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

Explanation:

Step1: Differentiate both sides with respect to x

Differentiate $x + 2xy - y^{2}=2$ term - by - term. The derivative of $x$ with respect to $x$ is $1$. For the term $2xy$, use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 2x$ and $v = y$. So its derivative is $2y + 2x\frac{dy}{dx}$. For the term $-y^{2}$, use the chain - rule. Its derivative is $-2y\frac{dy}{dx}$. The derivative of the constant $2$ is $0$. So, $1+2y + 2x\frac{dy}{dx}-2y\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$: [ \begin{align*} 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}(2x - 2y)&=-1 - 2y\ \frac{dy}{dx}&=\frac{-1 - 2y}{2x - 2y} \end{align*} ]

Step3: Substitute $x = 1$ and $y = 1$

Substitute $x = 1$ and $y = 1$ into the expression for $\frac{dy}{dx}$: [ \begin{align*} \frac{dy}{dx}&=\frac{-1-2\times1}{2\times1 - 2\times1}\ &=\frac{-1 - 2}{2 - 2}\ &=\text{nonexistent} \end{align*} ]

Answer:

E. nonexistent