Question

2y^{2}-x^{2}+x^{3}y = 2
find \frac{dy}{dx}.
choose 1 answer:
a \frac{4y + x^{3}}{2x - 3x^{2}y}
b \frac{2x - 4y}{3x^{2}}
c 2x - 3x^{2}y

Explanation:

Step1: Differentiate both sides

Differentiate \(2y^{2}-x^{2}+x^{3}y = 2\) with respect to \(x\).
Using the chain - rule \((u^{n})^\prime=nu^{n - 1}u^\prime\) for \(2y^{2}\) (where \(u = y\)), the power - rule \((x^{n})^\prime=nx^{n-1}\) for \(-x^{2}\), and the product - rule \((uv)^\prime = u^\prime v+uv^\prime\) for \(x^{3}y\) (where \(u=x^{3}\), \(v = y\)).
\(\frac{d}{dx}(2y^{2})-\frac{d}{dx}(x^{2})+\frac{d}{dx}(x^{3}y)=\frac{d}{dx}(2)\)
\(4y\frac{dy}{dx}-2x+(3x^{2}y+x^{3}\frac{dy}{dx}) = 0\)

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

Group the terms with \(\frac{dy}{dx}\) on one side:
\(4y\frac{dy}{dx}+x^{3}\frac{dy}{dx}=2x - 3x^{2}y\)
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(4y + x^{3})=2x-3x^{2}y\)
Then \(\frac{dy}{dx}=\frac{2x - 3x^{2}y}{4y+x^{3}}\) (by dividing both sides by \(4y + x^{3}\))

Answer:

None of the options A, B, C are correct. The correct derivative \(\frac{dy}{dx}=\frac{2x - 3x^{2}y}{4y + x^{3}}\)