Question: ASSIGNMENT: DUE: JULY 27, 2026
1.) w.r.t.x. y^3 = 2xy^2 + x^2y
2.) w.r.t.y. y/x = \sqrt{xy} + 3
3.) w.r.t.x. xy = x/y + y/x
4.) w.r.t.y. \left(\frac{x - 3}{y + 3}\right)^2 = 2xy - y
5.) required: y'' \sqrt{x} = \sqrt{y} - x
6.) required: y'' \frac{1}{xy} = 2 - x^2
answer that with complete solution
1. **Problem:** Differentiate implicitly with respect to $x$ the equation $$y^3 = 2xy^2 + x^2y.$$
2. **Formula and rules:** Use implicit differentiation. Recall the product rule: $$\frac{d}{dx}[uv] = u'v + uv'$$ and chain rule for $y$ as a function of $x$: $$\frac{d}{dx}[y] = y'.$$
3. **Differentiate both sides:**
$$\frac{d}{dx}[y^3] = \frac{d}{dx}[2xy^2 + x^2y]$$
Left side: $$3y^2 y'$$
Right side: Use product rule twice:
$$\frac{d}{dx}[2xy^2] = 2(y^2 + x \cdot 2y y') = 2y^2 + 4xy y'$$
$$\frac{d}{dx}[x^2 y] = 2x y + x^2 y'$$
So right side total: $$2y^2 + 4xy y' + 2x y + x^2 y'$$
4. **Set up the equation:**
$$3y^2 y' = 2y^2 + 4xy y' + 2x y + x^2 y'$$
5. **Group terms with $y'$ on one side:**
$$3y^2 y' - 4xy y' - x^2 y' = 2y^2 + 2x y$$
6. **Factor out $y'$:**
$$y'(3y^2 - 4xy - x^2) = 2y^2 + 2x y$$
7. **Solve for $y'$:**
$$y' = \frac{2y^2 + 2x y}{3y^2 - 4xy - x^2}$$
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**Final answer:**
$$\boxed{y' = \frac{2y^2 + 2x y}{3y^2 - 4xy - x^2}}$$
This completes the solution for the first problem.