Question

use implicit differentiation to find $\frac{dy}{dx}$. 7xy + y² = 6x + y $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate each term

Differentiate $7xy$ using product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 7x$ and $v = y$. So, $(7xy)^\prime=7y + 7x\frac{dy}{dx}$. Differentiate $y^{2}$ using chain - rule, $(y^{2})^\prime = 2y\frac{dy}{dx}$. Differentiate $6x$ to get $6$ and differentiate $y$ to get $\frac{dy}{dx}$. The equation $7xy + y^{2}=6x + y$ becomes $7y+7x\frac{dy}{dx}+2y\frac{dy}{dx}=6+\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side: $7x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=6 - 7y$.

Step3: Factor out $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(7x + 2y-1)=6 - 7y$.

Step4: Solve for $\frac{dy}{dx}$

Divide both sides by $7x + 2y - 1$ to get $\frac{dy}{dx}=\frac{6 - 7y}{7x+2y - 1}$.

Answer:

$\frac{6 - 7y}{7x+2y - 1}$