“Calculus: Differentiation”

🟢 Lite — Quick Review (1h–1d)

Rapid summary of differentiation for NABTEB mathematics.

Differentiation finds the rate of change of a function — the gradient of a curve at any point.

The Derivative:

If $y = f(x)$, the derivative is: $$f’(x) = \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

This gives the gradient of the tangent to the curve at any point.

Basic Derivatives:

Function	Derivative
$c$ (constant)	0
$x^n$	$nx^{n-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$e^x$	$e^x$
$\ln x$	$\frac{1}{x}$
$e^{kx}$	$ke^{kx}$

Differentiation Rules:

1. Sum Rule: $$\frac{d}{dx}[f(x) + g(x)] = f’(x) + g’(x)$$

2. Difference Rule: $$\frac{d}{dx}[f(x) - g(x)] = f’(x) - g’(x)$$

3. Product Rule: $$\frac{d}{dx}[f(x)g(x)] = f’(x)g(x) + f(x)g’(x)$$

4. Quotient Rule: $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f’(x)g(x) - f(x)g’(x)}{[g(x)]^2}$$

5. Chain Rule (Composite Functions): If $y = f(g(x))$, then: $$\frac{dy}{dx} = f’(g(x)) \cdot g’(x)$$

Or: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ where $u = g(x)$

⚡ NABTEB Exam Tip: For $y = (3x + 2)^5$, use the chain rule: $\frac{dy}{dx} = 5(3x+2)^4 \times 3 = 15(3x+2)^4$.

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🟡 Standard — Regular Study (2d–2mo)

For NABTEB students who want thorough understanding.

Differentiation by First Principles:

Find $f’(x)$ for $f(x) = x^2$:

$$f’(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} (2x + h) = 2x$$

Equations of Tangents and Normals:

Tangent to curve at point $(x_1, y_1)$: $$y - y_1 = f’(x_1)(x - x_1)$$

Normal (perpendicular to tangent): $$y - y_1 = -\frac{1}{f’(x_1)}(x - x_1)$$

Example: Find the equation of the tangent to $y = x^2$ at $(2, 4)$.

$f’(x) = 2x$, so $f’(2) = 4$ Equation: $y - 4 = 4(x - 2) \Rightarrow y = 4x - 4$

Stationary Points:

Where $f’(x) = 0$. These are maximum points, minimum points, or points of inflection.

Finding Stationary Points:

1. Find $f’(x)$
2. Set $f’(x) = 0$ and solve for $x$
3. Find the corresponding $y$ values
4. Use the second derivative to classify

Second Derivative Test:

• $f”(x) > 0$: Minimum point (curve is concave up)
• $f”(x) < 0$: Maximum point (curve is concave down)
• $f”(x) = 0$: May be point of inflection — check sign change of $f’$

Example: $f(x) = x^3 - 3x^2 + 2$ $f’(x) = 3x^2 - 6x = 3x(x-2)$ Stationary points when $x = 0$ or $x = 2$:

• At $x = 0$: $y = 2$; $f”(x) = 6x - 6 = -6 < 0$ → Maximum
• At $x = 2$: $y = -2$; $f”(x) = 6 > 0$ → Minimum

⚡ NABTEB Exam Tip: When asked for “turning points,” find both the coordinates AND classify them (maximum or minimum). When asked for “maximum/minimum values,” you need the y-coordinate only.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage of differentiation for thorough NABTEB preparation.

Derivatives of Trigonometric Functions:

$$\frac{d}{dx}(\sin x) = \cos x$$ $$\frac{d}{dx}(\cos x) = -\sin x$$ $$\frac{d}{dx}(\tan x) = \sec^2 x$$ $$\frac{d}{dx}(\cot x) = -\csc^2 x$$ $$\frac{d}{dx}(\sec x) = \sec x \tan x$$ $$\frac{d}{dx}(\csc x) = -\csc x \cot x$$

Derivatives of Exponential and Logarithmic Functions:

$$\frac{d}{dx}(e^x) = e^x$$ $$\frac{d}{dx}(e^{kx}) = ke^{kx}$$ $$\frac{d}{dx}(\ln x) = \frac{1}{x}$$ $$\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$$

Implicit Differentiation:

When $y$ is not isolated:

Example: $x^2 + y^2 = 25$ Differentiate both sides with respect to $x$: $$2x + 2y\frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = -\frac{x}{y}$$

For more complex implicit functions: Example: $x^3 + 2xy^2 - y^5 = x$ Differentiate term by term: $$3x^2 + 2y^2 + 4xy\frac{dy}{dx} - 5y^4\frac{dy}{dx} = 1$$ Collect $\frac{dy}{dx}$ terms: $$\frac{dy}{dx}(4xy - 5y^4) = 1 - 3x^2 - 2y^2$$ $$\frac{dy}{dx} = \frac{1 - 3x^2 - 2y^2}{4xy - 5y^4}$$

Parametric Differentiation:

If $x = f(t)$ and $y = g(t)$, then: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g’(t)}{f’(t)}$$

Example: $x = t^2$, $y = t^3$ $$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$$

Applications of Differentiation:

1. Motion:

• Position $s(t)$
• Velocity $v(t) = \frac{ds}{dt}$
• Acceleration $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

2. Rates of Change: $$\frac{dA}{dr} = 2\pi r \text{ (area of circle)}$$ $$\frac{dV}{dr} = 4\pi r^2 \text{ (volume of sphere)}$$

3. Optimisation: Find maximum or minimum values in practical problems.

Example: A rectangular field with 100m of fencing has one side against a wall. Maximise the area.

Let $x$ = length perpendicular to wall, $y$ = length parallel to wall. Constraint: $2x + y = 100 \Rightarrow y = 100 - 2x$ Area $A = xy = x(100-2x) = 100x - 2x^2$ $$\frac{dA}{dx} = 100 - 4x = 0 \Rightarrow x = 25$$ Maximum area: $A = 25(100-50) = 1250 \text{ m}^2$

Higher Derivatives:

$$f”(x) = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$$ $$f'''(x) = \frac{d^3y}{dx^3}$$

Maclaurin Series:

$$f(x) = f(0) + xf’(0) + \frac{x^2}{2!}f”(0) + \frac{x^3}{3!}f'''(0) + \ldots$$

For $e^x$: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$$

For $\sin x$: $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$$

⚡ NABTEB Quick Reference:

• $\frac{d}{dx}(x^n) = nx^{n-1}$
• $\frac{d}{dx}(\sin x) = \cos x$
• $\frac{d}{dx}(\cos x) = -\sin x$
• $\frac{d}{dx}(e^x) = e^x$
• $\frac{d}{dx}(\ln x) = \frac{1}{x}$
• Chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
• Product rule: $\frac{d}{dx}(uv) = u’v + uv’$
• Quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u’v - uv’}{v^2}$
• Tangent: $y - y_1 = f’(x_1)(x - x_1)$
• Normal: $y - y_1 = -\frac{1}{f’(x_1)}(x - x_1)$
• Stationary points: $f’(x) = 0$
• Maximum: $f”(x) < 0$; Minimum: $f”(x) > 0$
• $v = \frac{ds}{dt}$; $a = \frac{dv}{dt}$