Having trouble with trying to get any of the answers. I don’t know how to get 12? **Problem:**If \( y = \frac{4x}{x - 3} \), then \(\frac{dy}{dx} = \)- \( \frac{-12}{(x-3)^2} \)- \( \frac{12}{(x-3)} \)**Explanation:**This is a calculus problem where we need to find the derivative \(\frac{dy}{dx}\) of the function \( y = \frac{4x}{x - 3} \).### Steps to Solve:1. **Apply the Quotient Rule:** The quotient rule states that for a function \(\frac{u}{v}\), the derivative is given by: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] where \(u = 4x\) and \(v = x - 3\).2. **Find Derivatives of \(u\) and \(v\):** - \(u' = 4\) - \(v' = 1\)3. **Substitute in the Quotient Rule:** \[ \frac{dy}{dx} = \frac{(4)(x-3) - (4x)(1)}{(x-3)^2} \] \[ = \frac{4x - 12 - 4x}{(x-3)^2} \] \[ = \frac{-12}{(x-3)^2} \]Thus, the correct derivative is \( \frac{-12}{(x-3)^2} \).**Correct Answer:**- \( \frac{-12}{(x-3)^2} \)

Answered: If y = O 2-3' then -12 (2-3)² 12 dy dx

If y = O 2-3' then -12 (2-3)² 12 dy dx

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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