Application of Derivatives (AOD)

The Core Mission

Applying the tools of Calculus. In the previous module, we learned how to differentiate functions using core rules (Product, Quotient, Chain). Here we will apply those tools to physically analyze curves. We will find their Peaks and Valleys (Maxima/Minima), map out their exact angles (Tangents and Normals), and determine where the curve climbs or falls (Monotonicity).

Refresher: Advanced Methods of Differentiation

Before analyzing curves, you must be confident in deriving complex functions. These forms will show up constantly in AOD problems.

1. Implicit Differentiation

Functions where \(y\) and \(x\) are mixed (like \(x^3 + xy + y^2 = 0\)). Differentiate w.r.t \(x\) directly, and anytime you derive a \(y\) term, attach a \(\frac{dy}{dx}\).

2. Parametric Differentiation

When \(x = f(t)\) and \(y = g(t)\), calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Then, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Warning: For second derivative, \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}\).

3. Logarithmic Differentiation

Required essentially for variable-to-variable power, i.e., \(y = [f(x)]^{g(x)}\). Take \(\ln\) on both sides: \(\ln y = g(x) \ln f(x)\), then apply implicit differentiation.

Practice Illustrative Problems: MOD

Q1. If \(y = x^x\), find \(d^2y/dx^2\) at \(x=1\). [Show Solution]

Step 1: First Derivative. Take \(\ln\) on both sides: \(\ln y = x \ln x\).
Differentiate implicitly: \(\frac{1}{y} \frac{dy}{dx} = x(\frac{1}{x}) + \ln x(1) = 1 + \ln x\).
So, \(\frac{dy}{dx} = y(1 + \ln x) = x^x(1 + \ln x)\).
Step 2: Second Derivative. Apply product rule on \(y(1 + \ln x)\).
\(\frac{d^2y}{dx^2} = \frac{dy}{dx}(1 + \ln x) + y(\frac{1}{x})\).
Step 3: Evaluate at \(x=1\). \(y(1) = 1^1 = 1\). \(\ln(1) = 0\).
\(dy/dx\) at \(x=1\) is \(1(1+0) = 1\).
\(\frac{d^2y}{dx^2} = (1)(1+0) + (1)(1/1) = 1 + 1 = \mathbf{2}\).

PHASE 1: GEOMETRY & RATES

1. Tangents & Normals

The derivative \(\frac{dy}{dx}\) at any point \((x_1, y_1)\) represents the slope or gradient \(m\) of the tangent line to the curve at that exact point.

Equation of Tangent: \(y - y_1 = \left( \frac{dy}{dx} \right)_{(x_1, y_1)} \cdot (x - x_1)\)
Equation of Normal: \(y - y_1 = -\frac{1}{\left( \frac{dy}{dx} \right)_{(x_1, y_1)}} \cdot (x - x_1)\)

Key Conditions Common Triggers in Problems

Tangent is parallel to x-axis: \(\frac{dy}{dx} = 0\) (Horizontal Tangent, finding peaks/valleys)
Tangent is parallel to y-axis: \(\frac{dx}{dy} = 0\) or \(\frac{dy}{dx} \to \infty\) (Vertical Tangent)
Tangent makes angle \(\theta\) with x-axis: \(\frac{dy}{dx} = \tan \theta\)
Orthogonal Intersection of two curves: \((m_1)(m_2) = -1\) where \(m_1, m_2\) are derivatives of the two curves at the intersection point.

Practice Illustrative Problems: Tangent/Normal

Q2. Find the point on the curve \(y^2 = 4ax\) at which the normal is parallel to the line \(y = -2x\). [Show Solution]

The curve is \(y^2 = 4ax\). Differentiate w.r.t x: \(2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}\).
This is the slope of the tangent (\(m_T\)). The slope of normal (\(m_N\)) is \(-1/m_T = -\frac{y}{2a}\).
The normal is parallel to \(y = -2x\), which has slope -2.
So, \(-\frac{y}{2a} = -2 \implies \mathbf{y = 4a}\).
Substitute \(y\) back into curve equation: \((4a)^2 = 4ax \implies 16a^2 = 4ax \implies \mathbf{x = 4a}\).
The point is \((4a, 4a)\).

