Question

use implicit differentiation to find $\frac{dy}{dx}$
$x^{4}+y^{4}=12xy$
$\frac{dy}{dx}=\frac{- 14xy - 6y^{2}}{\text{blank}}$

Explanation:

Step1: Differentiate both sides

Differentiate $x^{4}+y^{4}=12xy$ with respect to $x$. The derivative of $x^{4}$ with respect to $x$ is $4x^{3}$ by the power - rule. For $y^{4}$, using the chain - rule, we get $4y^{3}\frac{dy}{dx}$. For the right - hand side, using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 12x$ and $v = y$, we have $12y + 12x\frac{dy}{dx}$. So, $4x^{3}+4y^{3}\frac{dy}{dx}=12y + 12x\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side: $4y^{3}\frac{dy}{dx}-12x\frac{dy}{dx}=12y - 4x^{3}$.

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

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(4y^{3}-12x)=12y - 4x^{3}$.

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

Divide both sides by $4y^{3}-12x$ to get $\frac{dy}{dx}=\frac{12y - 4x^{3}}{4y^{3}-12x}=\frac{3y - x^{3}}{y^{3}-3x}$.

Answer:

$\frac{3y - x^{3}}{y^{3}-3x}$