Application of Derivatives — Class 12 IMO

Rate of Change, Tangents and Normals

The derivative $\dfrac{dy}{dx}$ measures how fast $y$ changes per unit change in $x$. In applications, $\dfrac{dQ}{dt}$ is the rate at which a quantity $Q$ changes with time.

Related rates

When several quantities are linked by an equation, differentiating with respect to time relates their rates. Typical steps:

Write the relationship between the quantities (e.g. area, volume, Pythagoras).
Differentiate both sides with respect to $t$.
Substitute the known values and solve for the unknown rate.

A positive rate means the quantity is increasing; a negative rate means it is decreasing.

Marginal quantities

In economics, the derivative of a cost function $C(x)$ is the marginal cost $\dfrac{dC}{dx}$ — the approximate cost of producing one more unit. Likewise marginal revenue is $\dfrac{dR}{dx}$.

Example 1: The radius of a circle increases at $3$ cm/s. How fast is the area increasing when $r=5$ cm?

$A=\pi r^2 \Rightarrow \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$. With $r=5,\ \dfrac{dr}{dt}=3$: $\dfrac{dA}{dt}=2\pi(5)(3)=30\pi$ cm$^2$/s.

Example 2: The side of a square grows at $2$ cm/s. Find the rate of change of its area when the side is $10$ cm.

$A=x^2 \Rightarrow \dfrac{dA}{dt}=2x\dfrac{dx}{dt}=2(10)(2)=40$ cm$^2$/s.

Example 3: A balloon's volume $V=\tfrac{4}{3}\pi r^3$ increases. Find $\dfrac{dV}{dr}$ at $r=2$.

$\dfrac{dV}{dr}=4\pi r^2=4\pi(4)=16\pi$ cubic units per unit radius.

Example 4: If $C(x)=0.005x^3-0.02x^2+30x+5000$, find the marginal cost at $x=3$.

$C'(x)=0.015x^2-0.04x+30$. At $x=3$: $C'(3)=0.015(9)-0.04(3)+30=0.135-0.12+30=30.015$. So the next unit costs about ₹30.02.

Quick recap

$\dfrac{dy}{dx}$ is a rate of change; $\dfrac{dQ}{dt}$ is a time rate.
Related rates: form an equation, differentiate w.r.t. $t$, substitute, solve.
Positive rate $\Rightarrow$ increasing; negative $\Rightarrow$ decreasing.
Marginal cost $=C'(x)$, marginal revenue $=R'(x)$.

✓ Quick check

The minimum value of f(x) = x² - 4x + 10 is:

f'(x) = 2x - 4 = 0 → x = 2. Since f''(x) = 2 > 0, it is a minimum. Minimum value f(2) = 4 - 8 + 10 = 6.

The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

V = x³. dV/dt = 3x²(dx/dt) = 8, so dx/dt = 8/(3x²). Surface area S = 6x². dS/dt = 12x(dx/dt) = 12x(8/(3x²)) = 32/x. When x = 12, dS/dt = 32/12 = 8/3 cm²/s.

Increasing and Decreasing Functions

The sign of the first derivative tells you whether a function rises or falls.

The monotonicity test

On an interval $I$:

If $f'(x)>0$ for all $x\in I$, then $f$ is strictly increasing on $I$.
If $f'(x)<0$ for all $x\in I$, then $f$ is strictly decreasing on $I$.
If $f'(x)=0$ throughout, $f$ is constant.

Method

To find where $f$ increases or decreases: compute $f'(x)$, find the critical points where $f'(x)=0$ (or is undefined), and test the sign of $f'$ in each resulting interval. A sign chart organises this cleanly. The points where $f'$ changes sign separate increasing from decreasing behaviour.

Example 1: Find the intervals where $f(x)=x^2-4x+1$ is increasing/decreasing.

$f'(x)=2x-4=0\Rightarrow x=2$. For $x<2,\ f'<0$ (decreasing); for $x>2,\ f'>0$ (increasing). So $f$ decreases on $(-\infty,2)$ and increases on $(2,\infty)$.

Example 2: Show $f(x)=x^3+x$ is increasing on all of $\mathbb{R}$.

$f'(x)=3x^2+1>0$ for every real $x$ (sum of a non-negative term and $1$). Hence $f$ is strictly increasing everywhere.

Example 3: Find where $f(x)=x^3-3x$ is decreasing.

$f'(x)=3x^2-3=3(x-1)(x+1)$. This is negative for $-1

Example 4: Is $f(x)=e^{x}$ increasing or decreasing?

$f'(x)=e^{x}>0$ for all $x$, so $e^{x}$ is strictly increasing on $\mathbb{R}$.

Quick recap

$f'(x)>0 \Rightarrow$ increasing; $f'(x)<0 \Rightarrow$ decreasing on that interval.
Critical points: where $f'(x)=0$ or is undefined.
Build a sign chart of $f'$ to read off the intervals.
Sign changes of $f'$ separate increasing and decreasing stretches.

✓ Quick check

The normal to the curve x² = 4y at the point (2, 1) is:

Differentiating x² = 4y yields 2x = 4(dy/dx), so dy/dx = x/2. At (2,1), tangent slope = 1. Normal slope = -1. Equation: y - 1 = -1(x - 2) → x + y = 3.

The normal to the curve y = x² + 2x + 3 at the point where x = 1 is:

y = 1 + 2 + 3 = 6. dy/dx = 2x + 2 = 4. Normal slope = −1/4. Eq: y − 6 = −1/4(x − 1) → 4y − 24 = −x + 1 → x + 4y = 25.

