Answer:
[tex]\displaystyle \frac{d}{dx} = \frac{sin^2x + 1}{2cos^3x}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Pre-Calculus
Calculus
Derivatives
Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot 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]
Trig Derivative: [tex]\displaystyle \frac{d}{dx}[tanu] = u'sec^2u[/tex]
Trig Derivative:
[tex]\displaystyle \frac{d}{dx}[cosu] = -u'sinu[/tex]
Step-by-step explanation:
Step 1: Define
[tex]\displaystyle y = \frac{tanx}{2cosx}[/tex]
Step 2: Differentiate
- [Derivative] Quotient Rule: [tex]\displaystyle \frac{d}{dx} = \frac{\frac{d}{dx}[tanx](2cosx) - \frac{d}{dx}[2cosx](tanx)}{(2cosx)^2}[/tex]
- [Derivative] Simplify [Derivative Property - Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} = \frac{\frac{d}{dx}[tanx](2cosx) - 2\frac{d}{dx}[cosx](tanx)}{(2cosx)^2}[/tex]
- [Derivative] Evaluate [Trig Derivatives]: [tex]\displaystyle \frac{d}{dx} = \frac{sec^2x(2cosx) - (-2sinx)(tanx)}{(2cosx)^2}[/tex]
- [Derivative] Evaluate exponents: [tex]\displaystyle \frac{d}{dx} = \frac{sec^2x(2cosx) - (-2sinx)(tanx)}{4cos^2x}[/tex]
- [Derivative] Multiply: [tex]\displaystyle \frac{d}{dx} = \frac{2secx + \frac{2sin^2x}{cosx}}{4cos^2x}[/tex]
- [Derivative] Add: [tex]\displaystyle \frac{d}{dx} = \frac{\frac{2sin^2x + 2}{cosx}}{4cos^2x}[/tex]
- [Derivative] Divide: [tex]\displaystyle \frac{d}{dx} = \frac{sin^2x + 1}{2cos^3x}[/tex]
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Derivatives
Book: College Calculus 10e
