Question

question number 7. (4.00 points)
find \\(\frac{dy}{dx}\\) at \\(x = 0\\) given \\(y = \frac{1}{-3 + u^2}\\) and \\(u = 3x + 1\\).
\\(\circ \frac{dy}{dx} = 0\\)
\\(\circ \frac{dy}{dx} = -\frac{3}{2}\\)
\\(\circ \frac{dy}{dx} = -\frac{5}{2}\\)
\\(\circ \frac{dy}{dx} = -3\\)
\\(\circ \frac{dy}{dx} = -2\\)

Explanation:

Step1: Apply Chain Rule

The chain rule states that \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). First, find \(\frac{dy}{du}\) and \(\frac{du}{dx}\).

Given \(y = \frac{1}{u^{2}-3}=(u^{2}-3)^{-1}\), differentiate with respect to \(u\):
\(\frac{dy}{du}=-1\cdot(u^{2}-3)^{-2}\cdot2u=\frac{-2u}{(u^{2}-3)^{2}}\)

Given \(u = 3x + 1\), differentiate with respect to \(x\):
\(\frac{du}{dx}=3\)

Step2: Substitute \(u\) and \(x = 0\)

First, find \(u\) when \(x = 0\): \(u=3(0)+1 = 1\)

Now substitute \(u = 1\) into \(\frac{dy}{du}\):
\(\frac{dy}{du}\bigg|_{u = 1}=\frac{-2(1)}{(1^{2}-3)^{2}}=\frac{-2}{(-2)^{2}}=\frac{-2}{4}=-\frac{1}{2}\)

Step3: Calculate \(\frac{dy}{dx}\)

Using the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), substitute \(\frac{dy}{du}=-\frac{1}{2}\) and \(\frac{du}{dx}=3\):
\(\frac{dy}{dx}=-\frac{1}{2}\cdot3=-\frac{3}{2}\)

Answer:

\(\boldsymbol{\frac{dy}{dx}=-\frac{3}{2}}\) (corresponding to the option \(\frac{dy}{dx}=-\frac{3}{2}\))