Applications of Derivatives

The Hook: Finding the Perfect Shot

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Interactive: Explore Function Behavior

Try plotting $f(x) = x^3 - 3x$ and observe:

• Where the function is increasing/decreasing
• Local maximum and minimum points
• Points where slope = 0

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The Core Concept

Why Applications Matter

Derivatives tell us the rate of change of functions. This unlocks powerful applications:

1. Finding maxima/minima — Optimization problems
2. Checking if function is increasing/decreasing — Monotonicity
3. Finding equations of tangents/normals — Geometry
4. Solving rate of change problems — Physics, economics
5. Curve sketching — Understanding function behavior

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Application 1: Monotonicity (Increasing/Decreasing Functions)

Definition

A function $f(x)$ is:

• Increasing on $(a, b)$ if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$
• Decreasing on $(a, b)$ if $x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$

Derivative Test

$$\boxed{\begin{aligned} f'(x) > 0 \text{ on } (a,b) &\Rightarrow f \text{ is increasing on } (a,b) \\ f'(x) < 0 \text{ on } (a,b) &\Rightarrow f \text{ is decreasing on } (a,b) \\ f'(x) = 0 \text{ on } (a,b) &\Rightarrow f \text{ is constant on } (a,b) \end{aligned}}$$

Strict versions:

• Strictly increasing: $f'(x) \geq 0$ with equality at isolated points only
• Strictly decreasing: $f'(x) \leq 0$ with equality at isolated points only

Process to Find Intervals

Step 1: Find $f'(x)$

Step 2: Solve $f'(x) = 0$ to find critical points

Step 3: Make a sign chart for $f'(x)$

Step 4: Conclude:

• Where $f'(x) > 0$: Increasing
• Where $f'(x) < 0$: Decreasing

Example

Find intervals where $f(x) = x^3 - 3x^2 - 9x + 5$ is increasing/decreasing.

Solution:

$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$

Critical points: $x = -1, 3$

Sign chart:

Interval	$x < -1$	$x = -1$	$-1 < x < 3$	$x = 3$	$x > 3$
$f'(x)$	$+$	$0$	$-$	$0$	$+$
$f(x)$	Increasing	—	Decreasing	—	Increasing

Answer:

• Increasing on $(-\infty, -1) \cup (3, \infty)$
• Decreasing on $(-1, 3)$

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Application 2: Maxima and Minima

Definitions

Local Maximum: $f(a)$ is a local maximum if $f(a) \geq f(x)$ for all $x$ near $a$.

Local Minimum: $f(a)$ is a local minimum if $f(a) \leq f(x)$ for all $x$ near $a$.

Global (Absolute) Maximum/Minimum: Largest/smallest value on the entire domain.

Critical Points

Definition: $x = c$ is a critical point if:

• $f'(c) = 0$, OR
• $f'(c)$ does not exist

All local maxima and minima occur at critical points!

(But not every critical point is a max/min — could be an inflection point)

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First Derivative Test

Procedure

Step 1: Find all critical points (solve $f'(x) = 0$)

Step 2: Check sign of $f'(x)$ on either side of each critical point

Step 3: Conclude:

$f'(x)$ changes from	Conclusion
$+$ to $-$	Local maximum
$-$ to $+$	Local minimum
No sign change	Neither (inflection point)

Example

Find local maxima and minima of $f(x) = x^3 - 3x + 2$.

Solution:

$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)$

Critical points: $x = -1, 1$

Sign chart:

Interval	$x < -1$	$x = -1$	$-1 < x < 1$	$x = 1$	$x > 1$
$f'(x)$	$+$	$0$	$-$	$0$	$+$

At $x = -1$: $f'$ changes $+$ to $-$ → Local maximum

• $f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$

At $x = 1$: $f'$ changes $-$ to $+$ → Local minimum

• $f(1) = 1 - 3 + 2 = 0$

Answer: Local max at $(-1, 4)$, local min at $(1, 0)$

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Second Derivative Test

Procedure

Step 1: Find critical points (solve $f'(x) = 0$)

Step 2: Find $f''(x)$ (second derivative)

Step 3: For each critical point $c$:

$$\boxed{\begin{aligned} f''(c) > 0 &\Rightarrow \text{Local minimum at } x = c \\ f''(c) < 0 &\Rightarrow \text{Local maximum at } x = c \\ f''(c) = 0 &\Rightarrow \text{Test is inconclusive (use first derivative test)} \end{aligned}}$$

Intuition:

• $f''(c) > 0$ means function is concave up (like a cup) → minimum
• $f''(c) < 0$ means function is concave down (like a cap) → maximum

Example

Find local maxima and minima of $f(x) = x^3 - 6x^2 + 9x + 1$.

