Applications of Derivatives revision notes

Rate of Change of Quantities

The derivative $\frac{ds}{dt}$ represents the rate of change of distance $s$ with respect to time $t$. In general, whenever a quantity $y$ varies with another quantity $x$ via a rule $y = f(x)$, the derivative $\frac{dy}{dx}$ (or $f'(x)$) represents the instantaneous rate of change of $y$ with respect to $x$.

• Chain Rule in Rates: If two variables $x$ and $y$ are varying with respect to another variable $t$ (e.g., time), then $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$, provided $\frac{dx}{dt} \neq 0$.
• Sign Convention for Rates: $\frac{dy}{dx}$ is positive if $y$ increases as $x$ increases, and it is negative if $y$ decreases as $x$ increases. → [JEE TIP] Always substitute decreasing rates with a negative sign in equations.
• Marginal Cost (MC): The instantaneous rate of change of total cost $C(x)$ at any level of output $x$. $MC = \frac{dC}{dx}$.
• Marginal Revenue (MR): The rate of change of total revenue $R(x)$ with respect to the number of items sold $x$ at an instant. $MR = \frac{dR}{dx}$.

Tangents and Normals (Extended for JEE Advanced)

• Slope of Tangent & Normal: For a curve $y = f(x)$, the slope of the tangent at point $P(x_1, y_1)$ is $m = f'(x_1)$. The slope of the normal is $-\frac{1}{m} = -\frac{1}{f'(x_1)}$.
• Equations:
  • Tangent: $y - y_1 = m(x - x_1)$
  • Normal: $y - y_1 = -\frac{1}{m}(x - x_1)$
• Lengths of Tangent, Normal, Subtangent, and Subnormal: Let $P(x,y)$ be a point on the curve.
  • Length of Tangent ($PT$): $|y \sqrt{1 + (\frac{dx}{dy})^2}|$
  • Length of Normal ($PN$): $|y \sqrt{1 + (\frac{dy}{dx})^2}|$
  • Length of Subtangent: $|y \frac{dx}{dy}|$ → [JEE TIP] If subtangent is constant, the curve is an exponential function.
  • Length of Subnormal: $|y \frac{dy}{dx}|$ → [JEE TIP] If subnormal is constant, the curve is a parabola.
• Angle of Intersection of Two Curves: The angle $\theta$ between their tangents at the point of intersection. $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. If $m_1 m_2 = -1$, curves intersect orthogonally.

Increasing and Decreasing Functions (Monotonicity)

Let $I$ be an interval in the domain of a real-valued function $f$.

• Increasing: $x_1 < x_2 \implies f(x_1) \le f(x_2)$ for all $x_1, x_2 \in I$.
• Strictly Increasing: $x_1 < x_2 \implies f(x_1) < f(x_2)$ for all $x_1, x_2 \in I$.
• Decreasing: $x_1 < x_2 \implies f(x_1) \ge f(x_2)$ for all $x_1, x_2 \in I$.
• Strictly Decreasing: $x_1 < x_2 \implies f(x_1) > f(x_2)$ for all $x_1, x_2 \in I$.
• Constant: $f(x) = c$ for all $x \in I$.
• First Derivative Test for Monotonicity: Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$.
  • $f$ is strictly increasing in $[a, b]$ if $f'(x) > 0$ for each $x \in (a, b)$.
  • $f$ is strictly decreasing in $[a, b]$ if $f'(x) < 0$ for each $x \in (a, b)$.
• Generalized Theorem: If $f'(x) \ge 0$ for $x$ in an interval (and $f'(x)=0$ only at discrete points, not over a sub-interval), the function is strictly increasing. → [JEE TIP] Many students lose marks by incorrectly eliminating endpoints where $f'(x)=0$; discrete zero-derivative points do NOT destroy strict monotonicity (e.g., $f(x)=x^3$ at $x=0$).

