CALCULUS — DIFFERENTIATION

WHY THIS MATTERS

Calculus is the mathematical engine that drives all of physics. Differentiation answers the question: "At this exact instant, how fast is this quantity changing?". Velocity is the derivative of position. Acceleration is the derivative of velocity. Finding maxima and minima of functions — projectile range, maximum kinetic energy — all need differentiation.

§1. The Concept of Rate of Change

Secant line approaching tangent at $x=2$ on $y=x^2$

Average Rate vs. Instantaneous Rate

Average Rate of Change

Over a finite interval $[x_1, x_2]$:
$$\frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$ This is the slope of the secant line joining two points.

Instantaneous Rate (Derivative)

As $\Delta x \to 0$:
$$\frac{dy}{dx} = \lim_{\Delta x \to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$$ This is the slope of the tangent line at a single point.

⚛️ Physics Interpretation:
If $s(t)$ = displacement, then:
• $v = \frac{ds}{dt}$ = Instantaneous velocity (slope of s-t graph)
• $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ = Instantaneous acceleration (slope of v-t graph)

§2. First Principles of Differentiation

DEFINITION

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Derivation from First Principles: $f(x) = x^3$

$f'(x) = \lim_{h\to 0}\frac{(x+h)^3 - x^3}{h}$. Expanding gives $3x^2h + 3xh^2 + h^3$. Dividing by $h$ and letting $h \to 0$ results in $\mathbf{3x^2}$.

§3. Standard Derivative Formulas

These are the fundamental building blocks. Memorize every single one.

Function $f(x)$	Derivative $f'(x)$	Function $f(x)$	Derivative $f'(x)$
$c$ (constant)	$0$	$\sin x$	$\cos x$
$x^n$	$nx^{n-1}$	$\cos x$	$-\sin x$
$\sqrt{x} = x^{1/2}$	$\frac{1}{2\sqrt{x}}$	$\tan x$	$\sec^2 x$
$e^x$	$e^x$	$\ln x$	$1/x$
$\ln x$	$1/x$	$\sin^{-1} x$	$1/\sqrt{1-x^2}$

§4. The Golden Rules of Differentiation

1. Constant Multiple Rule

$\frac{d}{dx}[cf(x)] = c \cdot f'(x)$

2. Sum / Difference Rule

$\frac{d}{dx}[f \pm g] = f' \pm g'$

3. Product Rule (UV)

$\frac{d}{dx}[uv] = u'v + uv'$

4. Quotient Rule (U/V)

$\frac{d}{dx}[\frac{u}{v}] = \frac{u'v - uv'}{v^2}$

5. Chain Rule

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Product Rule Example: Differentiate $y = x^2\sin x$

$\frac{dy}{dx} = (2x)(\sin x) + (x^2)(\cos x) = \mathbf{2x\sin x + x^2\cos x}$

Chain Rule Example: Differentiate $y = (3x^2 + 5)^4$

$\frac{dy}{dx} = 4(3x^2+5)^3 \cdot (6x) = \mathbf{24x(3x^2+5)^3}$

§5. Higher Order Derivatives

The derivative of a derivative is the Second Derivative ($\frac{d^2y}{dx^2}$).

⚛️ Physics Meaning:
If $s$ is displacement: $v = \frac{ds}{dt}$ (velocity), $a = \frac{d^2s}{dt^2}$ (acceleration). In SHM: $\frac{d^2x}{dt^2} = -\omega^2 x$.

§6. Application — Maxima & Minima

Maxima and Minima of $y = x^3 - 3x$

METHOD

Step 1: Set $\frac{dy}{dx} = 0$. Solve for $x$ (Critical points).

Step 2: Check $\frac{d^2y}{dx^2}$ at those points:

If $\frac{d^2y}{dx^2} < 0$ → Maxima
If $\frac{d^2y}{dx^2} > 0$ → Minima

Practice Questions

DRILL 1 — STANDARD DERIVATIVES

1. Differentiate $y = 5x^4 - 3x^2 + 7x - 2$
2. Differentiate $y = \sqrt{x} + 1/\sqrt{x}$
3. Find $dy/dx$ for $y = 3e^x - 2\ln x$
4. Differentiate $y = \sin x + \cos x - \tan x$

DRILL 2 — PRODUCT, QUOTIENT & CHAIN RULES

5. Differentiate $y = x^3\cos x$
6. Differentiate $y = e^x \sin x$
7. Differentiate $y = \sin(3x^2)$
8. Differentiate $y = (2x^3 - 5)^6$

DRILL 3 — MAXIMA/MINIMA

9. Find the maxima/minima of $y = x^3 - 3x$
10. A ball height is $h = 40t - 5t^2$. When is $h$ maximum?

Answer Key

Q	Answer & Method
1	$20x^3 - 6x + 7$
2	$\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}$
3	$3e^x - 2/x$
5	$3x^2\cos x - x^3\sin x$
6	$e^x(\sin x + \cos x)$
7	$6x\cos(3x^2)$
9	Max at $x=-1$, Min at $x=1$
10	$dh/dt = 40 - 10t = 0 \implies t = \mathbf{4\text{ s}}$