Question

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

Explanation:

Step1: Differentiate both sides

Differentiate $x + 2xy - y^{2}=2$ with respect to $x$. Using sum - rule and product - rule. The derivative of $x$ with respect to $x$ is $1$, for $2xy$ using product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 2x$ and $v = y$, we get $2y + 2x\frac{dy}{dx}$, and the derivative of $-y^{2}$ with respect to $x$ is $-2y\frac{dy}{dx}$. 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 the point $(1,1)$

Substitute $x = 1$ and $y = 1$ into $\frac{dy}{dx}=\frac{-1 - 2y}{2x - 2y}$. We have $\frac{-1-2\times1}{2\times1 - 2\times1}=\frac{-3}{0}$, which is undefined.

Answer:

Undefined