Maxima and Minima

• Absolute Maximum/Minimum: A function $f$ has an absolute maximum at $c$ in interval $I$ if $f(c) \ge f(x)$ for all $x \in I$. Absolute minimum if $f(c) \le f(x)$.
• Local Maxima/Minima: A point $c$ is a local maximum if there exists an $h > 0$ such that $f(c) \ge f(x)$ for all $x \in (c-h, c+h)$. Local minimum if $f(c) \le f(x)$ in that neighborhood.
• Critical Point: A point $c$ in the domain of $f$ where either $f'(c) = 0$ or $f'(c)$ is undefined/not differentiable. → [JEE TIP] Sharp corners (like in $f(x) = |x|$) are critical points because the derivative is undefined, and extremum can occur here.
• First Derivative Test for Local Extrema: Let $c$ be a critical point.
  • If $f'(x)$ changes from positive to negative as $x$ increases through $c$, $c$ is a point of local maxima.
  • If $f'(x)$ changes from negative to positive as $x$ increases through $c$, $c$ is a point of local minima.
  • If $f'(x)$ does NOT change sign, $c$ is a point of inflection.
• Second Derivative Test: Let $f$ be twice differentiable at $c$.
  • $f'(c) = 0$ and $f''(c) < 0 \implies x=c$ is a local maxima.
  • $f'(c) = 0$ and $f''(c) > 0 \implies x=c$ is a local minima.
  • If $f''(c) = 0$, the test FAILS. You must revert to the First Derivative Test or Higher Order Derivative Test.
• Higher Order Derivative Test (JEE Advanced): If $f'(c) = f''(c) = ... = f^{n-1}(c) = 0$ and $f^n(c) \neq 0$:
  • If $n$ is EVEN and $f^n(c) < 0 \implies$ Local Maxima.
  • If $n$ is EVEN and $f^n(c) > 0 \implies$ Local Minima.
  • If $n$ is ODD $\implies$ Point of Inflection.

Approximations and Differentials (Extended for JEE Advanced)

• Differentials: If $y = f(x)$, the differential of $x$ is denoted by $dx = \Delta x$. The differential of $y$ is $dy = f'(x) dx$.
• Approximation Formula: $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
• Errors: Absolute error in $x$ is $\Delta x$. Relative error is $\frac{\Delta x}{x}$. Percentage error is $\frac{\Delta x}{x} \times 100$.

Rolle's Theorem and LMVT (Extended for JEE Advanced)

• Rolle's Theorem: If $f(x)$ is continuous in $[a,b]$, differentiable in $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$.
• Lagrange's Mean Value Theorem (LMVT): If $f(x)$ is continuous in $[a,b]$ and differentiable in $(a,b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Key Concepts & Definitions

Rate of Change:

The speed at which one variable changes in relation to another.

Monotonic Function:

A function that is exclusively increasing or exclusively decreasing in an interval.

Turning Point:

A point on a graph where the function changes its nature from decreasing to increasing or vice versa.

Extreme Value Theorem:

Every continuous function on a closed interval has both an absolute maximum and an absolute minimum.

Point of Inflection:

A point on a curve where the concavity changes (e.g., from concave upwards to concave downwards). The first derivative does not change sign across this point.

Conditions & Limitations

• Second Derivative Test Failure: When $f'(c) = 0$ AND $f''(c) = 0$, the Second Derivative Test cannot determine the nature of the critical point. You MUST use the First Derivative Test or the $n^{th}$ derivative test,.
• Differentiability Requirement for Tests: Standard tests (first and second derivative) require the function to be differentiable at the point of interest. If a function is not differentiable at $x=c$ (like sharp corners, e.g., $f(x)=|x|$), these derivative tests fail immediately. You must use basic definitions or graphing to find extrema,.
• Domain Restriction: Extreme Value Theorem only guarantees global extrema for continuous functions on closed intervals. In open intervals like $(0, 1)$, a continuous function might have neither a maximum nor a minimum.

Important Graphs & Diagrams

• $y = x^2$ (Parabola): Opens upwards, $f'(x) = 2x$. $x=0$ is the global and local minimum. Decreasing in $(-\infty, 0)$ and increasing in $(0, \infty)$,.
• $y = |x|$ (Modulus Function): V-shaped graph. Minimum at $x=0$. Not differentiable at $x=0$. Demonstrates that an extremum can exist without $f'(0) = 0$.
• $y = x^3$ (Cubic): Passes through origin. $f'(0) = 0$ but $x=0$ is neither a maxima nor a minima; it's a point of inflection. It strictly increases for all real $R$,.
• Turning Points (Hills and Valleys): At local maxima (hills), the function transitions from increasing to decreasing. At local minima (valleys), it transitions from decreasing to increasing.

