given x sin y - 2y^4 - 2 = 9x, use implicit differentiation to find dy/dx. dy/dx = □

Explanation:

Step1: Differentiate both sides

Differentiate $x\sin y - 2y^{4}-2$ and $9x$ with respect to $x$. $\frac{d}{dx}(x\sin y - 2y^{4}-2)=\frac{d}{dx}(9x)$

Step2: Apply product - rule and chain - rule on left side

The product - rule for $\frac{d}{dx}(x\sin y)$ is $\sin y + x\cos y\frac{dy}{dx}$, and for $\frac{d}{dx}(-2y^{4})=-8y^{3}\frac{dy}{dx}$, and $\frac{d}{dx}(-2) = 0$. The right side $\frac{d}{dx}(9x)=9$. $\sin y+x\cos y\frac{dy}{dx}-8y^{3}\frac{dy}{dx}=9$

Step3: Isolate $\frac{dy}{dx}$ terms

$x\cos y\frac{dy}{dx}-8y^{3}\frac{dy}{dx}=9 - \sin y$

Step4: Factor out $\frac{dy}{dx}$

$\frac{dy}{dx}(x\cos y - 8y^{3})=9 - \sin y$

Step5: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{9 - \sin y}{x\cos y - 8y^{3}}$

Answer:

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