y² - x²y + 3x³ = 4\nfind dy/dx.\nchoose 1 answer:\n(a) (2x - 2y)/(27x²)\n(b) (-27x²)/(2y - 2x)\n(c) (2xy…

Explanation:

Step1: Differentiate each term

Differentiate $y^{2}-x^{2}y + 3x^{3}=4$ term - by - term with respect to $x$. For $y^{2}$, using the chain - rule, we have $\frac{d(y^{2})}{dx}=2y\frac{dy}{dx}$. For $-x^{2}y$, using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=-x^{2}$ and $v = y$, we get $\frac{d(-x^{2}y)}{dx}=-2xy - x^{2}\frac{dy}{dx}$. For $3x^{3}$, $\frac{d(3x^{3})}{dx}=9x^{2}$, and $\frac{d(4)}{dx}=0$. So, $2y\frac{dy}{dx}-2xy - x^{2}\frac{dy}{dx}+9x^{2}=0$.

Step2: Isolate $\frac{dy}{dx}$

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

Answer:

C. $\frac{2xy - 9x^{2}}{2y - x^{2}}$