MA161 - Lesson 11: Rules of Differentiation

Graph Analysis: The handwritten notes include several comparative graphs showing the relationship between functions f(x), g(x), h(x), and j(x) and their corresponding derivatives, illustrating how the behavior of a function (increasing/decreasing) relates to the sign of its derivative (positive/negative).

Review: Sketching f'(x) from f(x) and Vice Versa

Key Relationships:

When \(f(x)\) is increasing, \(f'(x) > 0\)
When \(f(x)\) is decreasing, \(f'(x) < 0\)
When \(f'(x) > 0\), the function \(f(x)\) is increasing
When \(f'(x) < 0\), the function \(f(x)\) is decreasing

Multiple Function Analysis: The notes show a comparison table with functions f(x), h(x), g(x), and j(x), analyzing whether each function and its derivative are increasing, decreasing, positive, or negative at various points.

Basic Differentiation Rules

Fundamental Differentiation Rules:

Rule Name	Function	Derivative
Constant Rule	\(f(x) = c\)	\(f'(x) = 0\)
Power Rule	\(f(x) = x^n\)	\(f'(x) = nx^{n-1}\)
Constant Multiple Rule	\(f(x) = c \cdot g(x)\)	\(f'(x) = c \cdot g'(x)\)
Sum Rule	\(f(x) = g(x) + h(x)\)	\(f'(x) = g'(x) + h'(x)\)
Difference Rule	\(f(x) = g(x) - h(x)\)	\(f'(x) = g'(x) - h'(x)\)

Example: Applying Basic Rules

Find \(f'(x)\) if \(f(x) = 5x^3 + 8x - 9x^5\).

Solution:

Apply the rules term by term:

\(\frac{d}{dx}(5x^3) = 5 \cdot 3x^{3-1} = 15x^2\)
\(\frac{d}{dx}(8x) = 8 \cdot 1x^{1-1} = 8\)
\(\frac{d}{dx}(-9x^5) = -9 \cdot 5x^{5-1} = -45x^4\)

Therefore: \(f'(x) = 15x^2 + 8 - 45x^4 = -45x^4 + 15x^2 + 8\)

Derivative of Exponential Functions

Exponential Function Rule:

The most important exponential function is \(f(x) = e^x\):

\[\frac{d}{dx}(e^x) = e^x\]

This is the unique function that is its own derivative!

The Number e:

\(e\) is approximately 2.71828... and is defined as:

\[e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\]

or equivalently:

\[e = \lim_{h \to 0} (1 + h)^{1/h}\]

Example 1: Derivative of Exponential Functions

Find the derivatives of the following:

\(f(x) = e^x\)
\(g(x) = 3e^x\)
\(h(x) = e^x + x^2\)

Solutions:

\(f'(x) = e^x\)
\(g'(x) = 3e^x\)
\(h'(x) = e^x + 2x\)

Combining All Rules

Example 2

Find \(f'(x)\) if \(f(x) = 3x^4 - 2e^x + 7x^2 - 5 \cdot 2^x + 8\).

Solution:

Apply the rules term by term:

\(\frac{d}{dx}(3x^4) = 3 \cdot 4x^3 = 12x^3\)
\(\frac{d}{dx}(-2e^x) = -2e^x\)
\(\frac{d}{dx}(7x^2) = 7 \cdot 2x = 14x\)
\(\frac{d}{dx}(-5 \cdot 2^x) = -5 \cdot 2^x \ln(2)\)
\(\frac{d}{dx}(8) = 0\)

Therefore:

\[f'(x) = 12x^3 - 2e^x + 14x - 5 \cdot 2^x \ln(2)\]

Why is the derivative of \(e^x\) equal to \(e^x\)?

This comes from the limit definition. The number \(e\) is specifically chosen so that:

\[\lim_{h \to 0} \frac{e^h - 1}{h} = 1\]

This makes the derivative calculation clean:

\[\frac{d}{dx}(e^x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \cdot 1 = e^x\]

Common Mistake: Don't confuse \(\frac{d}{dx}(x^n) = nx^{n-1}\) with \(\frac{d}{dx}(a^x) = a^x \ln(a)\). The first is for variable base with constant exponent, the second is for constant base with variable exponent.