if $x^{2}+xy - 3y=3$, then at the point $(2,1),\frac{dy}{dx}=$\nq4 slope of an implicit function at a point

Explanation:

Step1: Differentiate both sides with respect to x

Differentiate $x^{2}+xy - 3y=3$ term - by - term. The derivative of $x^{2}$ with respect to $x$ is $2x$ using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$. For the term $xy$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = x$ and $v = y$. So $\frac{d}{dx}(xy)=x\frac{dy}{dx}+y$. The derivative of $-3y$ with respect to $x$ is $-3\frac{dy}{dx}$, and the derivative of the constant 3 is 0. We get $2x+x\frac{dy}{dx}+y - 3\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ on one side: $x\frac{dy}{dx}-3\frac{dy}{dx}=-2x - y$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x - 3)=-2x - y$. Then $\frac{dy}{dx}=\frac{-2x - y}{x - 3}$.

Step3: Substitute the point $(2,1)$

Substitute $x = 2$ and $y = 1$ into $\frac{dy}{dx}=\frac{-2x - y}{x - 3}$. $\frac{dy}{dx}\big|_{(2,1)}=\frac{-2\times2-1}{2 - 3}=\frac{-4 - 1}{-1}=5$.

Answer:

5