Question

use implicit differentiation to find $\frac{dy}{dx}$. $x^{5}+y^{5}=20xy$ $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $x^{5}+y^{5}=20xy$ with respect to $x$. The derivative of $x^{5}$ with respect to $x$ is $5x^{4}$ by the power - rule. For $y^{5}$, using the chain - rule, we get $5y^{4}\frac{dy}{dx}$. For the right - hand side, using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 20x$ and $v = y$, we have $20y + 20x\frac{dy}{dx}$. So, $5x^{4}+5y^{4}\frac{dy}{dx}=20y + 20x\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side: $5y^{4}\frac{dy}{dx}-20x\frac{dy}{dx}=20y - 5x^{4}$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(5y^{4}-20x)=20y - 5x^{4}$.

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

Divide both sides by $5y^{4}-20x$ to get $\frac{dy}{dx}=\frac{20y - 5x^{4}}{5y^{4}-20x}=\frac{4y - x^{4}}{y^{4}-4x}$.

Answer:

$\frac{4y - x^{4}}{y^{4}-4x}$