Differentiate Wrt Y 05E55C

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 with respect to $$y$$. - Right side: Use the chain rule on $$ (xy)^{\frac{1}{2}} $$: $$\frac{d}{dy} (xy)^{\frac{1}{2}} = \frac{1}{2} (xy)^{-\frac{1}{2}} \cdot \frac{d}{dy} (xy)$$ Since $$x$$ is constant, $$\frac{d}{dy} (xy) = x$$ So, $$\frac{d}{dy} (xy)^{\frac{1}{2}} = \frac{1}{2} (xy)^{-\frac{1}{2}} \cdot x = \frac{x}{2 \sqrt{xy}}$$ - The derivative of constant $$3$$ is $$0$$. 4. **Set derivatives equal:** $$\frac{1}{x} = \frac{x}{2 \sqrt{xy}}$$ 5. **Solve for $$x$$ or $$y$$ if needed:** Multiply both sides by $$2 \sqrt{xy}$$: $$2 \sqrt{xy} \cdot \frac{1}{x} = x$$ Simplify: $$\frac{2 \sqrt{xy}}{x} = x$$ Multiply both sides by $$x$$: $$2 \sqrt{xy} = x^2$$ Square both sides: $$4 xy = x^4$$ Divide both sides by $$x$$ (assuming $$x \neq 0$$): $$\cancel{4} \cancel{x} y = \cancel{x^4} \Rightarrow 4 y = x^3$$ 6. **Final relation:** $$4 y = x^3$$ or equivalently $$y = \frac{x^3}{4}$$ **Summary:** Differentiating $$\frac{y}{x} = \sqrt{xy} + 3$$ with respect to $$y$$ gives the relation $$4 y = x^3$$.