r} = 2 π r$ , which is exactly the circumference. The picture below shows why: a small increase $d r$ adds a thin ring, and the ring is almost a strip of length $2 π r$ .

Figure 2: When the radius grows by a small

d r

the circle gains a thin ring. Unrolled, the ring is almost a strip of length

2 π r

and width

d r

, so

d A \approx 2 π r d r

and

\frac{d A}{d r} = 2 π r

. At

r = 5 cm

this is

10 π cm^{2}

per cm.

The same circle can be described by its diameter $D$ : $A = \frac{π}{4} D^{2}$ , so $\frac{d A}{d D} = \frac{π D}{2}$ . The two rates are different numbers because they answer different questions (area gained per cm of radius, or per cm of diameter). They are linked by the chain rule: $\frac{d A}{d D} = \frac{d A}{d r} \cdot \frac{d r}{d D} = 2 π r \cdot \frac{1}{2} = π r$ .

Solid / figure	Formula	Rate	Note
Circle	$A = π r^{2}$	$\frac{d A}{d r} = 2 π r$	= circumference
Sphere	$V = \frac{4}{3} π r^{3}$ , $S = 4 π r^{2}$	$\frac{d V}{d r} = 4 π r^{2}$ , $\frac{d S}{d r} = 8 π r$	$\frac{d V}{d r}$ = surface area
Cube of edge $x$	$V = x^{3}$ , $S = 6 x^{2}$	$\frac{d V}{d x} = 3 x^{2}$ , $\frac{d S}{d x} = 12 x$	$\frac{d V}{d x} = \frac{S}{2}$
Cylinder, fixed height $h$	$V = π r^{2} h$	$\frac{d V}{d r} = 2 π r h$	= curved surface area
Cone, fixed height $h$	$V = \frac{1}{3} π r^{2} h$	$\frac{d V}{d r} = \frac{2}{3} π r h$

2.1 Linking two rates

When $y$ depends on $x$ and $x$ depends on $t$ , the chain rule gives

\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}

and, when both depend on a parameter $t$ , $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ (provided $\frac{d x}{d t} \neq = 0$ ). This single rule is behind every related-rates problem.

Exam Trick

Volume to surface-area shortcut. For a sphere $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ and $\frac{d S}{d t} = 8 π r \frac{d r}{d t}$ , so $\frac{d S}{d t} = \frac{2}{r} \cdot \frac{d V}{d t}$ . For a cube of edge $x$ : $\frac{d S}{d t} = \frac{4}{x} \cdot \frac{d V}{d t}$ . A cube whose volume grows at $7 cm^{3} /s$ has, at $x = 4 cm$ , $\frac{d S}{d t} = \frac{4}{4} \times 7 = 7 cm^{2} /s$ in one line.

Quick Recall: tap to check

What is

\frac{d A}{d r}

for a circle, and what does it equal geometrically?

2 π r

, the circumference.

A rate is negative. What does that tell you?

The quantity is decreasing as the other variable increases.

For a sphere, express

\frac{d S}{d t}

in terms of

\frac{d V}{d t}

\frac{d S}{d t} = \frac{2}{r} \cdot \frac{d V}{d t}

Key idea

Rates with respect to different variables are different numbers; the chain rule converts one into another.

3. Related Rates

In a related-rates problem two or more quantities change with time and are tied together by a geometric or physical relation. We know the rate of one and want the rate of another at a particular instant.

3.1 Method

Draw a figure. Name every quantity that changes with time ( $x$ , $y$ , $r$ , $h$ , $θ$ ...). Keep fixed quantities as numbers.
Write one equation linking the quantities (Pythagoras, area or volume formula, a trigonometric ratio).
If the equation has an extra variable, remove it with the fixed relation given in the problem (for example $h = \frac{r}{6}$ ).
Differentiate both sides with respect to $t$ .
Only now substitute the values at the given instant and solve for the unknown rate. State the unit and read the sign.

Figure 3: Method for related rates. The common error is substituting numbers (like

x = 4

) before differentiating: a constant has derivative zero and the rate is lost.

3.2 Typical set-ups

Two quantities through one variable. The circle and its inscribed square both depend on $r$ , so their rates are linked through $\frac{d r}{d t}$ :

Figure 4: Square inscribed in a circle of radius

r

. Its diagonal is the diameter

2 r

, so the side is

r 2

and

A_{2} = 2 r^{2}

. Both areas depend on the same

r

, which links their rates:

\frac{d A _{2}}{d t} = \frac{2}{π} \cdot \frac{d A _{1}}{d t}

Remove the extra variable first. For a cone, $V = \frac{1}{3} π r^{2} h$ has two variables. If the shape is fixed ( $h = \frac{r}{6}$ for the sand heap), substitute before differentiating so that $V$ depends on $h$ alone.

