given x sin y - 6y^5 - 2 = 8x, use implicit differentiation to find dy/dx. dy/dx = □

Explanation:

Step1: Differentiate each term

Differentiate $x\sin y$, $-6y^{5}$, $- 2$ and $8x$ with respect to $x$. For $x\sin y$, use the 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 $-6y^{5}$ is $-30y^{4}\frac{dy}{dx}$, the derivative of $-2$ is $0$, and the derivative of $8x$ is $8$. So we have $\sin y+x\cos y\frac{dy}{dx}-30y^{4}\frac{dy}{dx}=8$.

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

Group the terms with $\frac{dy}{dx}$ on one side: $x\cos y\frac{dy}{dx}-30y^{4}\frac{dy}{dx}=8 - \sin y$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x\cos y - 30y^{4})=8-\sin y$.

Step3: Solve for $\frac{dy}{dx}$

Divide both sides by $x\cos y - 30y^{4}$ to get $\frac{dy}{dx}=\frac{8 - \sin y}{x\cos y-30y^{4}}$.

Answer:

$\frac{8 - \sin y}{x\cos y-30y^{4}}$