Mathematics: Find the derivative: y={ (3x+1)cos(2x) } / e^2x

Find the derivative: y={ (3x+1)cos(2x) } / e^2x - id312211383109, orlandoreypot2491

Ответ:

m4167281

25.03.2021

Mathematics

$\displaystyle y' = \frac{3cos(2x) -2(3x + 1)[sin(2x) + cos(2x)]}{e^{2x}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Algebra I

FactoringExponential Rule [Dividing]: $\displaystyle \frac{b^m}{b^n} = b^{m - n}$ Exponential Rule [Powering]: $\displaystyle (b^m)^n = b^{m \cdot n}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

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: $\displaystyle \frac{d}{dx}[cos(u)] = -u'sin(u)$

eˣ Derivative: $\displaystyle \frac{d}{dx}[e^u] = u'e^u$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{(3x + 1)cos(2x)}{e^{2x}}$

Step 2: Differentiate

[Derivative] Quotient Rule: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - \frac{d}{dx}[e^{2x}](3x + 1)cos(2x)}{(e^{2x})^2}$ [Derivative] [Fraction - Numerator] eˣ derivative: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{(e^{2x})^2}$ [Derivative] [Fraction - Denominator] Exponential Rule - Powering: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$ [Derivative] [Fraction - Numerator] Product Rule: $\displaystyle y' = \frac{[\frac{d}{dx}[3x + 1]cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$ [Derivative] [Fraction - Numerator] [Brackets] Basic Power Rule: $\displaystyle y' = \frac{[(1 \cdot 3x^{1 - 1})cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$ [Derivative] [Fraction - Numerator] [Brackets] (Parenthesis) Simplify: $\displaystyle y' = \frac{[3cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$ [Derivative] [Fraction - Numerator] [Brackets] Trig derivative: $\displaystyle y' = \frac{[3cos(2x) -2sin(2x)(3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$ [Derivative] [Fraction - Numerator] Factor: $\displaystyle y' = \frac{e^{2x}[(3cos(2x) -2sin(2x)(3x + 1)) - 2(3x + 1)cos(2x)]}{e^{4x}}$ [Derivative] [Fraction] Simplify [Exponential Rule - Dividing]: $\displaystyle y' = \frac{3cos(2x) -2sin(2x)(3x + 1) - 2(3x + 1)cos(2x)}{e^{2x}}$ [Derivative] [Fraction - Numerator] Factor: $\displaystyle y' = \frac{3cos(2x) -2(3x + 1)[sin(2x) + cos(2x)]}{e^{2x}}$

Topic: AP Calculus AB/BC

Unit: Derivatives

Book: College Calculus 10e

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Ответ:

dylanolmeda9

19.02.2023

Mathematics

1,800 ft

Step-by-step explanation:

Rebecca will walk in the pattern of a regular polygon with 'n' sides and internal angles of 160 degrees (180 at a straight line - 20 degrees from each turn).

The equation that describes the internal angle of a regular polygon is:

$A = \frac{(n-2)*180}{n}$

For A = 160 degrees:

$160n=(n-2)*180\\(180-160)n=360\\n=18\ sides$

If each side is 100 ft long, the total distance that Rebecca has walked is:

$d = 18*100\\d=1,800\ ft$

She walked 1,800 ft.

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