Figure 5: Sand falls from a pipe and forms a flat cone whose height is always

\frac{1}{6}

of its base radius. Writing

r = 6 h

first gives

V = 12 π h^{3}

, a function of

h

alone; at

h = 4 cm

the height grows at

\frac{1}{48 π} \approx 0.0066 cm/s

The relation $V = 12 π h^{3}$ also shows how the height behaves in time. With a steady inflow $V = 12 t$ , so $h$ grows like a cube root: fast at first, then slower and slower, even though sand keeps arriving at the same rate.

Figure 6: Since sand arrives at

12 cm^{3} /s

12 π h^{3} = 12 t

, so

h = (\frac{t}{π})^{1/3}

. The graph rises steeply at first and then flattens: the same inflow spreads over an ever wider heap. At

h = 4 cm

the tangent slope is

\frac{d h}{d t} = \frac{1}{48 π} cm/s

Pythagoras. A ladder of fixed length sliding down a wall gives $x^{2} + y^{2} = L^{2}$ , so $x \frac{d x}{d t} + y \frac{d y}{d t} = 0$ . The two rates have opposite signs: the foot moves out while the top moves down.

Figure 7: A

5 m

ladder. Its foot is pulled away at

2 cm/s

; from

x^{2} + y^{2} = 25

x \frac{d x}{d t} + y \frac{d y}{d t} = 0

, so when

x = 4

y = 3

the top moves at

\frac{d y}{d t} = - \frac{8}{3} cm/s

(the minus sign means

y

decreases).

Angles. When an observer tracks a moving object, write a trigonometric ratio such as $tan θ = \frac{h}{d}$ and differentiate: $sec^{2} θ \frac{d θ}{d t} = \frac{1}{d} \frac{d h}{d t}$ .

Figure 8: Balloon tracked

500 ft

from its lift-off point. From

tan θ = \frac{h}{500}

\frac{d h}{d t} = 500 sec^{2} θ \frac{d θ}{d t} = 500 (2) (0.14) = 140 ft/min

θ = \frac{π}{4}

Liquid in a cone. In a funnel of semi-vertical angle $α$ , the water surface has radius $r = h tan α$ at every level, so the volume is a function of $h$ alone. For $α = 4 5^{\circ}$ , $r = h$ and the slant height of the water is $l = h 2$ .

Figure 9: Water in a funnel of semi-vertical angle

4 5^{\circ}

always has

r = h

, so

V = \frac{1}{3} π h^{3}

and the slant height is

l = h 2

. With

\frac{d V}{d t} = - 2 cm^{3} /s

h = 2 cm

\frac{d h}{d t} = - \frac{1}{π}

and

\frac{d l}{d t} = - \frac{2}{π} cm/s

Sign of a rate. "Water drains at $2 cm^{3} /s$ " means $\frac{d V}{d t} = - 2$ . "How fast is the level falling?" asks for the size of $\frac{d h}{d t}$ ; a negative value of $\frac{d h}{d t}$ simply confirms that the level is falling.

Exam Trick

Differentiate first, substitute later. If you put $x = 4$ into $S = 6 x^{2}$ before differentiating you get $S = 96$ , a constant, and its rate is $0$ . Substitute the instant's values only after the $\frac{d}{d t}$ step.

Key idea

Related rates: one equation, differentiate with respect to

t

, then substitute the instant's values.

4. Errors and Approximation

Let $y = f (x)$ . If $x$ has a small error (or change) $δ x$ , the corresponding error in $y$ is

δ y = f (x + δ x) - f (x) .

Since

δ x \to 0 lim \frac{f ( x + δ x ) - f ( x )}{δ x} = \frac{d y}{d x} = f^{'} (x),

for small $δ x$ we have $δ y \approx f^{'} (x) δ x$ .

4.1 Increment and differential

The differential of $y$ at $x$ , for the increment $δ x$ , is

d y = f^{'} (x) δ x, d x = δ x .

The actual change $δ y$ (also written $Δ y$ ) is often hard to compute, while $d y$ is easy. For small increments, $Δ y \approx d y$ .

