Differentiation

M2 Study Notes: Chapter 7 - Differentiation

Hello everyone! Welcome to your study notes for Differentiation. This is one of the most powerful and exciting topics in calculus. Think of it as getting a special lens to see how things change in an instant.

Why is this important? Differentiation helps us find the speed of a moving object at a specific moment, figure out the best price to sell a product for maximum profit, and even model the spread of diseases. It's all about understanding the rate of change.

Don't worry if it seems tricky at first. We'll break everything down into simple, manageable steps. Let's get started!

1. The Heart of the Matter: The Derivative

So, What is a Derivative?

Imagine you're driving a car. Your average speed for a 2-hour trip is the total distance divided by 2 hours. But your instantaneous speed is what you see on the speedometer at any given second.

The derivative is like the car's speedometer. It tells us the instantaneous rate of change of a function at a very specific point.

From a graphical point of view, the derivative gives us the slope of the tangent line to a curve at a point (the limiting slope of secant lines as the interval shrinks to zero).

Finding the Derivative from First Principles

This is the fundamental method to find a derivative, based on its definition. It's the "long way" of doing it, but it's crucial for understanding what's really happening. We find the slope of a line between two very close points on the curve.

The derivative of a function \(f(x)\), denoted as \(f'(x)\), is defined as:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

Breaking it down:

• \(h\) is a tiny change in the x-value.
• \(f(x+h) - f(x)\) is the corresponding change in the y-value.
• The fraction is the slope (rise over run) of the secant line connecting \((x, f(x))\) and \((x+h, f(x+h))\).
• \( \lim_{h \to 0} \) means we shrink the interval \(h\) towards zero to find the instantaneous slope.

Step-by-Step Example: Find the derivative of \(f(x) = x^2\) from first principles.

1. Write down the formula:
   \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
2. Substitute your function into the formula:
   Since \(f(x) = x^2\), then \(f(x+h) = (x+h)^2\).
   \( f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \)
3. Expand and simplify the numerator:
   \( (x+h)^2 = x^2 + 2xh + h^2 \)
   \( f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h} \)
   \( f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h} \)
4. Factor out h from the numerator and cancel:
   \( f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h} \)
   \( f'(x) = \lim_{h \to 0} (2x + h) \)
5. Now, evaluate the limit as \(h \to 0\):
   \( f'(x) = 2x + 0 = 2x \)

In HKDSE M2, you may also be asked to find derivatives from first principles for functions like \(C\), \(x^n\), \(\sqrt{x}\), \(\sin x\), \(\cos x\), and \(e^x\). For trigonometric and exponential functions, remember the standard limits:
\( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \quad \text{and} \quad \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \)

Notation, Notation, Notation!

You'll see the derivative written in a few different ways. If \(y = f(x)\):

• Lagrange's Notation: \(f'(x)\) or \(y'\) (Read as "f prime of x" or "y prime").
• Leibniz's Notation: \(\frac{dy}{dx}\) (Read as "dee y by dee x"). The operator \(\frac{d}{dx}\) means "differentiate with respect to x".

Key Takeaway for Section 1

The derivative is the instantaneous rate of change of a function, or the slope of its tangent line. We define and find it from the ground up using first principles.

2. The Differentiation Toolbox: Essential Rules

Using first principles every time would be very slow. Luckily, we have a set of differentiation rules that work as efficient tools.

The Basic Rules

• The Constant Rule: The derivative of any constant is zero.
  \( \frac{d}{dx}(C) = 0 \) Analogy: If an object is stationary (constant position), its velocity is 0.


• The Power Rule: For any real number \(n\):
  \( \frac{d}{dx}(x^n) = nx^{n-1} \) Memory Aid: "Bring the power to the front, then reduce the exponent by one."
  Example: \(\frac{d}{dx}(x^5) = 5x^{5-1} = 5x^4\)
  Example: \(\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\)


• The Constant Multiple Rule: Constants multiplying a function remain factored out.
  \( \frac{d}{dx}(c \cdot f(x)) = c \cdot f'(x) \) Example: \(\frac{d}{dx}(7x^3) = 7 \cdot (3x^2) = 21x^2\)


• The Sum/Difference Rule: Differentiate term by term.
  \( \frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x) \) Example: \(\frac{d}{dx}(x^2 + 5x - 3) = 2x + 5 - 0 = 2x+5\)

The Product Rule

Use this rule when differentiating two functions that are multiplied together (\(u\) and \(v\)):
\( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \quad \text{or} \quad (uv)' = u'v + uv' \) Memory Aid: "First times derivative of second, plus second times derivative of first."

Common Mistake: \((uv)' \neq u'v'\). Never simply multiply the derivatives.

Example: Find the derivative of \(y = x^2 \sin x\)
Let \(u = x^2\) and \(v = \sin x\). Then \(u' = 2x\) and \(v' = \cos x\).
\(\frac{dy}{dx} = (x^2)(\cos x) + (\sin x)(2x) = x^2\cos x + 2x\sin x\)

The Quotient Rule

Use this rule when differentiating one function divided by another (\(u/v\)):
\( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \quad \text{or} \quad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) Mnemonic: "Low d-high minus high d-low, over the square of what's below."

Common Mistake: The numerator order matters because of subtraction: \(v u'\) comes first!

