Application Of Derivatives Class 12 Maths: Chapter Notes

Key Terminology

Rate of Change

The derivative dy/dx represents the rate of change of y with respect to x.

Tangent

A straight line that touches a curve at a single point, the point of tangency. Its slope is given by dy/dx at that point.

Normal

A straight line perpendicular to the tangent at the point of tangency. Its slope is -1 / (dy/dx).

Increasing Function

A function f(x) is increasing on an interval if f'(x) ≥ 0 for all x in that interval.

Strictly Increasing Function

A function f(x) is strictly increasing on an interval if f'(x) > 0 for all x in that interval.

Critical Point

A point 'c' in the domain of a function f where either f'(c) = 0 or f'(c) is not defined. These are potential points for local maxima or minima.

Local Maximum

A point where the function's value is greater than or equal to the values at all nearby points. Condition: f'(c) = 0 and f''(c) < 0.

Local Minimum

A point where the function's value is less than or equal to the values at all nearby points. Condition: f'(c) = 0 and f''(c) > 0.

Absolute Maximum/Minimum

The largest/smallest value of a function over its entire domain or a given closed interval.

Point of Inflection

A point on a curve at which the concavity changes (from up to down, or down to up). Often occurs where f''(x) = 0.

Must Remember Formulas & Conditions

• Rate of Change: If y = f(x), then dy/dx is the rate of change of y w.r.t x. For related rates, use the chain rule: dy/dt = (dy/dx) * (dx/dt).
• Slope of Tangent: The slope (m) of the tangent to the curve y = f(x) at (x₁, y₁) is m = [dy/dx] at (x₁, y₁).
• Equation of Tangent: The equation of the tangent at (x₁, y₁) is y - y₁ = m(x - x₁).
• Slope of Normal: The slope of the normal to the curve y = f(x) at (x₁, y₁) is -1/m = -1 / [dy/dx] at (x₁, y₁).
• Equation of Normal: The equation of the normal at (x₁, y₁) is y - y₁ = (-1/m)(x - x₁).
• Increasing/Decreasing Test: For a function f(x) on an interval (a, b): If f'(x) > 0, it is strictly increasing. If f'(x) < 0, it is strictly decreasing. If f'(x) = 0, it is a constant function.
• First Derivative Test (for Local Maxima/Minima): At a critical point c, if f'(x) changes from +ve to -ve, c is a local maximum. If f'(x) changes from -ve to +ve, c is a local minimum.
• Second Derivative Test (for Local Maxima/Minima): Find points c where f'(c)=0. Then, if f''(c) < 0, c is a local maximum. If f''(c) > 0, c is a local minimum. If f''(c) = 0, the test fails.
• Absolute Maxima/Minima in [a, b]: Find all critical points in [a, b]. Evaluate the function at these critical points AND at the endpoints a and b. The largest value is the absolute maximum, and the smallest is the absolute minimum.
• Approximation: For a small change Δx, the corresponding change in y (Δy) is approximately dy. So, Δy ≈ (dy/dx)Δx. This gives the formula: f(x + Δx) ≈ f(x) + f'(x)Δx.

Understanding Rate of Change

One of the most fundamental applications of derivatives is to determine the rate at which a quantity changes. The derivative, dy/dx, literally means the instantaneous rate of change of y with respect to x. In many real-world problems, quantities change with respect to time. For example, the volume of a melting ice cube, the distance covered by a moving car, or the height of water in a filling tank. These are problems of related rates. The key is to use the chain rule. If a variable y depends on a variable x, which in turn depends on time t, then the rate of change of y with respect to t is given by dy/dt = (dy/dx) * (dx/dt). To solve these problems, you must first identify the quantities involved, write down an equation (often from geometry, like the volume of a sphere V = (4/3)πr³) that relates them, differentiate this equation with respect to time t, and then substitute the given values to find the unknown rate.

How to Find Local Maxima and Minima

1. —
2. —
3. —
4. —
5. —
6. —
7. —

Worked Examples

• {"header":"Example 1: Equation of a Tangent","body":"Find the equation of the tangent to the curve y = x⁴ – 6x³ + 13x² – 10x + 5 at the point (0, 5).","solution":"Solution:\n1. Find the derivative (slope function): dy/dx = 4x³ – 18x² + 26x – 10.\n2. Calculate the slope at the given point (0, 5): m = [dy/dx] at x=0 = 4(0)³ – 18(0)² + 26(0) – 10 = -10.\n3. Use the point-slope form y - y₁ = m(x - x₁): y - 5 = -10(x - 0) => y + 10x - 5 = 0."}
• {"header":"Example 2: Local Minima","body":"Find the local minimum value of the function f(x) = 2x³ - 3x² - 36x + 10.","solution":"Solution:\n1. First derivative: f'(x) = 6x² - 6x - 36.\n2. Find critical points by setting f'(x) = 0: 6(x² - x - 6) = 0 => 6(x-3)(x+2) = 0. Critical points are x = 3 and x = -2.\n3. Second derivative: f''(x) = 12x - 6.\n4. Test the points:\n - At x = 3: f''(3) = 12(3) - 6 = 30 > 0. So, x=3 is a point of local minimum.\n - At x = -2: f''(-2) = 12(-2) - 6 = -30 < 0. So, x=-2 is a point of local maximum.\n5. Find the local minimum value by putting x=3 in f(x): f(3) = 2(27) - 3(9) - 36(3) + 10 = 54 - 27 - 108 + 10 = -71."}

Board Exam Traps & Tips

Maxima vs. Minima Conditions: A very common mistake is to mix up the conditions for maxima and minima in the Second Derivative Test. Remember: f''(c) < 0 (negative) implies Maximum (think of a frowny face curve ∩), and f''(c) > 0 (positive) implies Minimum (think of a smiley face curve ∪).

Absolute vs. Local Extrema: For questions asking for absolute maximum or minimum in a closed interval [a, b], do not forget to check the function's value at the endpoints a and b. The absolute extremum can occur at a critical point or at an endpoint.

Word Problems: Read word problems carefully. Identify which quantity is constant and which are variables. Draw a diagram. Clearly state the formula (e.g., volume, surface area) you are starting with before differentiating. State what is given (e.g., dr/dt = 2 cm/s) and what needs to be found (e.g., find dV/dt when r = 5 cm).

Practice Questions with Solutions

• If f'(x) < 0 for all x in an interval (a, b), what can you conclude about the function f(x)? The function f(x) is strictly decreasing on the interval (a, b).
• What is the slope of the normal to the curve y = sin(x) at x = π/2? Slope of tangent: dy/dx = cos(x). At x = π/2, slope of tangent is cos(π/2) = 0. The tangent is horizontal. Therefore, the normal is a vertical line and its slope is undefined.
• For a function f(x), if f'(c) = 0 and f''(c) > 0, what is the point x=c? x=c is a point of local minimum.
• The radius 'r' of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? Circumference C = 2πr. dC/dt = 2π(dr/dt). Given dr/dt = 0.7 cm/s, so dC/dt = 2π(0.7) = 1.4π cm/s.

Frequently Asked Questions