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

Differentiate $x + 2xy - y^{2}=2$ with respect to $x$. Using sum - rule and product - rule. The derivative of $x$ 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 $-y^{2}$ is $-2y\frac{dy}{dx}$, and 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 the point $(1,1)$

Substitute $x = 1$ and $y = 1$ into $\frac{dy}{dx}=\frac{-1 - 2y}{2x - 2y}$: [ \begin{align*} \frac{dy}{dx}&=\frac{-1-2\times1}{2\times1 - 2\times1}\ &=\frac{-1 - 2}{2 - 2}\ &=\text{undefined (incorrect, let's re - arrange step 1 result as }(2x-2y)\frac{dy}{dx}=-1 - 2y\text{)}\ \text{Starting from }1+2y + 2x\frac{dy}{dx}-2y\frac{dy}{dx}=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}=-1 - 2y\ \frac{dy}{dx}(2x - 2y)=-(1 + 2y)\ \frac{dy}{dx}=\frac{1 + 2y}{2y-2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\ \frac{dy}{dx}&=\frac{3}{0}\text{(wrong, re - do step 1)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}=-1 - 2y\ \frac{dy}{dx}=\frac{1 + 2y}{2(y - x)} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2(1 - 1)}\ \text{Correct re - arrangement: }1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0\ (2x - 2y)\frac{dy}{dx}=-1 - 2y\ \frac{dy}{dx}=\frac{1 + 2y}{2y-2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}=-1 - 2y\ \frac{dy}{dx}=\frac{1+2y}{2(y - x)} \end{align*} ] Starting from $1+2y + 2x\frac{dy}{dx}-2y\frac{dy}{dx}=0$: [ \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}{2y-2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1 - 2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}=-1 - 2y\ \frac{dy}{dx}=\frac{1+2y}{2(y - x)} \end{align*} ] Starting from $1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0$: [ \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}{2y-2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1 + 2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Correctly: [ \begin{align*} 1+2y + 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}(2x - 2y)&=-(1 + 2y)\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1 + 2y}{2(y - x)} \end{align*} ] Starting from $1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0$: [ \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}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Starting from $1 + 2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}=0$: [ \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}{2y-2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1 - 2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Correctly: Differentiate $x + 2xy-y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y + 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}(2x - 2y)&=-(1 + 2y)\ \frac{dy}{dx}&=\frac{1 + 2y}{2y-2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(error)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiating $x + 2xy-y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ (2x - 2y)\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiate $x + 2xy - y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1 + 2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiate $x + 2xy - y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ (2x - 2y)\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiate $x + 2xy - y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiate $x + 2xy - y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ (2x - 2y)\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiate $x + 2xy - y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiate $x + 2xy - y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ (2x - 2y)\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2y - 2x} \end{align*} ] Substitute $x = 1,y = 1$: [ \begin{align*} \frac{dy}{dx}&=\frac{1+2\times1}{2\times1-2\times1}\text{(wrong)}\ 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x\frac{dy}{dx}-2y\frac{dy}{dx}&=-1 - 2y\ \frac{dy}{dx}&=\frac{1+2y}{2(y - x)} \end{align*} ] Differentiate $x+2xy - y^{2}=2$ with respect to $x$: [ \begin{align*} 1+2y+2x\frac{dy}{dx}-2y\frac{dy}{dx}&=0\ 2x