Problem
Let $\ln (x y)+y^{9}=x^{2}+3$. Find $\frac{d y}{d x}$. \[ \frac{d y}{d x}= \]
Solution
Step 1 ：\[\frac{d}{dx} (\ln(xy) + y^9) = \frac{d}{dx} (x^2 + 3)\]
Step 2 ：\[\frac{1}{xy}(y + x\frac{dy}{dx}) + 9y^8\frac{dy}{dx} = 2x\]
Step 3 ：\[\frac{dy}{dx}(x/y + 9y^8) = 2x - 1/x\]
Step 4 ：\[\frac{dy}{dx} = \frac{2x - 1/x}{x/y + 9y^8}\]
Step 5 ：\[\boxed{\frac{dy}{dx} = \frac{2x - 1/x}{x/y + 9y^8}}\]
