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Mathematics, 07.03.2021 05:10 limelight11

Differentiate (Tanx)÷(2cosx)

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Answer from: bsheepicornozj0gc

$\displaystyle \frac{d}{dx} = \frac{sin^2x + 1}{2cos^3x}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Pre-Calculus

Trigonometric Functions

Calculus

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Trig Derivative: $\displaystyle \frac{d}{dx}[tanu] = u'sec^2u$

Trig Derivative:

$\displaystyle \frac{d}{dx}[cosu] = -u'sinu$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{tanx}{2cosx}$

Step 2: Differentiate

[Derivative] Quotient Rule: $\displaystyle \frac{d}{dx} = \frac{\frac{d}{dx}[tanx](2cosx) - \frac{d}{dx}[2cosx](tanx)}{(2cosx)^2}$ [Derivative] Simplify [Derivative Property - Multiplied Constant]: $\displaystyle \frac{d}{dx} = \frac{\frac{d}{dx}[tanx](2cosx) - 2\frac{d}{dx}[cosx](tanx)}{(2cosx)^2}$ [Derivative] Evaluate [Trig Derivatives]: $\displaystyle \frac{d}{dx} = \frac{sec^2x(2cosx) - (-2sinx)(tanx)}{(2cosx)^2}$ [Derivative] Evaluate exponents: $\displaystyle \frac{d}{dx} = \frac{sec^2x(2cosx) - (-2sinx)(tanx)}{4cos^2x}$ [Derivative] Multiply: $\displaystyle \frac{d}{dx} = \frac{2secx + \frac{2sin^2x}{cosx}}{4cos^2x}$ [Derivative] Add: $\displaystyle \frac{d}{dx} = \frac{\frac{2sin^2x + 2}{cosx}}{4cos^2x}$ [Derivative] Divide: $\displaystyle \frac{d}{dx} = \frac{sin^2x + 1}{2cos^3x}$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

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Answer from: Quest

the answer is c.

step-by-step explanation:

ansver

Answer from: Quest

126 maybe. sorry

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