Solution:

$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$

Critical points: $x = 1, 3$

$f''(x) = 6x - 12$

At $x = 1$: $f''(1) = 6 - 12 = -6 < 0$ → Local maximum

• $f(1) = 1 - 6 + 9 + 1 = 5$

At $x = 3$: $f''(3) = 18 - 12 = 6 > 0$ → Local minimum

• $f(3) = 27 - 54 + 27 + 1 = 1$

Answer: Local max at $(1, 5)$, local min at $(3, 1)$

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Global Maxima/Minima on Closed Intervals

Extreme Value Theorem

If $f$ is continuous on $[a, b]$, then $f$ has both a global maximum and global minimum on $[a, b]$.

Procedure to Find Global Extrema on $[a, b]$

Step 1: Find all critical points in $(a, b)$

Step 2: Evaluate $f$ at:

• All critical points
• Both endpoints: $f(a)$ and $f(b)$

Step 3: The largest value is the global maximum, smallest is the global minimum

Example

Find absolute maximum and minimum of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$.

Solution:

$f'(x) = 3x^2 - 3 = 0 \Rightarrow x = \pm 1$

Both critical points are in $[-2, 2]$.

Evaluate at critical points and endpoints:

• $f(-2) = -8 + 6 + 1 = -1$
• $f(-1) = -1 + 3 + 1 = 3$
• $f(1) = 1 - 3 + 1 = -1$
• $f(2) = 8 - 6 + 1 = 3$

Global maximum: $3$ at $x = -1$ and $x = 2$ Global minimum: $-1$ at $x = -2$ and $x = 1$

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Application 3: Optimization Problems

General Strategy

Step 1: Understand the problem. Identify what to maximize/minimize.

Step 2: Draw a diagram if applicable.

Step 3: Write the objective function (what to optimize) in terms of variables.

Step 4: Use constraints to express everything in terms of one variable.

Step 5: Find derivative and critical points.

Step 6: Use first/second derivative test to find max/min.

Step 7: Check endpoints if domain is restricted.

Classic Problems

Problem Type 1: Geometric Optimization

Example: Find the dimensions of a rectangle with perimeter 100 m that has maximum area.

Solution:

Let length = $x$, width = $y$.

Constraint: $2x + 2y = 100 \Rightarrow y = 50 - x$

Objective: Maximize area $A = xy = x(50 - x) = 50x - x^2$

Domain: $0 < x < 50$

$\frac{dA}{dx} = 50 - 2x = 0 \Rightarrow x = 25$

$\frac{d^2A}{dx^2} = -2 < 0$ → Maximum at $x = 25$

When $x = 25$: $y = 50 - 25 = 25$

Answer: Square with side 25 m has maximum area.

Problem Type 2: Distance/Time Optimization

Example: A person is at point A, 3 km from a straight road. He wants to reach point B on the road, 5 km away from the foot of perpendicular from A. If he walks at 4 km/h on sand and 6 km/h on road, where should he hit the road to minimize time?

Solution:

Let him hit the road at distance $x$ from the foot of perpendicular.

Distance on sand: $\sqrt{9 + x^2}$ Distance on road: $5 - x$

Time: $T(x) = \frac{\sqrt{9 + x^2}}{4} + \frac{5 - x}{6}$

$\frac{dT}{dx} = \frac{x}{4\sqrt{9 + x^2}} - \frac{1}{6} = 0$

$\frac{x}{4\sqrt{9 + x^2}} = \frac{1}{6}$

$6x = 4\sqrt{9 + x^2}$

$36x^2 = 16(9 + x^2)$

$20x^2 = 144$

$x = \frac{12}{\sqrt{20}} = \frac{6}{\sqrt{5}}$ km

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Application 4: Tangents and Normals

Definitions

Tangent: Line that touches the curve at exactly one point (locally).

Normal: Line perpendicular to the tangent at the point of contact.

Equations

At point $(x_0, y_0)$ on curve $y = f(x)$:

Slope of tangent: $m_T = f'(x_0)$

Equation of tangent:

$$\boxed{y - y_0 = f'(x_0)(x - x_0)}$$

Slope of normal: $m_N = -\frac{1}{f'(x_0)}$ (negative reciprocal)

Equation of normal:

$$\boxed{y - y_0 = -\frac{1}{f'(x_0)}(x - x_0)}$$

Example

Find equations of tangent and normal to $y = x^2$ at $(1, 1)$.