Example: Find the derivative of \(y = \frac{e^x}{x^3}\)
\(\frac{dy}{dx} = \frac{(x^3)(e^x) - (e^x)(3x^2)}{(x^3)^2} = \frac{x^2 e^x(x - 3)}{x^6} = \frac{e^x(x - 3)}{x^4}\)

The Chain Rule

Use this for composite functions (a function inside another function):
\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) Steps:

1. Identify the outer function and inner function (\(u\)).
2. Differentiate the outer function with respect to \(u\).
3. Multiply by the derivative of the inner function with respect to \(x\).

Key Takeaway for Section 2

Master the core rules: Power Rule for \(x^n\), Product Rule for \(uv\), Quotient Rule for \(u/v\), and Chain Rule for composite functions.

3. Derivatives of Standard Functions

Here is the comprehensive list of standard function derivatives required in HKDSE M2:

Trigonometric and Reciprocal Trigonometric Functions

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

Exponential and Logarithmic Functions

• \( \frac{d}{dx}(e^x) = e^x \)
• \( \frac{d}{dx}(a^x) = a^x \ln a \quad (a > 0, a \neq 1) \)
• \( \frac{d}{dx}(\ln x) = \frac{1}{x} \quad (x > 0) \)
• \( \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} \quad (a > 0, a \neq 1, x > 0) \)

Quick Review Summary

Function \(f(x)\)     Derivative \(f'(x)\)
--------------------------------------------------
\(C\) (constant)              \(0\)
\(x^n\)                            \(nx^{n-1}\)
\(\sin x\)                       \(\cos x\)
\(\cos x\)                       \(-\sin x\)
\(\tan x\)                       \(\sec^2 x\)
\(\cot x\)                       \(-\csc^2 x\)
\(\sec x\)                       \(\sec x \tan x\)
\(\csc x\)                       \(-\csc x \cot x\)
\(e^x\)                           \(e^x\)
\(a^x\)                           \(a^x \ln a\)
\(\ln x\)                          \(\frac{1}{x}\)
\(\log_a x\)                     \(\frac{1}{x \ln a}\)

4. Advanced Techniques

Implicit Differentiation

When a relation between \(x\) and \(y\) is given implicitly (e.g. \(x^2 + y^2 = 25\)), we differentiate both sides with respect to \(x\). By the Chain Rule, whenever differentiating a term involving \(y\), multiply by \(\frac{dy}{dx}\).

Step-by-Step Example: Find \(\frac{dy}{dx}\) for \(x^2 + y^2 = 25\).

1. Differentiate both sides with respect to \(x\):
   \( \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25) \)
2. Apply differentiation rules:
   \( 2x + 2y\frac{dy}{dx} = 0 \)
3. Solve for \(\frac{dy}{dx}\):
   \( 2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y} \)

Logarithmic Differentiation

This technique is ideal for products, quotients with multiple factors, or functions where the variable appears in both the base and exponent (e.g. \(y = x^x\)).

Step-by-Step Example: Find \(\frac{dy}{dx}\) for \(y = \frac{(x+1)^3 \sqrt{x}}{(2x-1)^5}\)

1. Take natural logarithm on both sides:
   \( \ln y = \ln\left( \frac{(x+1)^3 \sqrt{x}}{(2x-1)^5} \right) \)
2. Expand using logarithm properties:
   \( \ln y = 3\ln(x+1) + \frac{1}{2}\ln x - 5\ln(2x-1) \)
3. Differentiate both sides implicitly:
   \( \frac{1}{y}\frac{dy}{dx} = \frac{3}{x+1} + \frac{1}{2x} - \frac{5(2)}{2x-1} = \frac{3}{x+1} + \frac{1}{2x} - \frac{10}{2x-1} \)
4. Multiply by \(y\) and substitute back:
   \( \frac{dy}{dx} = \frac{(x+1)^3 \sqrt{x}}{(2x-1)^5} \left( \frac{3}{x+1} + \frac{1}{2x} - \frac{10}{2x-1} \right) \)

5. The Second Derivative

What is it?

The second derivative is the derivative of the first derivative:

Physical Interpretation: If \(s(t)\) is position, \(v(t) = s'(t)\) is velocity, and \(a(t) = s''(t)\) is acceleration.

Notation: \(f''(x)\), \(y''\), or \(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)\).

Finding the Second Derivative

Example: Find \(f''(x)\) for \(f(x) = x^4 - 5x^2 + 8\)
Step 1: First derivative: \( f'(x) = 4x^3 - 10x \)
Step 2: Second derivative: \( f''(x) = \frac{d}{dx}(4x^3 - 10x) = 12x^2 - 10 \)

Applications: Concavity and the Second Derivative Test

• Concavity:
  • If \(f''(x) > 0\) on an interval, the curve is concave upwards (opens upward).
  • If \(f''(x) < 0\) on an interval, the curve is concave downwards (opens downward).
• The Second Derivative Test: For a stationary point where \(f'(c) = 0\):
  • If \(f''(c) < 0\), \(f(x)\) has a local maximum at \(x=c\).
  • If \(f''(c) > 0\), \(f(x)\) has a local minimum at \(x=c\).
  • If \(f''(c) = 0\), the test is inconclusive (use the first derivative test).

Key Takeaway for Section 5

The second derivative \(f''(x)\) provides information about the curve's concavity and helps determine whether stationary points are local maxima or minima.