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3.3 Rules of Differentiation

"to differentiate" — means to find the derivative of something

Definition of Derivative

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

geometric but not practical.

Let's find pattern in derivative of \( x^n \)

\( f(x) = 1 = x^0 \)

\[ f'(x) = \lim_{h \to 0} \frac{1 - 1}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \]

\( f(x) = x^1 \)

\[ f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = 1 = 1 \cdot x^0 \]

\( f(x) = x^2 \)

\[ f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \]\[ = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} \frac{2x + h}{1} = 2x = 2 \cdot x^1 \]

\( f(x) = \sqrt{x} = x^{1/2} \)

\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-1/2} \]
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note the pattern: deriv. of \( x^n \) is \( nx^{n-1} \)

This is called the Power Rule

\[ \frac{d}{dx}(x^n) = nx^{n-1} \]

\( n \): any real number

\( \frac{d}{dx} \) "derivative of ____" (Leibniz notation)

prime (\( ' \)) notation is called the Lagrange notation

build onto the power rule: what is the deriv. of \( C \cdot x^n \)

\( C \to \) some constant

if \( f(x) = C x^n \)

since the derivative is a limit process and we know:

\[ \lim_{x \to a} (C \cdot f(x)) = (\lim_{x \to a} C)(\lim_{x \to a} f(x)) = C \lim_{x \to a} f(x) \]
\[ \text{so, } \frac{d}{dx}(C \cdot x^n) = C \cdot \frac{d}{dx}(x^n) = C \cdot nx^{n-1} \]

\( C \): some number

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In fact,

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

Constant-Multiple Rule

Similarly, since \[ \lim_{x \to a} (f(x) \pm g(x)) = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \]

and since the derivative is a limit process,

\[ \frac{d}{dx} [f(x) + g(x)] = \frac{d}{dx} f(x) + \frac{d}{dx} g(x) \]

\[ \frac{d}{dx} [f(x) - g(x)] = \frac{d}{dx} f(x) - \frac{d}{dx} g(x) \]

Sum/Difference Rules

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Example

\[ f(x) = x^{2/3} - 5\sqrt{x} + \frac{1}{2x} + 10 \quad f'(x) = ? \]

rewrite in the form \( x^n \)

\[ f(x) = x^{2/3} - 5x^{1/2} + \frac{1}{2}x^{-1} + 10x^0 \]

\[ f'(x) = \frac{2}{3}x^{2/3-1} - 5 \cdot (\frac{1}{2}x^{1/2-1}) + \frac{1}{2} \cdot (-1 \cdot x^{-2}) + 10 \cdot (0 \cdot x^{0-1}) \]

\[ = \frac{2}{3}x^{-1/3} - \frac{5}{2}x^{-1/2} - \frac{1}{2}x^{-2} + 0 \]

\[ = \frac{2}{3x^{1/3}} - \frac{5}{2\sqrt{x}} - \frac{1}{2x^2} \]

\( f'(x) \) is itself a function, so we can evaluate it wherever it is defined

for example, \[ f'(1) = \frac{2}{3} - \frac{5}{2} - \frac{1}{2} = -\frac{7}{3} \]

in Leibniz notation: \[ \left. \frac{d}{dx} (x^{2/3} - 5\sqrt{x} + \frac{1}{2x} + 10) \right|_{x=1} = -\frac{7}{3} \]

