Question

use implicit differentiation to find $\frac{dy}{dx}$
$7x^{2}y + 6xy^{2}=1$
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate each term

Differentiate $7x^{2}y+6xy^{2}$ with respect to $x$ using product - rule $(uv)^\prime = u^\prime v+uv^\prime$.
For $7x^{2}y$: $(7x^{2}y)^\prime=7(2xy + x^{2}\frac{dy}{dx}) = 14xy+7x^{2}\frac{dy}{dx}$.
For $6xy^{2}$: $(6xy^{2})^\prime=6(y^{2}+2xy\frac{dy}{dx})=6y^{2}+12xy\frac{dy}{dx}$.
The derivative of the right - hand side, $\frac{d}{dx}(1) = 0$.
So, $14xy + 7x^{2}\frac{dy}{dx}+6y^{2}+12xy\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ on one side:
$7x^{2}\frac{dy}{dx}+12xy\frac{dy}{dx}=-14xy - 6y^{2}$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(7x^{2}+12xy)=-14xy - 6y^{2}$.
Then $\frac{dy}{dx}=\frac{-14xy - 6y^{2}}{7x^{2}+12xy}$.

Answer:

$\frac{-14xy - 6y^{2}}{7x^{2}+12xy}$