1. The problem is to differentiate the function given by $$dy/dn = -\left(4y(0.5mmy + \sin \cos n \sqrt{1 + y} + 4 \sqrt{1 + n})\right)$$ with respect to $n$. 2. To differentiate this expression, we need to apply the product rule and chain rule carefully because the function is a product of multiple terms involving $y$, $n$, and trigonometric and root functions. 3. The product rule states: $$\frac{d}{dn}[u(n)v(n)] = u'(n)v(n) + u(n)v'(n)$$ where $u$ and $v$ are functions of $n$. 4. The chain rule states: $$\frac{d}{dn}f(g(n)) = f'(g(n)) \cdot g'(n)$$ which is used when differentiating composite functions like $\sin(\cos n)$ or $\sqrt{1+n}$. 5. Let's denote: - $u = 4y$ - $v = 0.5mmy + \sin(\cos n) \sqrt{1 + y} + 4 \sqrt{1 + n}$ 6. Differentiate $u$ with respect to $n$: since $y$ may depend on $n$, $\frac{du}{dn} = 4 \frac{dy}{dn}$. 7. Differentiate $v$ with respect to $n$: - $\frac{d}{dn}(0.5mmy) = 0.5mm \frac{dy}{dn}$ (assuming $m$ and $m$ are constants) - $\frac{d}{dn}(\sin(\cos n) \sqrt{1 + y})$ requires product rule: - Let $a = \sin(\cos n)$ and $b = \sqrt{1 + y}$ - $a' = \cos(\cos n) \cdot (-\sin n)$ by chain rule - $b' = \frac{1}{2\sqrt{1 + y}} \frac{dy}{dn}$ - So $\frac{d}{dn}(ab) = a'b + ab'$ - $\frac{d}{dn}(4 \sqrt{1 + n}) = 4 \cdot \frac{1}{2\sqrt{1 + n}} = \frac{2}{\sqrt{1 + n}}$ 8. Combine all derivatives: $$\frac{dv}{dn} = 0.5mm \frac{dy}{dn} + \left(\cos(\cos n)(-\sin n) \sqrt{1 + y} + \sin(\cos n) \frac{1}{2\sqrt{1 + y}} \frac{dy}{dn}\right) + \frac{2}{\sqrt{1 + n}}$$ 9. Now apply product rule to original expression: $$\frac{d}{dn}(-u v) = -\left(u' v + u v'\right) = -\left(4 \frac{dy}{dn} v + 4y \frac{dv}{dn}\right)$$ 10. This gives the derivative of the original function in terms of $y$, $\frac{dy}{dn}$, and $n$. 11. If $y$ is an explicit function of $n$, substitute $\frac{dy}{dn}$ accordingly to get a full expression. Final answer: $$\frac{d}{dn} \left(-4y \left(0.5mmy + \sin(\cos n) \sqrt{1 + y} + 4 \sqrt{1 + n}\right)\right) = -\left(4 \frac{dy}{dn} \left(0.5mmy + \sin(\cos n) \sqrt{1 + y} + 4 \sqrt{1 + n}\right) + 4y \left(0.5mm \frac{dy}{dn} + \cos(\cos n)(-\sin n) \sqrt{1 + y} + \sin(\cos n) \frac{1}{2\sqrt{1 + y}} \frac{dy}{dn} + \frac{2}{\sqrt{1 + n}}\right)\right)$$