Question

use implicit differentiation to find $\frac{dy}{dx}$ for the following equation $7x^{3}+4y^{3}=11xy$

Explanation:

Step1: Differentiate both sides

Differentiate $7x^{3}+4y^{3}=11xy$ with respect to $x$.
The derivative of $7x^{3}$ using the power - rule is $21x^{2}$.
The derivative of $4y^{3}$ with respect to $x$ is $12y^{2}\frac{dy}{dx}$ (by the chain - rule).
The derivative of $11xy$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 11x$ and $v = y$ is $11y + 11x\frac{dy}{dx}$.
So, $21x^{2}+12y^{2}\frac{dy}{dx}=11y + 11x\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side:
$12y^{2}\frac{dy}{dx}-11x\frac{dy}{dx}=11y - 21x^{2}$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(12y^{2}-11x)=11y - 21x^{2}$.

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

$\frac{dy}{dx}=\frac{11y - 21x^{2}}{12y^{2}-11x}$.

Answer:

$\frac{11y - 21x^{2}}{12y^{2}-11x}$