given x sin y - 2y³ - 13 = 8x, use implicit differentiation to find dy/dx. dy/dx = □

Explanation:

Step1: Differentiate each term

Differentiate $x\sin y$, $-2y^{3}$, $- 13$ and $8x$ with respect to $x$. For $x\sin y$, use product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v=\sin y$. So its derivative is $\sin y+x\cos y\frac{dy}{dx}$. The derivative of $-2y^{3}$ is $-6y^{2}\frac{dy}{dx}$ (by chain - rule), the derivative of $-13$ is $0$, and the derivative of $8x$ is $8$. So we have $\sin y+x\cos y\frac{dy}{dx}-6y^{2}\frac{dy}{dx}-0 = 8$.

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

Group the terms with $\frac{dy}{dx}$ on one side: $x\cos y\frac{dy}{dx}-6y^{2}\frac{dy}{dx}=8 - \sin y$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x\cos y-6y^{2})=8 - \sin y$. Then $\frac{dy}{dx}=\frac{8 - \sin y}{x\cos y-6y^{2}}$.

Answer:

$\frac{8 - \sin y}{x\cos y-6y^{2}}$