Question

find $\frac{dy}{dx}$ by implicit differentiation.
$7x^{2}+9xy - y^{2}=7$
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate each term

Differentiate $7x^{2}$, $9xy$, $-y^{2}$ and $7$ with respect to $x$.
The derivative of $7x^{2}$ with respect to $x$ is $14x$ (using power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$).
For $9xy$, use the product - rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = 9x$ and $v = y$. So $\frac{d}{dx}(9xy)=9y + 9x\frac{dy}{dx}$.
For $-y^{2}$, use the chain - rule. Let $u = y$, then $\frac{d}{dx}(-y^{2})=-2y\frac{dy}{dx}$.
The derivative of the constant $7$ with respect to $x$ is $0$.
So, $14x+9y + 9x\frac{dy}{dx}-2y\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ on one side:
$9x\frac{dy}{dx}-2y\frac{dy}{dx}=-14x - 9y$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(9x - 2y)=-14x - 9y$.
Then $\frac{dy}{dx}=\frac{-14x - 9y}{9x - 2y}$.

Answer:

$\frac{-14x - 9y}{9x - 2y}$