Lesson 11 (3.3)

Lesson 11 (3.3)

Today:

• Review & sketching \( f(x) \) from \( f'(x) \) and vice versa
• Rules of Differentiation
• Derivative of exponential function.

Office Hours: MWF 130pm–230pm

Announcements:

• Exam 1: Next Wednesday
• Exam Reviews:
  • Monday 7:30pm–9:30pm (SI)
    ARMS B061
  • Tuesday in Recitation
  • Wednesday in Class

Function \(f(x)\)

\(f(x)\) increasing or Decreasing

\(f'(x)\) +ve or -ve

\(f'(x)\) increasing or Decreasing

Function \(h(x)\)

\(h(x)\) increasing or Decreasing

\(h'(x)\) +ve or -ve

\(h'(x)\) increasing or Decreasing

Function \(g(x)\)

\(g(x)\) increasing or Decreasing

\(g'(x)\) +ve or -ve

\(g'(x)\) increasing or Decreasing

Function \(j(x)\)

\(j(x)\) increasing or Decreasing

\(j'(x)\) +ve or -ve

\(j'(x)\) increasing or Decreasing

Use graph of \( f'(x) \) to sketch \( f(x) \)

\( -4 < x < -2 \rightsquigarrow f'(x) > 0 \)

\( \Rightarrow \text{increasing } f \)

\( -2 < x < 2 \rightsquigarrow f'(x) = 0 \)

\( \rightsquigarrow \text{function : constant} \)

\[ \left.\begin{array}{l} 2 < x < 4 \\ 4 < x < 6 \end{array}\right\} f'(x) < 0 \Rightarrow \text{decreasing} \]

Rules of Differentiation

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

other notation: \( \frac{df}{dx} \), \( y' \), \( \frac{dy}{dx} \)

1. \( f(x) = 8 \)

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{8 - 8}{h} = 0 \]

\[ \frac{d}{dx}[c] = 0 \quad \text{for any constant } c \]

2. \( f(x) = x \)

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

3.

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

\[ \frac{d}{dx}(x^2) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \] \[ = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \] \[ = \lim_{h \to 0} \frac{2xh + h^2}{h} = 2x \]

Example: \( f(x) = x^3 \)

\[ \frac{d}{dx}[x^3] = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} \]

Algebraic expansion: \[ (x+h)(x+h)(x+h) \] \[ (x^2 + 2xh + h^2)(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 \]

\[ = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} \] \[ = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} \] \[ = 3x^2 \]

Example: \( f(x) = \sqrt{x} = x^{1/2} \leadsto \text{Guess: } \frac{1}{2} x^{\frac{1}{2}-1} \)

\[ \begin{aligned} \frac{d}{dx} x^{1/2} &= \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \\ &= \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} \\ &= \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{2\sqrt{x}} \\ &= \frac{1}{2x^{1/2}} = \frac{1}{2} x^{-1/2} \end{aligned} \]

All 3 examples

\[ \frac{d}{dx} x^{\text{power}} = \text{power} \cdot x^{\text{power}-1} \] \[ \frac{d}{dx} x^n = n x^{n-1} \]

for any Real number \( n \)

4.

• \( \frac{d}{dx}[7x] = 7 \) (using the slope method)

  \[ \frac{d}{dx}[7x] = 7 \cdot \frac{d}{dx}(x) = 7 \]

• \[ \frac{d}{dx}[5x^2] = \lim_{h \to 0} \frac{5(x+h)^2 - 5x^2}{h} = 5 \cdot \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \]

  \[ = 5 \cdot \frac{d}{dx} x^2 \]

\[ \frac{d}{dx}(c f(x)) = c \frac{d}{dx} f(x), \quad \text{for any constant } c \]