2. Rate Measure & Increments

If two variables \(x\) and \(y\) vary with time \(t\), their rates of change \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are related by the Chain Rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

Errors/Approximations: Given \(y = f(x)\), a small change \(\Delta x\) in \(x\) causes a small change \(\Delta y\) in \(y\). The approximation formula is: \(\Delta y \approx \frac{dy}{dx} \Delta x\).

3. Monotonicity (Increasing and Decreasing Functions)

A function \(f(x)\) is monotonic if it only goes in one vertical direction (either strictly up or strictly down) as \(x\) moves to the right.

Strictly Increasing

\(f'(x) > 0\) for all \(x\) in interval \((a,b)\). The curve constantly climbs upwards.

Strictly Decreasing

\(f'(x) < 0\) for all \(x\) in interval \((a,b)\). The curve constantly slopes downwards.

Practice Illustrative Problems: Monotonicity Intervals

Q3. Find the intervals in which the function \(f(x) = 2x^3 - 3x^2 - 36x + 7\) is increasing. [Show Solution]

Step 1: Find \(f'(x)\).
\(f'(x) = 6x^2 - 6x - 36\).
Step 2: Factorize. \(6(x^2 - x - 6) = 6(x-3)(x+2)\).
Step 3: Use Wavy Curve Method. The critical points are \(x=3\) and \(x=-2\).
- For \(x > 3\), \(f'(x) > 0\) (+)
- For \(-2 < x < 3\), \(f'(x) < 0\) (-)
- For \(x < -2\), \(f'(x)> 0\) (+)
Answer: The function is strictly increasing in \((-\infty, -2) \cup (3, \infty)\).

PHASE 2: MAXIMA, MINIMA & MVT

4. Maxima and Minima (Optimization)

A point where a function transitions from increasing to decreasing is a Local Maximum. A point where it goes from decreasing to increasing is a Local Minimum. These turning points exclusively occur where \(f'(x) = 0\) or is undefined.

First Derivative Test

Check the sign change of \(f'(x)\) around critical points:

If \(f'(x)\) changes \(+ \to -\): Local Maximum
If \(f'(x)\) changes \(- \to +\): Local Minimum
No sign change: Point of Inflection.

Second Derivative Test (Faster)

Find critical points where \(f'(x) = 0\). Let \(x=c\) be one.

If \(f''(c) < 0\): Local Maximum (Curve is concave down ∩)
If \(f''(c) > 0\): Local Minimum (Curve is concave up ∪)

Global Extrema Absolute Max/Min on Closed Intervals [a, b]

When asked for the absolute maximum or minimum in a closed interval \([a,b]\), the answer is the largest (or smallest) value among:

The values at all critical points inside \((a,b)\).
The value at the left boundary \(x=a\).
The value at the right boundary \(x=b\).

Practice Illustrative Problems: Maxima/Minima

Q4. A window is in the form of a rectangle surmounted by a semi-circle. The total perimeter is 10m. Find the dimensions allowing maximum light. [Show Solution]

Let rectangle base = \(2x\) (radius of semicircle = \(x\)) and height = \(y\).
Perimeter = base + 2*height + arc of semi-circle = \(2x + 2y + \pi x = 10\).
From this, \(y = 5 - x - \frac{\pi x}{2}\).
Area \(A = \text{Area}_{\text{rect}} + \text{Area}_{\text{semi}}\) = \((2x \cdot y) + \frac{1}{2} \pi x^2\).
Substitute \(y\): \(A = 2x(5 - x - \frac{\pi x}{2}) + \frac{\pi x^2}{2} = 10x - 2x^2 - \pi x^2 + \frac{\pi x^2}{2} = 10x - 2x^2 - \frac{\pi x^2}{2}\).
To maximize, \(\frac{dA}{dx} = 10 - 4x - \pi x = 0\).
So, \(x = \frac{10}{4+\pi}\). The base is \(2x = \frac{20}{4+\pi}\).
Second derivative \(\frac{d^2A}{dx^2} = -4 - \pi < 0\) (Confirms Maximum!).

5. Mean Value Theorems

These theorems form fundamental existence proofs linking average rates with instantaneous rates.

Rolle's Theorem

If \(f(x)\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a) = f(b)\), then there exists AT LEAST one \(c \in (a,b)\) such that \(f'(c) = 0\).

Geometric Meaning: If a curve starts and ends at the same height, it must have at least one horizontal peak or valley.