Take $P (x, f (x))$ and $Q (x + δ x, f (x + δ x))$ on the curve, $S$ on the horizontal through $P$ below $Q$ , and $R$ where the tangent at $P$ meets $QS$ . Then $δ x = P S$ , $δ y = QS$ (the curve's rise) and $d y = R S$ (the tangent's rise).

Figure 10: For a step

Δ x = P S

, the curve rises by the true change

Δ y = QS

while the tangent rises by the differential

d y = R S = f^{'} (x) Δ x

. The gap

QR

shrinks much faster than

Δ x

, which is why

Δ y \approx d y

for small steps.

Δ y

(increment)

Actual change along the curve: $f (x + Δ x) - f (x)$ .

Exact, often hard to compute.

d y

(differential)

Change along the tangent: $f^{'} (x) Δ x$ .

An estimate; the gap $Δ y - d y$ is of order $(Δ x)^{2}$ .

4.2 Linear approximation

Writing $x = a$ and $Δ x = h$ in $Δ y \approx d y$ gives the working formula

f (a + h) \approx f (a) + h f^{'} (a) .

Geometrically we replace the curve near $a$ by its tangent line $y = f (a) + f^{'} (a) (x - a)$ .

Write the number as $f (a + h)$ : choose $f$ , and choose $a$ close to the number with $f (a)$ and $f^{'} (a)$ easy (a perfect square, a perfect cube, a standard angle).
Find $h$ (it may be negative) and $f^{'} (a)$ .
Compute $f (a) + h f^{'} (a)$ .

Figure 11: Near

x = 25

the tangent

y = 5 + \frac{x - 25}{10}

hugs

y = x

: it gives

25.3 \approx 5.0300

(true value

5.0299

). Farther away it drifts: at

x = 36

it gives

6.1

against

6

. Choose the base point

a

as close as possible to

x

Function	Approximation for small $h$	Example
$x$ near $a^{2}$	$a^{2} + h \approx a + \frac{h}{2 a}$	$25.3 \approx 5 + \frac{0.3}{10} = 5.03$
$3 x$ near $a^{3}$	$3 a^{3} + h \approx a + \frac{h}{3 a ^{2}}$	$3127 \approx 5 + \frac{2}{75} = 5.0267$
$\frac{1}{x}$ near $a$	$\frac{1}{a + h} \approx \frac{1}{a} - \frac{h}{a ^{2}}$	$\frac{1}{1.02} \approx 1 - 0.02 = 0.98$
$sin x$ near $a$	$sin (a + h) \approx sin a + h cos a$ ( $h$ in radians)	$sin 3 1^{\circ} \approx 0.5 + \frac{π}{180} \cdot \frac{3}{2} = 0.5151$
$ln x$ near $1$	$ln (1 + h) \approx h$	$ln 1.04 \approx 0.04$
$e^{x}$ near $0$	$e^{h} \approx 1 + h$	$e^{0.01} \approx 1.01$

4.3 Absolute, relative and percentage error

If $Δ x$ is the error in measuring $x$ :

Absolute error $= Δ x$
Relative error $= \frac{Δ x}{x}$
Percentage error $= \frac{Δ x}{x} \times 100$

For $y = f (x)$ the error in $y$ is estimated by $d y = f^{'} (x) Δ x$ , and the relative error in $y$ by $\frac{d y}{y}$ .

For a power law the relative errors are simply proportional. If $y = k x^{n}$ , then $d y = k n x^{n - 1} d x$ and

\frac{d y}{y} = n \frac{d x}{x} .

A cube makes this visible: growing the edge by $d x$ adds three thin slabs, so $d V \approx 3 x^{2} d x$ and the relative error in the volume is three times that in the edge.

Figure 12: Growing a cube of edge

x

d x

adds three slabs of volume

x^{2} d x

(the small edge pieces are of order

(d x)^{2}

), so

d V \approx 3 x^{2} d x

. Relative error:

\frac{d V}{V} = 3 \frac{d x}{x}

, so a

1%

error in the edge gives a

3%

error in the volume.

Exam Trick

Percentage error multiplies by the power. $1%$ error in the radius gives about $2%$ error in the area of a circle ( $π r^{2}$ ) and $3%$ in the volume of a sphere ( $\frac{4}{3} π r^{3}$ ). For products, percentage errors add: for $V = π r^{2} h$ , the maximum percentage error is about $2 (% in r) + (% in h)$ .

