Implicit Differentiation 5360E1

1. **State the problem:** We need to find $\frac{dy}{dx}$ for the implicit equation $$x^2y^3 - 8xy^2 = 4$$ using implicit differentiation. 2. **Recall the rule:** When differentiating terms involving $y$, treat $y$ as a function of $x$ (i.e., $y = y(x)$) and apply the chain rule: $$\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}$$. 3. **Differentiate each term:** - Differentiate $x^2 y^3$ using the product rule: $$\frac{d}{dx}[x^2 y^3] = \frac{d}{dx}[x^2] \cdot y^3 + x^2 \cdot \frac{d}{dx}[y^3] = 2x y^3 + x^2 \cdot 3 y^2 \frac{dy}{dx} = 2x y^3 + 3 x^2 y^2 \frac{dy}{dx}$$ - Differentiate $-8 x y^2$ using the product rule: $$\frac{d}{dx}[-8 x y^2] = -8 \left( \frac{d}{dx}[x] \cdot y^2 + x \cdot \frac{d}{dx}[y^2] \right) = -8 (1 \cdot y^2 + x \cdot 2 y \frac{dy}{dx}) = -8 y^2 - 16 x y \frac{dy}{dx}$$ - Differentiate the constant $4$: $$\frac{d}{dx}[4] = 0$$ 4. **Write the differentiated equation:** $$2x y^3 + 3 x^2 y^2 \frac{dy}{dx} - 8 y^2 - 16 x y \frac{dy}{dx} = 0$$ 5. **Group terms with $\frac{dy}{dx}$ on one side:** $$3 x^2 y^2 \frac{dy}{dx} - 16 x y \frac{dy}{dx} = 8 y^2 - 2x y^3$$ 6. **Factor out $\frac{dy}{dx}$:** $$\frac{dy}{dx} (3 x^2 y^2 - 16 x y) = 8 y^2 - 2 x y^3$$ 7. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{8 y^2 - 2 x y^3}{3 x^2 y^2 - 16 x y}$$ 8. **Simplify by factoring out common terms:** $$\frac{dy}{dx} = \frac{2 y^2 (4 - x y)}{x y (3 x y - 16)}$$ **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{2 y^2 (4 - x y)}{x y (3 x y - 16)}}$$