Standard Derivations & Step-by-Step Problem Solving

Algorithm for finding absolute maximum and minimum in $[a, b]$:

1. Step 1: Find all critical points in the open interval $(a, b)$ by setting $f'(x) = 0$ and identifying points where $f'(x)$ does not exist.
2. Step 2: Identify the endpoints $x = a$ and $x = b$.
3. Step 3: Calculate the function values $f(x)$ at all points identified in Steps 1 and 2.
4. Step 4: The greatest value computed is the absolute maximum; the least value is the absolute minimum.

Algorithm for Optimization (e.g., maximizing volume):

1. Identify the primary quantity to be maximized/minimized (e.g., Volume $V$),.
2. Express this quantity as a function of a single variable $x$, using given constraints (e.g., $V(x)$),.
3. Find $V'(x)$ and set it to $0$ to find critical points,.
4. Verify using the Second Derivative Test: evaluate $V''(x)$ at the critical point. If $< 0$, it's a maximum; if $> 0$, it's a minimum.

Previous Year JEE Topics

• Geometrical problems with optimization: Finding the maximum area of a rectangle inscribed in an ellipse/circle, or volume of maximum cone in a sphere (Result: Volume is $\frac{8}{27}$ of sphere volume).
• Shortest Distance: Finding minimum distance between a point and a curve (e.g., parabola) or between two curves. Typically requires equating the slopes of normals or maximizing/minimizing the distance formula,.
• Roots of equations: Using Rolle's theorem to prove the existence of roots or find the maximum number of real roots for a given polynomial.
• Strict monotonicity in Trigonometry: Identifying intervals of increase/decrease for composite trigonometric functions (e.g., $\sin x + \cos x$, $\tan^{-1}(\sin x + \cos x)$) by analyzing the sign of trigonometric factors in specific quadrants,.
• Inflection Points & Concavity: Checking conditions for graphs shifting from upwards to downwards opening (important for matching-list questions).

Memory Aids & JEE Traps

• [JEE TIP] Trap 1 - The Undefined Derivative Blindspot:

  • Misconception: Critical points of a function $f(x)$ only occur where the first derivative equals zero ($f'(x) = 0$).
  • Correct Understanding: Critical points also include coordinates where the function is defined but its derivative is undefined or non-differentiable. For instance, $f(x) = x^{2/3}$ is perfectly continuous at $x=0$, but its derivative $f'(x) = \frac{2}{3x^{1/3}}$ goes to infinity. This point $x=0$ is a valid critical point and represents the absolute minimum of the function.

• [JEE TIP] Trap 2 - Closed Interval Monotonicity Limits:

  • Misconception: If a function satisfies $f'(x) > 0$ strictly within an open interval $(a, b)$, the interval of strict increase must be written with open brackets as $(a, b)$.
  • Correct Understanding: If a function is strictly increasing on the open interval $(a, b)$ and is continuous at the boundary endpoints $a$ and $b$, the interval of strict increase is mathematically maximized and denoted using closed brackets as $[a, b]$. Overlooking the boundary inclusion is a frequent cause of errors in multiple-choice questions.

• [JEE TIP] Trap 3 - The Inflection Point Illusion:

  • Misconception: Finding a coordinate where $f'(x) = 0$ is a sufficient guarantee that the function achieves a local maximum or a local minimum at that point.
  • Correct Understanding: A zero derivative is a necessary condition but not a sufficient one. For example, the function $f(x) = x^3$ has $f'(0) = 0$, yet $x=0$ is neither a maximum nor a minimum. You must explicitly verify a sign change in $f'(x)$ across the point, or use the second derivative test to confirm it is not a point of inflection.

• [JEE TIP] Trap 4 - The Related Rates Sign Deficit:

  • Misconception: When a word problem states that a quantity is changing or decreasing at a given rate, the value can be substituted into related rates equations directly as a positive scalar.
  • Correct Understanding: The algebraic sign must match the physical direction of change. If water is draining or a volume is leaking at a rate of $5\text{ ml/s}$, you must explicitly assign a negative value: $\frac{dV}{dt} = -5$. Omitting this negative sign causes catastrophic errors in subsequent differentiation chains.

• [JEE TIP] Trap 5 - Common Normal Geometric Monopoly:

  • Misconception: The shortest distance between two distinct, non-intersecting curves can be evaluated by choosing arbitrary random points on each curve and applying the standard distance formula.
  • Correct Understanding: The shortest geometric distance between two smooth, non-intersecting curves is strictly measured along their common normal line. To solve these problems efficiently, equate the slope of the normal (or tangent) at a parametric point on the first curve to the slope of the normal (or tangent) on the second curve.