Lagrange's MVT (LMVT)

If \(f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\), there exists AT LEAST one \(c \in (a,b)\) such that \(f'(c) = \frac{f(b)-f(a)}{b-a}\).

Geometric Meaning: The instantaneous slope somewhere inside equals the average slope between endpoints (tangent parallel to secant).

Practice Illustrative Problems: LMVT

Q5. Use Lagrange's MVT to prove that \(x < \tan x\) for \(x \in (0, \pi/2)\). [Show Solution]

Let \(f(t) = \tan t\) in the interval \([0, x]\) where \(0 < x < \pi/2\).
\(f(t)\) is continuous in \([0,x]\) and differentiable in \((0,x)\).
By LMVT, there exists \(c \in (0, x)\) such that \(f'(c) = \frac{f(x) - f(0)}{x - 0}\).
\(\sec^2 c = \frac{\tan x - 0}{x} \implies \frac{\tan x}{x} = \sec^2 c\).
Since \(c \in (0, \pi/2)\), \(\sec c > 1 \implies \sec^2 c > 1\).
Therefore, \(\frac{\tan x}{x} > 1\). Since \(x>0\), multiplying both sides by \(x\) gives \(\mathbf{\tan x > x}\).

PHASE 6: The Master Practice Matrix

Show detailed steps for every problem. Calculus requires systematic algebraic neatness.

Drill 1 Tangents, Normals & Monotonicity (1-10)

Find the slope of tangent to curve \(y = x^3 - 3x + 2\) at point where x coordinate is 3.
Find equation of normal to \(y = \sin x\) at \((0, 0)\).
Prove that curve \(y = e^x\) is strictly increasing for all real x.
Find interval where \(f(x) = x^2 - 4x + 6\) is strictly decreasing.
Find tangents to \(y = x^2 - 2x + 7\) parallel to line \(2x-y+9=0\).

Drill 2 Maxima, Minima & Absolute Extrema (6-10)

Find local max and min for \(f(x) = x^3 - 3x\).
Find maximum value of \(f(x) = \sin x + \cos x\).
Find absolute extrema of \(f(x) = x^2 - 1\) on interval \([-2, 2]\).
A square piece of tin of side 18cm is to be made into a box without top, by cutting a square from each corner. Find size of square cut.
Find two positive numbers x and y such that x + y = 60 and \(xy^3\) is maximum.

AOD Logic Answer Key

Q	Derived Solution / Method
1	\(y' = 3x^2 - 3\). At x = 3, \(y' = 3(3^2) - 3 = 27-3 = \mathbf{24}\).
2	\(y' = \cos x\). At (0,0), slope of tangent = 1. Slope of normal = -1. Eq: \(y-0 = -1(x-0) \Rightarrow \mathbf{y = -x}\).
3	\(y' = e^x\). Since \(e^x > 0\) for all real x, the function is strictly increasing.
4	\(f'(x) = 2x - 4 < 0 \Rightarrow 2x < 4 \Rightarrow x < 2\). Interval: \(\mathbf{(-\infty, 2)}\).
5	Slope of line is 2. \(y' = 2x - 2 = 2 \Rightarrow 2x = 4 \Rightarrow x = 2\). At x=2, y=7. Eq: \(y - 7 = 2(x - 2) \Rightarrow \mathbf{y = 2x + 3}\).
6	\(y' = 3x^2 - 3 = 0 \Rightarrow x = \pm 1\). \(y'' = 6x\). At x=1, \(y''>0\) (Local Min = -2). At x=-1, \(y''<0\) (Local Max=2).
7	\(y' = \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \pi/4\). Maximum is \(\sin(\pi/4) + \cos(\pi/4) = \mathbf{\sqrt{2}}\).
8	\(y' = 2x = 0 \Rightarrow x = 0\). Check endpoints and critical pt: f(0)=-1. f(-2)=3. f(2)=3. Absolute max is 3, absolute min is -1.
9	Vol = \(x(18-2x)^2\). \(V' = 0 \Rightarrow x = 3\) or \(x = 9\). \(x=9\) gives Vol=0. Thus, side of square = \(\mathbf{3cm}\).
10	\(x = 60-y\). Maximize \(P = (60-y)y^3 = 60y^3 - y^4\). \(P' = 180y^2 - 4y^3 = 0 \Rightarrow y^2(180-4y) = 0 \Rightarrow \mathbf{y=45, x=15}\).