JEE Advanced

How large is the approximation error? Taylor's theorem gives $f (a + h) = f (a) + h f^{'} (a) + \frac{h ^{2}}{2} f^{''} (a + θ h)$ for some $0 < θ < 1$ . So the error of the linear approximation is about $\frac{h ^{2}}{2} f^{''} (a)$ : it is second order in $h$ , and its sign tells whether the tangent over- or under-estimates.

For $x$ , $f^{''} (x) = - \frac{1}{4} x^{- 3/2} < 0$ , so the tangent lies above the curve and over-estimates. At $a = 25$ , $h = 0.3$ : $f^{''} (25) = - \frac{1}{500}$ , error $\approx \frac{0.09}{2} (- \frac{1}{500}) = - 0.00009$ , giving $25.3 \approx 5.03 - 0.00009 = 5.02991$ (true value $5.029911$ ).

Quick Recall: tap to check

Write the linear approximation formula.

f (a + h) \approx f (a) + h f^{'} (a)

Which is the exact change and which is the estimate:

Δ y

d y

Δ y

is exact (along the curve);

d y = f^{'} (x) Δ x

is the estimate (along the tangent).

Edge of a cube measured with

2%

error. Error in volume?

About

3 \times 2 = 6%

Why choose

a = 25

to approximate

25.3

Because

25

and

f^{'} (25) = \frac{1}{10}

are exact and

25

is very close to

25.3

, so

h

is small.

Key idea

Near

x = a

the tangent line stands in for the curve:

f (a + h) \approx f (a) + h f^{'} (a)

, and relative errors scale with the power of

x

Figure 13: Mind map of rate of change and approximation: the derivative as a rate, related rates, standard results, differentials and error estimates.

5. Solved Examples

Solved Example 1

How fast does the area of a circle increase when its radius is

5 cm

: (i) with respect to the radius (ii) with respect to the diameter?

Solution:

(i) $A = π r^{2}$ , so $\frac{d A}{d r} = 2 π r$ . At $r = 5$ : $\frac{d A}{d r} = 10 π cm^{2} /cm$ .
(ii) $A = \frac{π}{4} D^{2}$ , so $\frac{d A}{d D} = \frac{π}{2} D$ . Radius $5 cm$ means $D = 10 cm$ : $\frac{d A}{d D} = \frac{π}{2} \times 10 = 5 π cm^{2} /cm$ .

Answer: (i) $10 π cm^{2} /cm$ (ii) $5 π cm^{2} /cm$ .

Solved Example 2

The area of a circle increases at the rate of

2 cm^{2} /s

. Find the rate at which the area of the inscribed square increases.

Solution:

Circle: $A_{1} = π r^{2}$ . The square's diagonal is $2 r$ , so its area is $A_{2} = \frac{( 2 r ) ^{2}}{2} = 2 r^{2}$ (Figure 4).
$\frac{d A _{1}}{d t} = 2 π r \frac{d r}{d t} = 2$ , so $r \frac{d r}{d t} = \frac{1}{π}$ .
$\frac{d A _{2}}{d t} = 4 r \frac{d r}{d t} = 4 \times \frac{1}{π}$ .

Answer: the area of the square increases at $\frac{4}{π} cm^{2} /s$ .

Solved Example 3

The volume of a cube increases at the rate of

7 cm^{3} /s

. How fast is the surface area increasing when the edge is

4 cm

Solution:

Let the edge be $x$ at time $t$ . $V = x^{3}$ , so $\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t} = 7$ and $\frac{d x}{d t} = \frac{7}{3 x ^{2}}$ .
$S = 6 x^{2}$ , so $\frac{d S}{d t} = 12 x \frac{d x}{d t} = 12 x \cdot \frac{7}{3 x ^{2}} = \frac{28}{x}$ .
At $x = 4$ : $\frac{d S}{d t} = 7$ .

Answer: $7 cm^{2} /s$ .

Solved Example 4

Sand pours from a pipe at the rate of

12 cm^{3} /s

and forms a cone on the ground whose height is always one-sixth of the radius of its base. How fast is the height of the cone increasing when the height is

4 cm

Solution:

$V = \frac{1}{3} π r^{2} h$ with $h = \frac{r}{6}$ , so $r = 6 h$ and $V = \frac{1}{3} π (6 h)^{2} h = 12 π h^{3}$ (Figure 5).
$\frac{d V}{d t} = 36 π h^{2} \frac{d h}{d t}$ .
With $\frac{d V}{d t} = 12$ and $h = 4$ : $\frac{d h}{d t} = \frac{12}{36 π ( 4 ) ^{2}} = \frac{1}{48 π}$ .

