if (3x^{2}+2xy + y^{2}=1), then (\frac{dy}{dx}=) \na (-\frac{3x + y}{y}) \nb (-\frac{3x + y}{x + y}) \nc…

Explanation:

Step1: Differentiate each term

Differentiate $3x^{2}+2xy + y^{2}=1$ with respect to $x$. The derivative of $3x^{2}$ is $6x$ (using power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$). For $2xy$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = 2x$ and $v = y$. So $\frac{d}{dx}(2xy)=2y + 2x\frac{dy}{dx}$. The derivative of $y^{2}$ with respect to $x$ is $2y\frac{dy}{dx}$ (using chain - rule $\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}$), and the derivative of the constant $1$ is $0$. So we have $6x+2y + 2x\frac{dy}{dx}+2y\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ together: $2x\frac{dy}{dx}+2y\frac{dy}{dx}=-6x - 2y$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2x + 2y)=-6x - 2y$. Then $\frac{dy}{dx}=\frac{-6x - 2y}{2x + 2y}=-\frac{3x + y}{x + y}$.

Answer:

B. $-\frac{3x + y}{x + y}$