use implicit differentiation to find $\frac{dy}{dx}$ for the following equation.\n$3x^{3}+5y^{3}=8xy$

Explanation:

Step1: Differentiate both sides

Differentiate $3x^{3}+5y^{3}=8xy$ with respect to $x$. The derivative of $3x^{3}$ with respect to $x$ is $9x^{2}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$. For $5y^{3}$, by the chain - rule $\frac{d}{dx}(y^{3}) = 3y^{2}\frac{dy}{dx}$, so its derivative is $15y^{2}\frac{dy}{dx}$. For $8xy$, by the product - rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ where $u = 8x$ and $v = y$, we get $8y+8x\frac{dy}{dx}$. So, $9x^{2}+15y^{2}\frac{dy}{dx}=8y + 8x\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side: $15y^{2}\frac{dy}{dx}-8x\frac{dy}{dx}=8y - 9x^{2}$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(15y^{2}-8x)=8y - 9x^{2}$. Then, $\frac{dy}{dx}=\frac{8y - 9x^{2}}{15y^{2}-8x}$.

Answer:

$\frac{8y - 9x^{2}}{15y^{2}-8x}$