Implicit Differentiation 19F82D

1. **State the problem:** Find $\frac{dx}{dy}$ given the implicit equation $$x = \sqrt{xy} + 3$$. 2. **Recall the formula and rules:** We will differentiate both sides of the equation with respect to $y$ using implicit differentiation. - Remember that $x$ is a function of $y$, so when differentiating $x$ with respect to $y$, use $\frac{dx}{dy}$. - Use the chain rule and product rule as needed. 3. **Rewrite the equation:** $$x = \sqrt{xy} + 3 = (xy)^{\frac{1}{2}} + 3$$ 4. **Differentiate both sides with respect to $y$:** $$\frac{d}{dy}[x] = \frac{d}{dy}[(xy)^{\frac{1}{2}} + 3]$$ This gives: $$\frac{dx}{dy} = \frac{1}{2}(xy)^{-\frac{1}{2}} \cdot \frac{d}{dy}[xy] + 0$$ 5. **Differentiate $xy$ with respect to $y$ using the product rule:** $$\frac{d}{dy}[xy] = x \cdot \frac{d}{dy}[y] + y \cdot \frac{dx}{dy} = x + y \frac{dx}{dy}$$ 6. **Substitute back:** $$\frac{dx}{dy} = \frac{1}{2}(xy)^{-\frac{1}{2}} (x + y \frac{dx}{dy})$$ 7. **Isolate $\frac{dx}{dy}$ terms:** $$\frac{dx}{dy} = \frac{1}{2}(xy)^{-\frac{1}{2}} x + \frac{1}{2}(xy)^{-\frac{1}{2}} y \frac{dx}{dy}$$ Bring the $\frac{dx}{dy}$ terms to one side: $$\frac{dx}{dy} - \frac{1}{2}(xy)^{-\frac{1}{2}} y \frac{dx}{dy} = \frac{1}{2}(xy)^{-\frac{1}{2}} x$$ Factor out $\frac{dx}{dy}$: $$\frac{dx}{dy} \left(1 - \frac{1}{2} y (xy)^{-\frac{1}{2}} \right) = \frac{1}{2} x (xy)^{-\frac{1}{2}}$$ 8. **Solve for $\frac{dx}{dy}$:** $$\frac{dx}{dy} = \frac{\frac{1}{2} x (xy)^{-\frac{1}{2}}}{1 - \frac{1}{2} y (xy)^{-\frac{1}{2}}}$$ 9. **Simplify the expression:** Rewrite $(xy)^{-\frac{1}{2}} = \frac{1}{\sqrt{xy}}$: $$\frac{dx}{dy} = \frac{\frac{1}{2} x \frac{1}{\sqrt{xy}}}{1 - \frac{1}{2} y \frac{1}{\sqrt{xy}}} = \frac{\frac{x}{2 \sqrt{xy}}}{1 - \frac{y}{2 \sqrt{xy}}}$$ Multiply numerator and denominator by $2 \sqrt{xy}$ to clear fractions: $$\frac{dx}{dy} = \frac{x}{2 \sqrt{xy} - y}$$ **Final answer:** $$\boxed{\frac{dx}{dy} = \frac{x}{2 \sqrt{xy} - y}}$$