1. **State the problem:** Differentiate the expression $$b + wL + F - c - q(y - x) - \frac{Rb}{a} - \frac{pq}{2} \left(\frac{y - x}{x}\right)^2 x$$ with respect to $x$. 2. **Recall differentiation rules:** - The derivative of a constant is 0. - Use the product rule: $\frac{d}{dx}[uv] = u'v + uv'$. - Use the chain rule for composite functions. 3. **Identify constants and variables:** - $b, w, L, F, c, q, y, R, a, p$ are constants. - $x$ is the variable. 4. **Differentiate term by term:** - $\frac{d}{dx}[b] = 0$ - $\frac{d}{dx}[wL] = 0$ - $\frac{d}{dx}[F] = 0$ - $\frac{d}{dx}[-c] = 0$ - $\frac{d}{dx}[-q(y - x)] = -q \frac{d}{dx}(y - x) = -q(0 - 1) = q$ - $\frac{d}{dx}\left[-\frac{Rb}{a}\right] = 0$ 5. **Differentiate the last term:** Let $$u = \left(\frac{y - x}{x}\right)^2, \quad v = x$$ Then $$\frac{d}{dx}\left[-\frac{pq}{2} u v\right] = -\frac{pq}{2} \left(u' v + u v'\right)$$ 6. **Find $u'$:** $$u = \left(\frac{y - x}{x}\right)^2 = \left(\frac{y}{x} - 1\right)^2$$ Let $$w = \frac{y}{x} - 1$$ Then $$u = w^2 \implies u' = 2w w'$$ Calculate $$w' = \frac{d}{dx}\left(\frac{y}{x} - 1\right) = -\frac{y}{x^2}$$ So $$u' = 2 \left(\frac{y}{x} - 1\right) \left(-\frac{y}{x^2}\right) = -\frac{2y}{x^2} \left(\frac{y}{x} - 1\right)$$ 7. **Calculate $v'$:** $$v = x \implies v' = 1$$ 8. **Substitute back:** $$\frac{d}{dx}\left[-\frac{pq}{2} u v\right] = -\frac{pq}{2} \left(u' x + u \cdot 1\right) = -\frac{pq}{2} \left(-\frac{2y}{x^2} \left(\frac{y}{x} - 1\right) x + \left(\frac{y - x}{x}\right)^2\right)$$ Simplify inside the parentheses: $$-\frac{2y}{x^2} \left(\frac{y}{x} - 1\right) x = -\frac{2y}{x} \left(\frac{y}{x} - 1\right)$$ So $$= -\frac{pq}{2} \left(-\frac{2y}{x} \left(\frac{y}{x} - 1\right) + \left(\frac{y - x}{x}\right)^2\right)$$ 9. **Final derivative:** $$\frac{d}{dx} = q - \frac{pq}{2} \left(-\frac{2y}{x} \left(\frac{y}{x} - 1\right) + \left(\frac{y - x}{x}\right)^2\right)$$ This is the derivative of the given expression with respect to $x$.