given the equation x^4y^9 - x^9y^4 = 2, find dy/dx by implicit differentiation.

Explanation:

Step1: Differentiate both sides

Differentiate $x^{4}y^{9}-x^{9}y^{4}=2$ with respect to $x$. Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, we have: $\frac{d}{dx}(x^{4}y^{9})-\frac{d}{dx}(x^{9}y^{4})=\frac{d}{dx}(2)$. For $\frac{d}{dx}(x^{4}y^{9})$, let $u = x^{4}$ and $v = y^{9}$. Then $\frac{d}{dx}(x^{4}y^{9})=4x^{3}y^{9}+9x^{4}y^{8}\frac{dy}{dx}$. For $\frac{d}{dx}(x^{9}y^{4})$, let $u = x^{9}$ and $v = y^{4}$. Then $\frac{d}{dx}(x^{9}y^{4})=9x^{8}y^{4}+4x^{9}y^{3}\frac{dy}{dx}$. And $\frac{d}{dx}(2) = 0$. So, $4x^{3}y^{9}+9x^{4}y^{8}\frac{dy}{dx}-(9x^{8}y^{4}+4x^{9}y^{3}\frac{dy}{dx})=0$.

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

Expand the left - hand side: $4x^{3}y^{9}+9x^{4}y^{8}\frac{dy}{dx}-9x^{8}y^{4}-4x^{9}y^{3}\frac{dy}{dx}=0$. Group the terms with $\frac{dy}{dx}$ together: $9x^{4}y^{8}\frac{dy}{dx}-4x^{9}y^{3}\frac{dy}{dx}=9x^{8}y^{4}-4x^{3}y^{9}$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(9x^{4}y^{8}-4x^{9}y^{3})=9x^{8}y^{4}-4x^{3}y^{9}$. Then $\frac{dy}{dx}=\frac{9x^{8}y^{4}-4x^{3}y^{9}}{9x^{4}y^{8}-4x^{9}y^{3}}$.

Answer:

$\frac{9x^{8}y^{4}-4x^{3}y^{9}}{9x^{4}y^{8}-4x^{9}y^{3}}$