find dy.\n2y^{\frac{9}{2}}+xy - x = 0\ndy = ( ) dx

Explanation:

Step1: Differentiate each term

Differentiate $2y^{\frac{9}{2}}+xy - x=0$ term - by - term with respect to $x$. Using the power rule $\frac{d}{dx}(u^n)=nu^{n - 1}\frac{du}{dx}$ for $2y^{\frac{9}{2}}$, we get $2\times\frac{9}{2}y^{\frac{9}{2}-1}\frac{dy}{dx}=9y^{\frac{7}{2}}\frac{dy}{dx}$. Using the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ for $xy$, we get $x\frac{dy}{dx}+y$. The derivative of $-x$ with respect to $x$ is $-1$. So, $9y^{\frac{7}{2}}\frac{dy}{dx}+x\frac{dy}{dx}+y - 1 = 0$.

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

Factor out $\frac{dy}{dx}$ from the terms involving $\frac{dy}{dx}$: $\frac{dy}{dx}(9y^{\frac{7}{2}}+x)=1 - y$. Then, $\frac{dy}{dx}=\frac{1 - y}{9y^{\frac{7}{2}}+x}$. Since $dy=\frac{dy}{dx}dx$, we have $dy=\frac{1 - y}{9y^{\frac{7}{2}}+x}dx$.

Answer:

$\frac{1 - y}{9y^{\frac{7}{2}}+x}$