x + 2xy - y^2 = 2\nfind the value of dy/dx at the point (2, 4).\nchoose 1 answer:\na 3/2\nb -9/4\nc 9/4\nd 1/2

Explanation:

Step1: Differentiate both sides

Differentiate $x + 2xy - y^{2}=2$ with respect to $x$. Using sum - rule and product - rule and chain - rule: The derivative of $x$ with respect to $x$ is $1$. For $2xy$, by product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 2x$ and $v = y$, we have $(2xy)^\prime=2y + 2x\frac{dy}{dx}$. For $-y^{2}$, by chain - rule, its derivative is $-2y\frac{dy}{dx}$. The derivative of the right - hand side (a constant 2) is 0. So, $1+2y + 2x\frac{dy}{dx}-2y\frac{dy}{dx}=0$.

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

Rearrange the terms: $2x\frac{dy}{dx}-2y\frac{dy}{dx}=-1 - 2y$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2x - 2y)=-1 - 2y$. Then $\frac{dy}{dx}=\frac{-1 - 2y}{2x - 2y}$.

Step3: Substitute the point $(2,4)$

Substitute $x = 2$ and $y = 4$ into $\frac{dy}{dx}=\frac{-1 - 2y}{2x - 2y}$. $\frac{dy}{dx}=\frac{-1-2\times4}{2\times2 - 2\times4}=\frac{-1 - 8}{4 - 8}=\frac{-9}{-4}=\frac{9}{4}$.

Answer:

C. $\frac{9}{4}$