Let $\ln (x y)+y^{9}=x^{2}+3$. Find $\frac{d y}{d x}$.
\[
\frac{d y}{d x}=
\]

Answer

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Answer

\[\boxed{\frac{dy}{dx} = \frac{2x - 1/x}{x/y + 9y^8}}\]

Steps

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 ：\[\boxed{\frac{dy}{dx} = \frac{2x - 1/x}{x/y + 9y^8}}\]