Application of Derivatives

🎯 Key Points

• f'(x)>0 → strictly increasing; f'(x)<0 → strictly decreasing; f'(x)=0 throughout → constant
• First Derivative Test: sign change +→- at c means local MAX; -→+ means local MIN; no sign change means inflection (neither)
• Linear approximation: f(x+Δx) ≈ f(x)+f'(x)·Δx — the basis for quickly estimating √25.2, cube roots, etc. near a known value
• Absolute extrema on [a,b]: check ALL critical points INSIDE (a,b) AND both endpoints a, b — missing the endpoints is a common error since the true global max/min can occur there even if f' never vanishes there
• Rolle's theorem (f(a)=f(b) case) is a SPECIAL CASE of the Mean Value Theorem (general case) — MVT's conclusion reduces to Rolle's when the right side [f(b)-f(a)]/(b-a) becomes 0

Application of Derivatives

Derivatives are not just abstract formulas, they have powerful real-world and geometric applications. This chapter uses the derivative as a rate of change and as a tool to study the shape and behavior of curves, find tangents and normals, locate maximum and minimum values, and solve optimization problems.

Rate of Change of Quantities

• If y = f(x), the derivative dy/dx represents the instantaneous rate of change of y with respect to x.
• If two quantities x and y are both functions of time t, and are related, then dy/dt = (dy/dx) × (dx/dt) (chain rule), which is the basis of "related rates" problems.
• Example: For a circle, Area A = πr². Then dA/dt = 2πr × dr/dt, relating the rate of change of area to the rate of change of radius.
• Example: For a sphere, Volume V = (4/3)πr³. Then dV/dt = 4πr² × dr/dt.

Increasing and Decreasing Functions

• A function f is strictly increasing on an interval if f'(x) > 0 for all x in that interval.
• A function f is strictly decreasing on an interval if f'(x) < 0 for all x in that interval.
• If f'(x) = 0 throughout an interval, f is constant there.
• Method: Find f'(x), set f'(x) = 0 to get critical points, then test the sign of f'(x) in each interval formed by these points on the number line.

Tangents and Normals

• The slope of the tangent to y = f(x) at point (x₁, y₁) is m = f'(x₁), the derivative evaluated at that point.
• Equation of tangent: y - y₁ = f'(x₁)(x - x₁)
• The normal is perpendicular to the tangent at the point of contact, so its slope is -1/f'(x₁) (when f'(x₁) ≠ 0).
• Equation of normal: y - y₁ = -1/f'(x₁) × (x - x₁)
• If f'(x₁) = 0, the tangent is horizontal (parallel to the x-axis) and the normal is vertical (x = x₁).
• If f'(x₁) is undefined (tangent vertical), the tangent is x = x₁ and the normal is horizontal.

Approximations Using Derivatives

• For a small change Δx in x, the corresponding small change in y is approximated as Δy ≈ f'(x) × Δx.
• This gives the approximation f(x + Δx) ≈ f(x) + f'(x) × Δx, useful for estimating values like square roots or cube roots near a known point.
• Example: To approximate √25.2, take f(x) = √x, x = 25, Δx = 0.2. Then f'(x) = 1/(2√x) = 1/10, so √25.2 ≈ 5 + 0.1 × 0.2 = 5.02.

Maxima and Minima

• Critical point: A point c in the domain where f'(c) = 0 or f'(c) does not exist.
• Local maximum: f(c) is greater than or equal to f(x) for all x in some neighborhood of c.
• Local minimum: f(c) is less than or equal to f(x) for all x in some neighborhood of c.
• First Derivative Test: At a critical point c, if f'(x) changes sign from positive to negative as x increases through c, f has a local maximum at c. If it changes from negative to positive, f has a local minimum. If there is no sign change, c is a point of inflection (neither maximum nor minimum).
• Second Derivative Test: At a critical point c where f'(c) = 0: if f''(c) < 0, f has a local maximum at c. If f''(c) > 0, f has a local minimum at c. If f''(c) = 0, the test fails and the first derivative test must be used.
• Absolute (global) maximum/minimum: On a closed interval [a, b], the absolute extrema occur either at critical points inside (a, b) or at the endpoints a and b. Evaluate f at all of these and compare.

Curve Sketching Basics

• Find intervals of increase and decrease using f'(x).
• Locate local maxima and minima using the first or second derivative test.
• Find points of inflection where f''(x) = 0 and concavity changes (concave up when f''(x) > 0, concave down when f''(x) < 0).
• Combine this information with intercepts and asymptotes to sketch the general shape of the curve.

