1. The problem is to differentiate the function $$y = \left(\frac{x - 1}{3 + x^2}\right)^4$$ with respect to $x$. 2. This is a composite function where an inner function $$u = \frac{x - 1}{3 + x^2}$$ is raised to the power 4. The chain rule applies: $$\frac{dy}{dx} = 4u^3 \cdot \frac{du}{dx}$$. 3. First, differentiate the inner function $u$ using the quotient rule: $$u = \frac{f(x)}{g(x)} = \frac{x - 1}{3 + x^2}$$. 4. The quotient rule states: $$\frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$$. 5. Compute derivatives: - $$f'(x) = \frac{d}{dx}(x - 1) = 1$$ - $$g'(x) = \frac{d}{dx}(3 + x^2) = 2x$$ 6. Substitute into quotient rule: $$\frac{du}{dx} = \frac{1 \cdot (3 + x^2) - (x - 1) \cdot 2x}{(3 + x^2)^2} = \frac{3 + x^2 - 2x(x - 1)}{(3 + x^2)^2}$$ 7. Simplify numerator: $$3 + x^2 - 2x^2 + 2x = 3 - x^2 + 2x$$ 8. So, $$\frac{du}{dx} = \frac{3 - x^2 + 2x}{(3 + x^2)^2}$$ 9. Now apply the chain rule: $$\frac{dy}{dx} = 4 \left(\frac{x - 1}{3 + x^2}\right)^3 \cdot \frac{3 - x^2 + 2x}{(3 + x^2)^2}$$ 10. Rewrite the derivative as a single fraction: $$\frac{dy}{dx} = \frac{4 (x - 1)^3 (3 - x^2 + 2x)}{(3 + x^2)^5}$$ This is the derivative of the given function. Yes, it is easier to differentiate the inside function first ignoring the power 4, then multiply by 4 times the inside function to the power 3, which is exactly the chain rule method shown here.