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\[ \frac{d}{dx} [ f(x) \pm g(x) ] = f'(x) \pm g'(x) \]

but this is NOT true for multiplication or division

Example

\[ f(x) = (2x - 7)^2 \]\[ = (2x - 7)(2x - 7) \]
\[ \frac{d}{dx} [ (2x - 7)(2x - 7) ] \neq \frac{d}{dx} (2x - 7) \cdot \frac{d}{dx} (2x - 7) \]

we can work around that

rewrite:

\[ f(x) = (2x - 7)(2x - 7) \]\[ = 4x^2 - 14x - 14x + 49 \]\[ = 4x^2 - 28x + 49 \]

now use sum/difference rules

\[ f'(x) = 4(2x^1) - 28(1 \cdot x^0) + 49(0 \cdot x^{0-1}) \]\[ = \boxed{8x - 28} \]
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Example

\[ f(x) = \frac{6x^9 - 7x^4}{5x^4} \]
again, \[ f'(x) \neq \frac{\frac{d}{dx}(6x^9 - 7x^4)}{\frac{d}{dx}(5x^4)} \]

workaround: rewrite as combinations of \( x^n \)

\[ f(x) = \frac{6x^9}{5x^4} - \frac{7x^4}{5x^4} = \frac{6}{5}x^5 - \frac{7}{5} \]\[ f'(x) = \frac{6}{5}(5x^4) - 0 = \boxed{6x^4} \]

unfortunately, we can't work around all quotients

for example, if \[ f(x) = \frac{5x^4}{6x^9 - 7x^4} \neq \frac{5x^4}{6x^9} - \frac{5x^4}{7x^4} \]

just like \[ \frac{5}{2+3} \neq \frac{5}{2} + \frac{5}{3} \]

this is a case that requires another rule that we will see later

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Derivative of the Exponential Function

Next, let's find the derivative of \( e^x \).

There are many ways to define \( e \).

Defining \( e \) via Tangent Slope

One way: define \( e \) such that \( f(x) = e^x \) has a tangent line with slope of 1 at \( x = 0 \).

The graph shows the exponential function \( e^x \) and its tangent line at the y-intercept.

Coordinate graph showing the curve e^x in green and a red tangent line at x=0 with slope 1.

Deriving the Limit Property

This means: if \( f(x) = e^x \) then \( f'(0) = 1 \).

\[ f'(0) = \lim_{h \to 0} \frac{e^{0+h} - e^0}{h} = \lim_{h \to 0} \frac{e^0 e^h - e^0}{h} = \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \]

Important Limit:

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

We will come back to this.

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General Derivative of \( e^x \)

Starting with the definition of the derivative for \( f(x) = e^x \):

\[ f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^x e^h - e^x}{h} \]\[ = \lim_{h \to 0} e^x \left( \frac{e^h - 1}{h} \right) = \left( \lim_{h \to 0} e^x \right) \left( \lim_{h \to 0} \frac{e^h - 1}{h} \right) \]

From the previous page, we know that the second limit is equal to 1.

\[ = \lim_{h \to 0} e^x = e^x \]

This means:

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

\( e^x \) is its own derivative.

Example

Find the derivative of:

\[ f(x) = \frac{8e^{2x} + 7e^x}{e^x} \]

Rewrite the function first:

\[ f(x) = \frac{8e^{2x}}{e^x} + \frac{7e^x}{e^x} = 8e^x + 7 \]

Now, differentiate:

\[ f'(x) = 8(e^x) + 0 = 8e^x \]
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Higher-Order Derivatives

The derivative is a function, so we can take the derivative of a derivative (of a derivative...) over and over again.

\[ y = f(x) \]
\[ y' = f'(x) = \frac{dy}{dx} = \frac{df}{dx} \]
first derivative

deriv. of \( y' \)

\[ y'' = f''(x) = \frac{d^2y}{dx^2} = \frac{d^2f}{dx^2} \]
second deriv.
\[ y''' = f'''(x) = \frac{d^3y}{dx^3} = \frac{d^3f}{dx^3} \]
3rd deriv.

Sometimes \( y^{(3)} = f^{(3)}(x) \)

Beyond 3, we always write \( y^{(n)} = f^{(n)}(x) \)

For example, the \( 15^{\text{th}} \) deriv. of \( y \) is \[ y^{(15)} = f^{(15)}(x) = \frac{d^{15}y}{dx^{15}} \]