2. **if y = xy + x² + 1, then when x = -1, dy/dx is (a) 1/2 (b) -1/2 (c) -1 (d) -2 (e) nonexistent

Explanation:

Step1: Differentiate both sides

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

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

Rearrange the equation $\frac{dy}{dx}-x\frac{dy}{dx}=y + 2x$. Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(1 - x)=y + 2x$. Then $\frac{dy}{dx}=\frac{y + 2x}{1 - x}$.

Step3: Find $y$ when $x=-1$

Substitute $x = - 1$ into the original equation $y=xy+x^{2}+1$. We get $y=-y + 1+1$. Add $y$ to both sides: $2y=2$, so $y = 1$.

Step4: Substitute $x$ and $y$ values

Substitute $x=-1$ and $y = 1$ into $\frac{dy}{dx}=\frac{y + 2x}{1 - x}$. Then $\frac{dy}{dx}=\frac{1+2\times(-1)}{1-(-1)}=\frac{1 - 2}{2}=-\frac{1}{2}$.

Answer:

B. $-\frac{1}{2}$