Lesson 14 (09/28/20)

Lesson 14 (09/28/20)

Today:

• Derivatives as Rate of Change (3.6)
• Chain Rule (3.7)

Office Hours:

MWF 1:30 – 2:30pm

Announcements:

• Exam 1 is Graded
  • Average \(71\% \approx 12.75 / 18\)
  • Median \(75\% \approx 13.5 / 18\)
• Exam 2: Monday, 10/19, 8pm – 9pm
• Final Exam: Tuesday, 12/15, 10:30am – 12:30pm
• Feasting with Faculty: Wednesday 6:45pm – 7:45pm, Windsor

Review: Rules of Differentiation

\[ \frac{d}{dx}(c) = 0 \quad \text{for any constant } c \] \[ \frac{d}{dx}(x^n) = n \cdot x^{n-1} \quad \text{for any } n \] \[ \frac{d}{dx}(e^x) = e^x \] \[ \frac{d}{dx}\left[c \cdot f(x)\right] = c \cdot \frac{d}{dx}f(x) \] \[ \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)] \] \[ \frac{d}{dx}[f(x) - g(x)] = \frac{d}{dx}[f(x)] - \frac{d}{dx}[g(x)] \] \[ \frac{d}{dx}[f(x) g(x)] = f'(x) g(x) + f(x) g'(x) \] \[ \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2} \] \[ \frac{d}{dx}(\sin x) = \cos x \] \[ \frac{d}{dx}(\cos x) = -\sin x \] \[ \frac{d}{dx}(\tan x) = \sec^2 x \] \[ \frac{d}{dx}(\sec x) = \sec x \tan x \] \[ \frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x \] \[ \frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x \]

Derivatives as Rate of Change

\( s(t) = \text{position at time } t \)

Average velocity on \( [a, b] \)

\[ = \frac{s(b) - s(a)}{b - a} \]

instantaneous velocity at \( t = a \)

\[ = \lim_{b \to a} \frac{s(b) - s(a)}{b - a} \] \[ = s'(a) \]

\( v(t) = s'(t) \rightsquigarrow \text{velocity}, \quad |v(t)| = \text{Speed} \)

\( a(t) = v'(t) = s''(t) \rightsquigarrow \text{acceleration.} \)

eg: \( s(t) = t^3 - 9t^2 + 15t + 25 \), find \( v(t) \), \( a(t) \)

\[ v(t) = s'(t) = \frac{d}{dt}\left[t^3 - 9t^2 + 15t + 25\right] \] \[ = 3t^2 - 18t + 15 \]

\( v(0) = 15 \rightsquigarrow \text{initial velocity} = 15 \)

\[ v(t) = 3\left[t^2 - 6t + 5\right] = 3[t - 1][t - 5] \]

\[ v(1) = v(5) = 0 \]

\[ a(t) = v'(t) = \frac{d}{dt}\left[3t^2 - 18t + 15\right] = 6t - 18 \]

\( a(0) = -18 \)

\[ a(3) = 0 \]

Sketching \( s(t), v(t), a(t) \)

\[ a(t) = 6t - 18 \]

\[ v(t) = 3t^2 - 18t + 15 \]

\[ v(0) = 15 \]

\[ v(1) = v(5) = 0 \]

\( t = 3 \), velocity has horizontal tangent

\[ s(t) = t^3 - 9t^2 + 15t + 25 \]

\[ s(5) = 5^3 - 9(25) + 15(5) + 25 \]

\[ = 0 \]

Chain Rule

eg:

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

use Product Rule

\[= \left[\frac{d}{dx} \sin x\right] \sin x + \sin x \frac{d}{dx} [\sin x]\] \[= \cos x \sin x + \sin x \cos x = 2 \sin x \cos x .\]

How about \(\frac{d}{dx} \left[\sin^{75} x\right] \leadsto\) if we use product Rule, Apply it 75 times Not feasible

idea: treat it as two functions

\((\sin x)^{75} \longrightarrow\)

1. \(u = \sin x\)
2. \(u^{75}\)

\( f \circ g(x) = f(g(x)) \)

want: \( \frac{d}{dx}\left[f(g(x))\right] \)

inside: \( u = g(x) \)    outside: \( f(u) \)

\[ \frac{d}{dx}\left[f(g(x))\right] = \frac{d}{dx} f(u) \underset{\substack{\uparrow \\ \text{Chain} \\ \text{Rule}}}{=} \frac{d f(u)}{du} \cdot \frac{du}{dx} \] \[ = f'(u) \cdot \frac{du}{dx} \] \[ = f'(g(x)) \cdot g'(x). \] \[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]

Example

\[ \frac{d}{dx}\left[\sin^2 x\right] = 2\sin x \cos x \quad (\text{using product rule}) \]

Using Chain Rule:

\[ \sin^2 x = (\sin x)^2, \quad \text{Let } \sin x = u \]

\[ \frac{d}{dx}(\sin x)^2 = \frac{d}{dx} u^2 = \frac{d u^2}{du} \cdot \frac{du}{dx} \]

\[ = 2u \cdot \frac{du}{dx} \]

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

\[ = 2\sin x \cos x \]

Example:

\[ \frac{d}{dx}\left[(\sin x)^{75}\right] \]

inside function \( = \sin x = u \)

\[ \begin{aligned} = \frac{d}{dx} u^{75} &= \frac{d}{du} u^{75} \cdot \frac{du}{dx} \\ &= 75 \cdot u^{74} \cdot \frac{du}{dx} \\ &= 75 (\sin x)^{74} \cdot \frac{d}{dx}\sin x \\ &= 75 (\sin x)^{74} \cos x. \end{aligned} \]

eg: \(\frac{d}{dx}(\sin(x^{75}))\)

\(\text{inside} = u = x^{75} \leadsto \text{inside}' = 75x^{74}\)
\(\text{outside} = \sin(u) \leadsto \text{outside}' = \cos(u)\)

\[ \begin{aligned} \frac{d}{dx}(\sin(x^{75})) &= \text{outside}' \cdot \text{inside}' \\ &= \cos(u) \cdot 75x^{74} \\ &= \cos(x^{75}) \cdot 75x^{74} \end{aligned} \]

eg: \( f(x) = (7x^3 + 2x - 9)^{5/4} \)

inside: \( u = 7x^3 + 2x - 9 \rightsquigarrow (\text{inside})' = 21x^2 + 2 \)

outside: \( u^{5/4} \rightsquigarrow (\text{outside})' = \frac{5}{4} u^{5/4-1} = \frac{5}{4} u^{1/4} \)

\[ \begin{aligned} f'(x) &= (\text{outside})' \cdot (\text{inside})' \\ &= \frac{5}{4} u^{1/4} \cdot (21x^2 + 2) \\ &= \frac{5}{4} (7x^3 + 2x - 9)^{1/4} (21x^2 + 2) \end{aligned} \]

Example

\[ f(x) = \tan(8x^{-11} - 23x + 1) \]

\[ \text{inside} = u = 8x^{-11} - 23x + 1, \quad (\text{inside})' = -8 \times 11 \times x^{-11-1} - 23 \]

\[ = -88x^{-12} - 23 \]

\[ \text{outside} = \tan(u), \quad (\text{outside})' = \sec^2 u \]

\[ \begin{aligned} f'(x) &= (\text{outside})' \cdot (\text{inside})' \\ &= \sec^2 u \cdot (-88x^{-12} - 23) \\ &= \sec^2(8x^{-11} - 23x + 1) \cdot (-88x^{-12} - 23) \end{aligned} \]