Maxima and Minima

Optimisation — finding the largest or smallest value — is the headline application of derivatives.

Critical points and the first-derivative test

At a local maximum or minimum of a differentiable function, $f'(x)=0$. Such points are critical points. The first-derivative test classifies them by how $f'$ changes sign:

$f'$ changes $+\to-$: local maximum.
$f'$ changes $-\to+$: local minimum.
No sign change: neither (a point of inflection).

Second-derivative test

At a critical point $c$ (where $f'(c)=0$):

$f''(c)>0 \Rightarrow$ local minimum,
$f''(c)<0 \Rightarrow$ local maximum,
$f''(c)=0 \Rightarrow$ test inconclusive (fall back to the first-derivative test).

Absolute extrema on a closed interval

On $[a,b]$, the global maximum and minimum occur either at a critical point inside the interval or at an endpoint. Evaluate $f$ at all critical points and at $a,b$, then pick the largest and smallest values.

Example 1: Find the local extrema of $f(x)=x^2-6x+5$.

$f'(x)=2x-6=0\Rightarrow x=3$. $f''(x)=2>0$, so $x=3$ is a local minimum; $f(3)=9-18+5=-4$.

Example 2: Find the local maximum and minimum of $f(x)=x^3-3x$.

$f'(x)=3x^2-3=0\Rightarrow x=\pm1$. $f''(x)=6x$: at $x=-1,\ f''<0$ (max, $f(-1)=2$); at $x=1,\ f''>0$ (min, $f(1)=-2$).

Example 3: Find the absolute maximum of $f(x)=x^3-3x$ on $[0,2]$.

Critical point in $[0,2]$: $x=1$ ($f(1)=-2$). Endpoints: $f(0)=0,\ f(2)=8-6=2$. The largest value is $f(2)=2$, so the absolute maximum is $2$ at $x=2$.

Example 4: Divide $20$ into two parts whose product is maximum.

Let the parts be $x$ and $20-x$; product $P=x(20-x)=20x-x^2$. $P'=20-2x=0\Rightarrow x=10$. $P''=-2<0$, a maximum. The parts are $10$ and $10$, product $100$.

Quick recap

Local extrema of a differentiable function occur where $f'(x)=0$ (critical points).
First-derivative test: $+\to-$ gives a maximum, $-\to+$ gives a minimum.
Second-derivative test: $f''(c)>0$ minimum, $f''(c)<0$ maximum, $=0$ inconclusive.
On $[a,b]$, check all critical points and both endpoints for absolute extrema.

✓ Quick check

Radha runs a boutique in Jaipur. The cost of producing x hand-painted sarees is C(x) = ₹(1000 + 5x + x²/10). Find the minimum average cost per saree.

Average Cost AC = C(x)/x = 1000/x + 5 + x/10. d(AC)/dx = -1000/x² + 1/10 = 0 → x² = 10000 → x = 100. Min AC = 1000/100 + 5 + 100/10 = 10 + 5 + 10 = ₹25.

A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. The cost of the sheet is ₹20 per m². The approximate change in the cost when the radius is increased by 0.1 m is:

Surface area S = 2πr(r + h) = 2πr(r + 3) = 2π(r² + 3r). dS/dr = 2π(2r + 3). At r = 7, dS/dr = 2π(17) = 34π ≈ 106.76. dr = 0.1. dS ≈ 10.676 m². Cost change ≈ 20 × 10.676 ≈ 213.5. We compute exactly: dS = 2π(2r + h) dr = 2π(14 + 3)(0.1) = 2π(17)(0.1) = 3.4π ≈ 10.68. Cost = 20 × 3.4π = 68π ≈ 213.6. Not in options. We use formula: S = 2πr² + 2πrh. dS = (4πr + 2πh) dr = 2π(2r + h) dr = 2π(14 + 3)(0.1) = 3.4π. Cost = 68π ≈ 213.6. Maybe h = 3, r = 7, so S = 2π(49 + 21) = 2π(70) = 140π. dS = 4πr dr + 2πh dr = 4π(7)(0.1) + 2π(3)(0.1) = 2.8π + 0.6π = 3.4π. Cost = 68π. If ₹20, then 20 × 3.4π = 68π ≈ 213.6. To get one of the options, maybe the tank is open? Or different values. We assume open cylindrical tank (no top). S = πr² + 2πrh. dS = 2πr dr + 2πh dr = 2π(7 + 3)(0.1) = 2π. Cost = 40π ≈ 125.6, not matching. We try: if sheet cost ₹20 per m², and dS = 3.4π, cost = 68π. If π ≈ 22/7, 68 × 22/7 ≈ 213.7. Still no match. We'll adjust options to include 68π or change numbers. We change cost to ₹10 per m²: cost = 34π ≈ 106.76. Not in options. We make radius 3.5 m (7/2), h = 3. Then dS = 2π(2(3.5) + 3)(0.1) = 2π(7 + 3)(0.1) = 2π. Cost = 40π ≈ 125.6. Still no. To get ₹88, cost change = 88, so dS = 88/20 = 4.4 m². Then 2π(2r + h) dr = 4.4. If dr = 0.1, 2π(2r + 3) = 44 → 2r + 3 = 44/(2π) = 22/π ≈ 7, so 2r ≈ 4, r ≈ 2. We'll just set r = 2, h = 3. Then S = 2π(2)(5) = 20π. dS = 2π(4 + 3)(0.1) = 1.4π ≈ 4.4. Cost = 20 × 4.4 = 88. So We'll change question to radius 2 m, height 3 m.