Optimization Word Problems

• General method: write the quantity to be maximized/minimized as a function of one variable using the given constraint, differentiate, set the derivative to zero, and apply the second derivative test to confirm a maximum or minimum.
• Classic example (maximum area): Among all rectangles with a fixed perimeter, the square encloses the maximum area. If x + y = 20, area A = xy = x(20 - x) is maximized at x = y = 10, giving maximum area 100.
• Classic example (minimum cost/material): Among all closed cylinders of a given volume, the one with height equal to the diameter (h = 2r) has minimum surface area.
• Classic example (maximum volume box): Cutting equal squares of side x from the corners of a square sheet of side a and folding up the sides gives an open box of volume V = x(a - 2x)², maximized at x = a/6.

Rolle's Theorem and Mean Value Theorem

• Rolle's Theorem: If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists at least one point c in (a, b) such that f'(c) = 0.
• Geometric meaning: somewhere between two points of equal height on the curve, the tangent must be horizontal.
• Lagrange's Mean Value Theorem (MVT): If f is continuous on [a, b] and differentiable on (a, b), then there exists at least one point c in (a, b) such that f'(c) = [f(b) - f(a)] / (b - a).
• Geometric meaning: there is some point on the curve where the tangent is parallel to the chord joining (a, f(a)) and (b, f(b)).
• Rolle's theorem is a special case of the Mean Value Theorem where f(a) = f(b), making the right-hand side zero.

Angle of Intersection of Two Curves

• The angle between two curves at a point of intersection is defined as the angle between their tangents at that common point.
• If m₁ and m₂ are the slopes of the tangents to the two curves at the point, the acute angle θ between them satisfies tan θ = |(m₁ - m₂) / (1 + m₁m₂)|.
• Orthogonal curves: the curves cut at right angles when m₁ × m₂ = -1 (so 1 + m₁m₂ = 0 and θ = 90°).
• The curves touch each other (are tangential) when m₁ = m₂ at the point of intersection, giving θ = 0.
• Method: find the point(s) of intersection, compute dy/dx for each curve there to get m₁ and m₂, then apply the formula.

Differentials, Errors and Approximation of Errors

• The differential dy is defined by dy = f'(x) dx, where dx = Δx is the change in x. It measures the change along the tangent line, while the actual change Δy = f(x + Δx) - f(x) is along the curve.
• For small Δx, dy ≈ Δy, which is what makes differentials useful for approximating values and estimating errors.
• Absolute error in y is approximately |dy| = |f'(x)| |Δx|.
• Relative error = Δy/y and percentage error = (Δy/y) × 100.
• Example: if the radius of a sphere is measured as 9 cm with a possible error of 0.03 cm, then V = (4/3)πr³ gives dV = 4πr² dr = 4π(81)(0.03) = 9.72π cm³, the approximate error in the computed volume.

🚀 JEE Advanced Edge

Using AM-GM instead of calculus for optimization: Many "maximize/minimize" problems that look like they need derivatives can be solved faster with the AM-GM inequality (AM≥GM, equality when all terms are equal) — e.g. minimizing x+1/x for x>0 gives minimum value 2 instantly (since AM of x and 1/x ≥ GM = 1, so x+1/x≥2), without setting up and solving f'(x)=0. Recognizing when a problem is an AM-GM problem in disguise saves significant time.

Why the second derivative test can fail and when to fall back to the first: If f''(c)=0 at a critical point, the second derivative test gives NO information — the point could be a max, min, or inflection point (e.g. f(x)=x⁴ at x=0 has f''(0)=0 but is actually a minimum; f(x)=x³ at x=0 has f''(0)=0 and is an inflection point). Whenever f''(c)=0, you MUST revert to the first derivative test (sign change analysis) to classify the point correctly.

Worked problem: A wire of length 28m is cut into two pieces, one bent into a circle and the other into a square, to minimize the combined area. If x is the length used for the circle, find the value of x that minimizes total area. Approach: Circle circumference=x, so radius=x/(2π), area₁=πr²=x²/(4π). Square side=(28-x)/4, area₂=(28-x)²/16. Total A=x²/(4π)+(28-x)²/16. dA/dx = x/(2π) - (28-x)/8 = 0. Solving: 8x = 2π(28-x) → 8x+2πx=56π → x(8+2π)=56π → x=56π/(8+2π)=28π/(4+π).

Worked Example: Maxima and Minima

Find the local maximum and minimum values of f(x) = 2x³ − 9x² + 12x + 1.

f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x−1)(x−2). Critical points: x = 1 and x = 2.

f″(x) = 12x − 18. At x = 1: f″(1) = −6 < 0 → local max, f(1) = 6. At x = 2: f″(2) = 6 > 0 → local min, f(2) = 5. The second-derivative test decides max/min at each critical point.

Worked Example: Equation of Tangent

Find the equation of the tangent to y = x³ at the point (1, 1).

dy/dx = 3x². At x = 1, slope m = 3. Tangent: y − 1 = 3(x − 1) → y = 3x − 2. For NEET questions: find the derivative, substitute the given x-value for slope, then use point-slope form.