Question

use implicit differentiation to find $\frac{dy}{dx}$ for the following equation.
$4x^{5}+3y^{5}=7xy$
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $4x^{5}+3y^{5}=7xy$ with respect to $x$. The derivative of $4x^{5}$ using the power - rule is $20x^{4}$. For $3y^{5}$, by the chain - rule, it is $15y^{4}\frac{dy}{dx}$. For $7xy$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 7x$ and $v = y$, we get $7y + 7x\frac{dy}{dx}$. So, $20x^{4}+15y^{4}\frac{dy}{dx}=7y + 7x\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side: $15y^{4}\frac{dy}{dx}-7x\frac{dy}{dx}=7y - 20x^{4}$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(15y^{4}-7x)=7y - 20x^{4}$.
Then solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{7y - 20x^{4}}{15y^{4}-7x}$.

Answer:

$\frac{7y - 20x^{4}}{15y^{4}-7x}$