Answer:
a) [tex]\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1[/tex]
b) [tex]\displaystyle \frac{dy}{dx} \bigg| \limits_{x = \frac{\pi}{2}} = -1[/tex]
General Formulas and Concepts:
Pre-Calculus
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Trigonometric Differentiation
Logarithmic Differentiation
Step-by-step explanation:
a)
Step 1: Define
Identify
[tex]\displaystyle y = ln \bigg( \frac{1 - x}{\sqrt{1 + x^2}} \bigg)[/tex]
Step 2: Differentiate
- Logarithmic Differentiation [Chain Rule]: [tex]\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 - x}{\sqrt{1 + x^2}}} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}][/tex]
- Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}][/tex]
- Quotient Rule: [tex]\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{(1 - x)'\sqrt{1 + x^2} - (1 - x)(\sqrt{1 + x^2})'}{(\sqrt{1 + x^2})^2}[/tex]
- Basic Power Rule [Chain Rule]: [tex]\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{-\sqrt{1 + x^2} - (1 - x)(\frac{x}{\sqrt{x^2 + 1}})}{(\sqrt{1 + x^2})^2}[/tex]
- Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \bigg( \frac{x(x - 1)}{(x^2 + 1)^\bigg{\frac{3}{2}}} - \frac{1}{\sqrt{x^2 + 1}} \bigg)[/tex]
- Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{x + 1}{(x - 1)(x^2 + 1)}[/tex]
Step 3: Find
- Substitute in x = 0 [Derivative]: [tex]\displaystyle \frac{dy}{dx} \bigg| \limit_{x = 0} = \frac{0 + 1}{(0 - 1)(0^2 + 1)}[/tex]
- Evaluate: [tex]\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1[/tex]
b)
Step 1: Define
Identify
[tex]\displaystyle y = ln \bigg( \frac{1 + sinx}{1 - cosx} \bigg)[/tex]
Step 2: Differentiate
- Logarithmic Differentiation [Chain Rule]: [tex]\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 + sinx}{1 - cosx}} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}][/tex]
- Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}][/tex]
- Quotient Rule: [tex]\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{(1 + sinx)'(1 - cosx) - (1 + sinx)(1 - cosx)'}{(1 - cosx)^2}[/tex]
- Trigonometric Differentiation: [tex]\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{cos(x)(1 - cosx) - sin(x)(1 + sinx)}{(1 - cosx)^2}[/tex]
- Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - sin(x) - 1]}{[sin(x) + 1][cos(x) - 1]}[/tex]
Step 3: Find
- Substitute in x = π/2 [Derivative]: [tex]\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = \frac{-[cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) - 1]}{[sin(\frac{\pi}{2}) + 1][cos(\frac{\pi}{2}) - 1]}[/tex]
- Evaluate [Unit Circle]: [tex]\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = -1[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Book: College Calculus 10e