Answer: $\frac{1}{48 π} cm/s \approx 0.0066 cm/s$ .

Solved Example 5

The total cost of producing

x

items is

C (x) = 0.005 x^{3} - 0.02 x^{2} + 30 x + 5000

rupees. Find the marginal cost when

3

items are produced.

Solution:

Marginal cost $= C^{'} (x) = 0.015 x^{2} - 0.04 x + 30$ .
$C^{'} (3) = 0.015 (9) - 0.04 (3) + 30 = 0.135 - 0.12 + 30$ .

Answer: marginal cost $= 30.015$ rupees per item (the approximate cost of the 4th item).

Solved Example 6

Use differentials to find the approximate value of

25.3

Solution:

Take $f (x) = x$ , $a = 25$ , $h = 0.3$ . Then $f (25) = 5$ and $f^{'} (x) = \frac{1}{2 x}$ , so $f^{'} (25) = \frac{1}{10}$ .
$f (25.3) \approx f (25) + h f^{'} (25) = 5 + 0.3 \times 0.1$ .

Answer: $25.3 \approx 5.03$ (calculator: $5.0299$ ).

Solved Example 7

Find the approximate value of

(3.968)^{3/2}

Solution:

Take $f (x) = x^{3/2}$ , $a = 4$ , $h = - 0.032$ . Then $f (4) = 8$ and $f^{'} (x) = \frac{3}{2} x^{1/2}$ , so $f^{'} (4) = 3$ .
$f (3.968) \approx 8 + (- 0.032) (3) = 8 - 0.096$ .

Answer: $(3.968)^{3/2} \approx 7.904$ (calculator: $7.9042$ ).

Solved Example 8

The radius of a sphere is measured as

9 cm

with an error of

0.03 cm

. Find the approximate error and the percentage error in its volume.

Solution:

$V = \frac{4}{3} π r^{3}$ , so $d V = 4 π r^{2} d r = 4 π (81) (0.03) = 9.72 π cm^{3}$ .
Percentage error $= \frac{d V}{V} \times 100 = 3 \frac{d r}{r} \times 100 = 3 \times \frac{0.03}{9} \times 100$ .

Answer: error $\approx 9.72 π \approx 30.5 cm^{3}$ ; percentage error $= 1%$ .

Solved Example 9

A spherical balloon is inflated so that its volume increases at

25 cm^{3} /s

. When its radius is

5 cm

, its surface area increases at
(A)

5 cm^{2} /s

(B)

10 cm^{2} /s

(C)

20 cm^{2} /s

(D)

10 π cm^{2} /s

Solution:

Answer: (B). $\frac{d S}{d t} = \frac{2}{r} \cdot \frac{d V}{d t} = \frac{2}{5} \times 25 = 10 cm^{2} /s$ . (Long way: $25 = 4 π (25) \frac{d r}{d t}$ gives $\frac{d r}{d t} = \frac{1}{4 π}$ , and $\frac{d S}{d t} = 8 π (5) \cdot \frac{1}{4 π} = 10$ .)

Solved Example 10

Using differentials, the approximate value of

3127

is
(A)

5.0267

(B)

5.0333

(C)

4.9733

(D)

5.0083

Solution:

Answer: (A). $f (x) = x^{1/3}$ , $a = 125$ , $h = 2$ : $f^{'} (125) = \frac{1}{3} (125)^{- 2/3} = \frac{1}{75}$ , so $3127 \approx 5 + \frac{2}{75} = 5.0267$ (calculator: $5.0265$ ).

Practice Questions

The radius of a circle increases at $3 cm/s$ . Find the rate at which its area increases when the radius is $10 cm$ .Answer: $60 π cm^{2} /s$
A ladder $5 m$ long leans against a wall. Its foot is pulled along the ground away from the wall at $2 cm/s$ . How fast does the top slide down the wall when the foot is $4 m$ from the wall?Answer: $\frac{8}{3} cm/s$ (Figure 7)
Water drips out of a conical funnel of semi-vertical angle $4 5^{\circ}$ at $2 cm^{3} /s$ . Find the rate at which the slant height of the water decreases when the height of the water is $2 cm$ .Answer: $\frac{2}{π} cm/s$ (Figure 9)
A hot-air balloon rising straight up from a level field is tracked by a range finder $500 ft$ from the lift-off point. At the moment the elevation angle is $\frac{π}{4}$ , the angle increases at $0.14 rad/min$ . How fast is the balloon rising?Answer: $140 ft/min$ (Figure 8)
Using differentials, find the approximate value of $0.037$ .Answer: $0.2 - \frac{0.003}{0.4} = 0.1925$
The edge of a cube is measured with an error of $2%$ . Find the approximate percentage error in its volume.Answer: $6%$
The revenue from selling $x$ units is $R (x) = 3 x^{2} + 36 x + 5$ rupees. Find the marginal revenue when $x = 5$ .Answer: $R^{'} (5) = 66$ rupees per unit

