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2 years ago

Perform the division and leave the result in trigonometric form.

Mathematics

1 answer:

Anettt [7]2 years ago

8 0

Answer:

$\frac{1}{2}$ ( cos40° + isin40°)

Step-by-step explanation:

To divide

$\frac{r_{1(cosx_{1}+isinx_{1}) } }{r_{2}(cosx_{2}+isinx_{2}) }$

= $\frac{r_{1} }{r_{2} }$ [ cos(x₁ - x₂) + isin(x₁ - x₂)

Given

$\frac{2(cos90+isin90}{4(cos50+isin50)}$

= $\frac{2}{4}$ ( cos(90 - 50)° + isin(90 - 50)°

= $\frac{1}{2}$ ( cos40° + isin40° )

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DochEvi [55]

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

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring
Exponential 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) = cxⁿ
f’(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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3 years ago