Question

find $\frac{dy}{dx}$ of $x^{2}+9y^{2}-4x + 3y=0$

Explanation:

Step1: Differentiate each term

Differentiate $x^{2}+9y^{2}-4x + 3y=0$ term - by - term with respect to $x$.
The derivative of $x^{2}$ with respect to $x$ is $2x$ (using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$).
The derivative of $9y^{2}$ with respect to $x$ is $18y\frac{dy}{dx}$ (using the chain - rule $\frac{d}{dx}(u^{2}) = 2u\frac{du}{dx}$, where $u = y$).
The derivative of $-4x$ with respect to $x$ is $-4$.
The derivative of $3y$ with respect to $x$ is $3\frac{dy}{dx}$. So we have:
$2x+18y\frac{dy}{dx}-4 + 3\frac{dy}{dx}=0$

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

Group the terms with $\frac{dy}{dx}$ on one side:
$18y\frac{dy}{dx}+3\frac{dy}{dx}=4 - 2x$
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(18y + 3)=4 - 2x$

Step3: Solve for $\frac{dy}{dx}$

Divide both sides by $18y + 3$:
$\frac{dy}{dx}=\frac{4 - 2x}{18y+3}=\frac{2(2 - x)}{3(6y + 1)}$

Answer:

$\frac{2(2 - x)}{3(6y + 1)}$