Question

homework7: problem 1
(2 points)
for the equation given below, evaluate $\frac{dy}{dx}$ at the point $(-1, -2)$.
$4y^{3}+y^{2}-4x^{2}=-32$
$\frac{dy}{dx}$ at $(-1, -2)=$
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Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(4y^{3}+y^{2}-4x^{2}=-32\) term - by - term.
Using the chain rule \((u^{n})^\prime = nu^{n - 1}u^\prime\) (where \(u = y\) and \(u^\prime=\frac{dy}{dx}\)) and \((x^{n})^\prime=nx^{n - 1}\):
\(\frac{d}{dx}(4y^{3})+\frac{d}{dx}(y^{2})-\frac{d}{dx}(4x^{2})=\frac{d}{dx}(-32)\)
\(4\times3y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-4\times2x = 0\)
\(12y^{2}\frac{dy}{dx}+2y\frac{dy}{dx}-8x = 0\)

Step2: Solve for \(\frac{dy}{dx}\)

Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(12y^{2}+2y)=8x\)
Then \(\frac{dy}{dx}=\frac{8x}{12y^{2}+2y}=\frac{4x}{6y^{2}+y}\)

Step3: Substitute \(x=-1\) and \(y = - 2\)

\(\frac{dy}{dx}\mid_{x=-1,y = - 2}=\frac{4\times(-1)}{6\times(-2)^{2}+(-2)}\)
First, calculate the denominator: \(6\times(-2)^{2}+(-2)=6\times4-2=24 - 2=22\)
Then \(\frac{4\times(-1)}{22}=-\frac{2}{11}\)

Answer:

\(-\frac{2}{11}\)