Solution:

$f(x) = x^2 \Rightarrow f'(x) = 2x$

At $x = 1$: $f'(1) = 2$

Tangent: $y - 1 = 2(x - 1) \Rightarrow y = 2x - 1$

Normal: $y - 1 = -\frac{1}{2}(x - 1) \Rightarrow y = -\frac{1}{2}x + \frac{3}{2}$

Angle Between Two Curves

Two curves $y = f(x)$ and $y = g(x)$ intersect at point $P$.

Angle of intersection = Angle between their tangents at $P$

$$\boxed{\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|}$$

where $m_1 = f'(x_0)$ and $m_2 = g'(x_0)$ at intersection point $(x_0, y_0)$.

Orthogonal (perpendicular) curves: $m_1 \cdot m_2 = -1$

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Application 5: Rate of Change

Related Rates

When two or more quantities are related and changing with time, their rates of change are related by differentiation.

Method: Differentiate the relation with respect to time $t$.

Example: Expanding Circle

The radius of a circle is increasing at 2 cm/s. How fast is the area increasing when radius is 5 cm?

Solution:

Given: $\frac{dr}{dt} = 2$ cm/s

Find: $\frac{dA}{dt}$ when $r = 5$

Relation: $A = \pi r^2$

Differentiate w.r.t. $t$:

$$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$

When $r = 5$:

$$\frac{dA}{dt} = 2\pi (5)(2) = 20\pi \text{ cm}^2/\text{s}$$

Example: Ladder Sliding Down Wall

A 10 m ladder leans against a wall. The bottom slides away at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall?

Solution:

Let $x$ = distance of bottom from wall, $y$ = height of top on wall.

Relation: $x^2 + y^2 = 100$ (Pythagoras)

Given: $\frac{dx}{dt} = 1$ m/s

Find: $\frac{dy}{dt}$ when $x = 6$

Differentiate:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$

When $x = 6$: $y = \sqrt{100 - 36} = 8$

$$2(6)(1) + 2(8)\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4} \text{ m/s}$$

(Negative means sliding down)

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Application 6: Mean Value Theorem (MVT)

Statement

If $f$ is:

1. Continuous on $[a, b]$
2. Differentiable on $(a, b)$

Then there exists at least one $c \in (a, b)$ such that:

$$\boxed{f'(c) = \frac{f(b) - f(a)}{b - a}}$$

Geometric meaning: There exists a point where the tangent is parallel to the chord joining $(a, f(a))$ and $(b, f(b))$.

Physical meaning: At some instant, instantaneous velocity equals average velocity.

Rolle’s Theorem (Special Case)

If additionally $f(a) = f(b)$, then:

$$\boxed{f'(c) = 0 \text{ for some } c \in (a, b)}$$

Meaning: Between two points with same height, there’s a point with horizontal tangent.

Example

Verify MVT for $f(x) = x^2$ on $[1, 3]$.

Solution:

$f$ is polynomial → continuous and differentiable everywhere.

$\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4$

$f'(x) = 2x$

We need $f'(c) = 4$:

$$2c = 4 \Rightarrow c = 2$$

Since $2 \in (1, 3)$, MVT is verified.

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Practice Problems

Level 1: Foundation (NCERT)

Level 2: JEE Main

Level 3: JEE Advanced

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Common Mistakes to Avoid

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Quick Revision Box

Application	Key Formula/Method
Increasing	$f'(x) > 0$
Decreasing	$f'(x) < 0$
Local Max	$f'$ changes $+$ to $-$ OR $f'' < 0$
Local Min	$f'$ changes $-$ to $+$ OR $f'' > 0$
Global Max/Min	Check critical points + endpoints
Tangent slope	$m = f'(x_0)$
Normal slope	$m = -\frac{1}{f'(x_0)}$
Related Rates	Differentiate relation w.r.t. time
MVT	$f'(c) = \frac{f(b) - f(a)}{b - a}$

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JEE Strategy Tips

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Teacher’s Summary

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Related Topics

Within Limits, Continuity & Differentiability

• Differentiability — Foundation for applications
• Differentiation Rules — Computing derivatives efficiently
• Continuity — Mean Value Theorem requires continuity

Mathematical Foundations

• Functions Basics — Understanding domain, range
• Coordinate Geometry — Tangents, normals, curves
• Trigonometry — Optimization with trig functions

Applications

• Integral Calculus — Area, volume optimization
• Differential Equations — Rate equations
• Sequences & Series — Maxima/minima of sequences

Physics Connections

• Kinematics — Velocity/acceleration optimization
• Projectile Motion — Maximum range, height
• Work, Energy, Power — Potential energy minimization
• Simple Harmonic Motion — Amplitude, phase analysis

Real-World Applications

• Economics: Profit maximization, cost minimization
• Engineering: Structural optimization, efficiency
• Biology: Population growth rates
• Business: Revenue optimization, inventory management

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