Mathematics, 17.07.2021 05:10 123abc321cbaabc

Help with b please. thank you

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Answer from: gracethegreat1

See explanation.

General Formulas and Concepts:

Algebra I

Terms/CoefficientsFactoring

Algebra II

Polynomial Long Division

Pre-Calculus

Parametrics

Calculus

Differentiation

DerivativesDerivative Notation

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

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

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

Parametric Differentiation: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle x = 2t - \frac{1}{t}$

$\displaystyle y = t + \frac{4}{t}$

Step 2: Find Derivative

[x] Differentiate [Basic Power Rule and Quotient Rule]: $\displaystyle \frac{dx}{dt} = 2 + \frac{1}{t^2}$ [y] Differentiate [Basic Power Rule and Quotient Rule]: $\displaystyle \frac{dy}{dt} = 1 - \frac{4}{t^2}$ Substitute in variables [Parametric Derivative]: $\displaystyle \frac{dy}{dx} = \frac{1 - \frac{4}{t^2}}{2 + \frac{1}{t^2}}$ [Parametric Derivative] Simplify: $\displaystyle \frac{dy}{dx} = \frac{t^2 - 4}{2t^2 + 1}$ [Parametric Derivative] Polynomial Long Division: $\displaystyle \frac{dy}{dx} = \frac{1}{2} - \frac{7}{2(2t^2 - 1)}$ [Parametric Derivative] Factor: $\displaystyle \frac{dy}{dx} = \frac{1}{2} \bigg( 1 - \frac{9}{2t^2 + 1} \bigg)$

Here we see that if we increase our values for t, our derivative would get smaller and smaller. Hence $\displaystyle \frac{dy}{dx} < \frac{1}{2}$ .

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametrics

Book: College Calculus 10e

Answer from: Quest

answer: i oofed

step-by-step explanation:

oofed=wow

Answer from: Quest

I’m not sure i’m sorry

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Help with b please. thank you

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