Common Mistakes to Avoid

Watch out

Substituting the instant's values before differentiating. A number has derivative zero; substitute only after the $\frac{d}{d t}$ step.
Dropping the sign: a draining tank has $\frac{d V}{d t} < 0$ and a sliding ladder top has $\frac{d y}{d t} < 0$ . Keep the sign in the equation, then answer the size if the question says 'how fast is it decreasing'.
Confusing rates with respect to different variables: $\frac{d A}{d r} = 2 π r$ is not $\frac{d A}{d t}$ , and $\frac{d A}{d D}$ is not $\frac{d A}{d r}$ .
Putting the radius into a diameter formula (or the reverse): a circle of radius $5$ has $D = 10$ .
Leaving two variables in a cone formula. Use the fixed shape ( $r = 6 h$ , $r = h tan α$ ) to reduce $V$ to one variable before differentiating.
Choosing a far-away base point in approximation. Pick $a$ as close as possible to the number (for $25.3$ use $25$ , not $16$ ).
Using degrees in $sin (a + h) \approx sin a + h cos a$ : $h$ must be in radians ( $1^{\circ} = \frac{π}{180}$ ).
Adding the power instead of multiplying: for $y = x^{3}$ the percentage error is $3$ times that of $x$ , not $3$ more.

Frequently Asked Questions

What does the derivative mean as a rate of change?

The derivative $f^{'} (a)$ is the instantaneous rate at which $f (x)$ changes as $x$ passes through $a$ . Its unit is the unit of $f$ per unit of $x$ , and its sign shows whether $f$ is rising or falling. Velocity, marginal cost and the growth of an area with radius are all derivatives.

What is the difference between average and instantaneous rate of change?

The average rate over $[a, a + h]$ is the change in $f$ divided by $h$ , the slope of a secant. The instantaneous rate is its limit as $h$ tends to zero, the slope of the tangent, which is $f^{'} (a)$ . Average depends on the interval; instantaneous belongs to a single point.

How do you solve related rates problems step by step?

Draw a figure and name the changing quantities, write one equation that links them, remove any extra variable using a fixed relation, differentiate both sides with respect to time, and only then substitute the values at the given instant. Finally state the unit and interpret the sign.

What is the difference between delta y and dy?

$Δ y$ is the actual change $f (x + Δ x) - f (x)$ along the curve. The differential $d y = f^{'} (x) Δ x$ is the change along the tangent line. For small $Δ x$ the two are nearly equal, which is the basis of approximation by differentials.

How do you approximate a square root using derivatives?

Write the number as $a + h$ with $a$ a nearby perfect square, and use $f (a + h) \approx f (a) + h f^{'} (a)$ with $f (x) = x$ . For example $25.3 \approx 5 + \frac{0.3}{10} = 5.03$ . The closer $a$ is to the number, the better the estimate.

How is percentage error calculated using differentials?

Find $d y = f^{'} (x) d x$ and divide by $y$ , then multiply by $100$ . For a power law $y = k x^{n}$ , the relative error is $n$ times the relative error in $x$ , so a $1%$ error in a radius gives about $3%$ error in the volume of a sphere.

How are rate of change questions asked in JEE Main?

JEE Main asks short related-rates problems with spheres, cubes, cones, ladders and shadows, often as a single numerical answer, and occasional error-estimate questions. Writing one relation, differentiating with respect to time and substituting values at the end solves almost all of them.

Is the derivative as a rate of change useful for JEE Advanced?

Yes. JEE Advanced rarely asks a plain related-rates question, but rates appear inside multi-step problems: a changing angle or length in geometry, a rate step in a curve-sketching problem, and approximation ideas in limits and inequality proofs. A clear grip on dy and the tangent-line estimate helps there.

Previous year questions on Some Basic Applications Of Derivatives

4 questions from past papers, each with a step-by-step solution.

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