5. Derivative of Sums and Differences

eg: \(\frac{d}{dx}[7x+9] = 7\) (using slope)

\[ = \lim_{h \to 0} \frac{7(x+h)+9 - (7x+9)}{h} \] \[ = \lim_{h \to 0} \frac{7(x+h) - (7x)}{h} + \lim_{h \to 0} \frac{9-9}{h} \] \[ = \frac{d}{dx}[7x] + \frac{d}{dx}[9] \]

eg:

\[ \frac{d}{dx}[11x - x^2] = \lim_{h \to 0} \frac{11(x+h) - (x+h)^2 - 11x + x^2}{h} \] \[ = \lim_{h \to 0} \frac{11(x+h) - 11x}{h} - \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \] \[ = \frac{d}{dx}[11x] - \frac{d}{dx}[x^2] = 11 - 2x \] \[ \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}f(x) + \frac{d}{dx}g(x) \] \[ \frac{d}{dx}[f(x) - g(x)] = \frac{d}{dx}f(x) - \frac{d}{dx}g(x) \]

eg:

\[ \frac{d}{dx}\left[5x^4 - 9x^{1/2}\right] \] \[ = \frac{d}{dx}\left[5x^4\right] - \frac{d}{dx}\left[9x^{1/2}\right] \] \[ = 5 \frac{d}{dx}x^4 - 9 \frac{d}{dx}x^{1/2} \]

power rule

\[ = 5 * 4x^{4-1} - 9 * \frac{1}{2}x^{1/2-1} \] \[ = 20x^3 - \frac{9}{2}x^{-1/2} \]

Example: \( f(x) = \frac{1}{\sqrt{x}} + 8x^7 - \frac{2}{x^3} + 95 \), find \( f'(x) \).

\[ f(x) = x^{-1/2} + 8x^7 - 2x^{-3} + 95 \] \[ f'(x) = \frac{d}{dx} \left[ x^{-1/2} + 8x^7 - 2x^{-3} + 95 \right] \] \[ = \frac{d}{dx} x^{-1/2} + 8\frac{d}{dx} x^7 - 2\frac{d}{dx} x^{-3} + \require{cancel}\cancel{\frac{d}{dx} 95}^{\,0} \] \[ = \left(-\frac{1}{2}\right)x^{-1/2 - 1} + 8 \times 7 \times x^{7-1} - 2(-3)x^{-3-1} \] \[ = -\frac{1}{2}x^{-3/2} + 56x^6 + 6x^{-4} \]

eg: \(f(x) = (3x-5)(2x+3)\), find \(f'(x)\)

\[ \frac{d}{dx}(3x-5) = 3, \quad \frac{d}{dx}(2x+3) = 2 \]

\[ \frac{d}{dx}[(3x-5)(2x+3)] \neq \left[\frac{d}{dx}(3x-5)\right]\left[\frac{d}{dx}(2x+3)\right] = 3 \times 2 = 6\]

\[ \begin{aligned} f(x) &= (3x-5)(2x+3) = 6x^2 - 10x + 9x - 15 \\ &= 6x^2 - x - 15 \end{aligned} \]

\[ \begin{aligned} \frac{d}{dx} f(x) &= \frac{d}{dx} [6x^2 - x - 15] \\ &= 12x - 1 \end{aligned} \]

Derivative of \( e^x \)

\[ f(x) = e^x \]

\[ \begin{aligned} f'(0) &= \lim_{h \to 0} \frac{f(h) - f(0)}{h} \\ &= \lim_{h \to 0} \frac{e^h - 1}{h} \\ &= 1 \quad \text{(see later)} \end{aligned} \]

but looking at graph, \( \frac{d}{dx} e^x \) has similar properties as \( e^x \)

in fact

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

Derivative of \( e^x \)

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

\( e^{a+b} = e^a \cdot e^b \)

\[ = \lim_{h \to 0} \frac{e^x \cdot e^h - e^x}{h} = \lim_{h \to 0} \frac{e^x [e^h - 1]}{h} \] \[ = e^x \cdot \left[ \lim_{h \to 0} \frac{e^h - 1}{h} \right] \]

Since \( \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \):

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