Question

homework7: problem 2
(2 points)
for the equation given below, evaluate $\frac{dy}{dx}$ at the point $(2,4)$.
$4x^{2}-xy + 4y^{3}=264$
$\frac{dy}{dx}$ at $(2,4)=$
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Explanation:

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

Differentiate \(4x^{2}-xy + 4y^{3}\) term - by - term.
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), \(\frac{d}{dx}(4x^{2})=8x\).
For the term \(-xy\), use the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u=-x\) and \(v = y\). So \(\frac{d}{dx}(-xy)=-y - x\frac{dy}{dx}\).
For the term \(4y^{3}\), use the chain rule \(\frac{d}{dx}(f(g(x)))=f^\prime(g(x))\cdot g^\prime(x)\). Let \(u = y\), then \(\frac{d}{dx}(4y^{3})=12y^{2}\frac{dy}{dx}\).
Differentiating the right - hand side \(\frac{d}{dx}(264) = 0\).
So, \(8x-y - x\frac{dy}{dx}+12y^{2}\frac{dy}{dx}=0\).

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

Group the terms with \(\frac{dy}{dx}\) together:
\((-x + 12y^{2})\frac{dy}{dx}=y - 8x\).
Then \(\frac{dy}{dx}=\frac{y - 8x}{-x + 12y^{2}}\).

Step3: Substitute \(x = 2\) and \(y = 4\)

Substitute \(x = 2\) and \(y = 4\) into \(\frac{dy}{dx}=\frac{y - 8x}{-x + 12y^{2}}\).
\(\frac{dy}{dx}=\frac{4-8\times2}{-2 + 12\times4^{2}}=\frac{4 - 16}{-2+192}=\frac{-12}{190}=-\frac{6}{95}\).

Answer:

\(-\frac{6}{95}\)