Question

homework7: problem 5
(2 points)
find \\( \frac{d y}{d x} \\) by implicit differentiation.
\\( 3 x^{3}+x^{2} y-x y^{3}=-2 \\)
answer: \\( \frac{d y}{d x}= \\)
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page generated october 19, 2025, 8:39:39 pm cdt
webwork \\( \odot 1996-2024 \\) | theme: math4_ttu | ww_version: 2.19 | pg_version 2.1
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Explanation:

Step1: Differentiate each term

Differentiate \(3x^{3}+x^{2}y - xy^{3}=-2\) term - by - term with respect to \(x\).
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), the product rule \(\frac{d}{dx}(uv)=u'v + uv'\) (where \(u\) and \(v\) are functions of \(x\)), and the chain rule \(\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}\) (since \(y\) is a function of \(x\)).

• \(\frac{d}{dx}(3x^{3})=9x^{2}\)
• For \(\frac{d}{dx}(x^{2}y)\), let \(u = x^{2}\) and \(v = y\). Then \(\frac{d}{dx}(x^{2}y)=2xy+x^{2}\frac{dy}{dx}\)
• For \(\frac{d}{dx}(xy^{3})\), let \(u = x\) and \(v = y^{3}\). Then \(\frac{d}{dx}(xy^{3})=y^{3}+3xy^{2}\frac{dy}{dx}\)
• \(\frac{d}{dx}(-2)=0\)

The differentiated equation is:
\(9x^{2}+2xy + x^{2}\frac{dy}{dx}-y^{3}-3xy^{2}\frac{dy}{dx}=0\)

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

Group the terms with \(\frac{dy}{dx}\) on one side:
\(x^{2}\frac{dy}{dx}-3xy^{2}\frac{dy}{dx}=y^{3}-9x^{2}-2xy\)
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(x^{2}-3xy^{2})=y^{3}-9x^{2}-2xy\)
Then \(\frac{dy}{dx}=\frac{y^{3}-9x^{2}-2xy}{x^{2}-3xy^{2}}\)

Answer:

\(\frac{y^{3}-9x^{2}-2xy}{x^{2}-3xy^{2}}\)