Differentiate Wrt Y 8064Ba

1. **State the problem:** Differentiate the equation $$\frac{y}{x} = \sqrt{xy} + 3$$ with respect to $$y$$. 2. **Rewrite the equation:** $$\frac{y}{x} = (xy)^{\frac{1}{2}} + 3$$ 3. **Differentiate both sides with respect to $$y$$:** - Left side: $$\frac{d}{dy}\left(\frac{y}{x}\right) = \frac{1}{x}$$ since $$x$$ is treated as a constant. - Right side: Use the chain rule for $$ (xy)^{\frac{1}{2}} $$: $$\frac{d}{dy} (xy)^{\frac{1}{2}} = \frac{1}{2}(xy)^{-\frac{1}{2}} \cdot \frac{d}{dy}(xy)$$ Since $$x$$ is constant, $$\frac{d}{dy}(xy) = x$$ So, $$\frac{d}{dy} (xy)^{\frac{1}{2}} = \frac{1}{2}(xy)^{-\frac{1}{2}} \cdot x = \frac{x}{2\sqrt{xy}}$$ - Derivative of constant $$3$$ is $$0$$. 4. **Set derivatives equal:** $$\frac{1}{x} = \frac{x}{2\sqrt{xy}}$$ 5. **Simplify:** Multiply both sides by $$2\sqrt{xy}$$: $$2\sqrt{xy} \cdot \frac{1}{x} = x$$ $$\frac{2\sqrt{xy}}{x} = x$$ Multiply both sides by $$x$$: $$2\sqrt{xy} = x^2$$ 6. **Square both sides:** $$4xy = x^4$$ 7. **Solve for $$y$$:** $$y = \frac{x^3}{4}$$ **Final answer:** $$\frac{dy}{dy} = 1$$ (since differentiating $$y$$ with respect to $$y$$ is 1), but the problem was to differentiate the original equation w.r.t. $$y$$, which we did stepwise. **Summary:** The derivative of the left side w.r.t. $$y$$ is $$\frac{1}{x}$$, and the derivative of the right side w.r.t. $$y$$ is $$\frac{x}{2\sqrt{xy}}$$. Hence, the differentiation w.r.t. $$y$$ is: $$\frac{d}{dy}\left(\frac{y}{x}\right) = \frac{d}{dy}\left(\sqrt{xy} + 3\right)$$ or $$\frac{1}{x} = \frac{x}{2\sqrt{xy}}$$