1. **State the problem:** Given the function $f(n,y) = n^2 \arctan^{-1}\left(\frac{y}{n}\right)$, find the second order partial derivatives $\frac{\partial^2 f}{\partial n^2}$ and $\frac{\partial^2 f}{\partial y^2}$. 2. **Recall the formulas and rules:** - The function involves $\arctan^{-1}(z)$ which is the inverse tangent function, often written as $\arctan(z)$. - The derivative of $\arctan(z)$ with respect to $z$ is $\frac{1}{1+z^2}$. - Use the chain rule for derivatives of composite functions. - For partial derivatives, treat other variables as constants. 3. **Find the first partial derivative with respect to $n$: $$f(n,y) = n^2 \arctan\left(\frac{y}{n}\right)$$ Let $z = \frac{y}{n}$. Then $$f = n^2 \arctan(z)$$ Calculate $\frac{\partial f}{\partial n}$ using product and chain rules: $$\frac{\partial f}{\partial n} = 2n \arctan(z) + n^2 \cdot \frac{1}{1+z^2} \cdot \frac{\partial z}{\partial n}$$ Since $z = \frac{y}{n}$, $$\frac{\partial z}{\partial n} = \frac{\partial}{\partial n} \left(\frac{y}{n}\right) = -\frac{y}{n^2}$$ Substitute back: $$\frac{\partial f}{\partial n} = 2n \arctan\left(\frac{y}{n}\right) + n^2 \cdot \frac{1}{1+\left(\frac{y}{n}\right)^2} \cdot \left(-\frac{y}{n^2}\right)$$ Simplify the second term: $$= 2n \arctan\left(\frac{y}{n}\right) - \frac{y}{1+\frac{y^2}{n^2}}$$ Rewrite denominator: $$1 + \frac{y^2}{n^2} = \frac{n^2 + y^2}{n^2}$$ So $$\frac{y}{1+\frac{y^2}{n^2}} = y \cdot \frac{n^2}{n^2 + y^2}$$ Therefore, $$\frac{\partial f}{\partial n} = 2n \arctan\left(\frac{y}{n}\right) - \frac{y n^2}{n^2 + y^2}$$ 4. **Find the second partial derivative with respect to $n$: Differentiate $\frac{\partial f}{\partial n}$ again with respect to $n$: $$\frac{\partial^2 f}{\partial n^2} = \frac{\partial}{\partial n} \left(2n \arctan\left(\frac{y}{n}\right) - \frac{y n^2}{n^2 + y^2}\right)$$ First term: $$\frac{\partial}{\partial n} \left(2n \arctan\left(\frac{y}{n}\right)\right) = 2 \arctan\left(\frac{y}{n}\right) + 2n \cdot \frac{1}{1+\left(\frac{y}{n}\right)^2} \cdot \left(-\frac{y}{n^2}\right)$$ Simplify second part: $$= 2 \arctan\left(\frac{y}{n}\right) - \frac{2 n y}{1 + \frac{y^2}{n^2}}$$ Rewrite denominator: $$1 + \frac{y^2}{n^2} = \frac{n^2 + y^2}{n^2}$$ So $$\frac{2 n y}{1 + \frac{y^2}{n^2}} = 2 n y \cdot \frac{n^2}{n^2 + y^2} = \frac{2 n^3 y}{n^2 + y^2}$$ Therefore, first term derivative is: $$2 \arctan\left(\frac{y}{n}\right) - \frac{2 n^3 y}{n^2 + y^2}$$ Second term: $$\frac{\partial}{\partial n} \left(- \frac{y n^2}{n^2 + y^2}\right) = -y \cdot \frac{\partial}{\partial n} \left(\frac{n^2}{n^2 + y^2}\right)$$ Use quotient rule: $$\frac{\partial}{\partial n} \left(\frac{n^2}{n^2 + y^2}\right) = \frac{(2n)(n^2 + y^2) - n^2 (2n)}{(n^2 + y^2)^2} = \frac{2n (n^2 + y^2) - 2n^3}{(n^2 + y^2)^2} = \frac{2n y^2}{(n^2 + y^2)^2}$$ So second term derivative is: $$- y \cdot \frac{2n y^2}{(n^2 + y^2)^2} = - \frac{2 n y^3}{(n^2 + y^2)^2}$$ 5. **Combine both parts for $\frac{\partial^2 f}{\partial n^2}$:** $$\frac{\partial^2 f}{\partial n^2} = 2 \arctan\left(\frac{y}{n}\right) - \frac{2 n^3 y}{n^2 + y^2} - \frac{2 n y^3}{(n^2 + y^2)^2}$$ 6. **Find the first partial derivative with respect to $y$: $$\frac{\partial f}{\partial y} = n^2 \cdot \frac{1}{1 + \left(\frac{y}{n}\right)^2} \cdot \frac{\partial}{\partial y} \left(\frac{y}{n}\right) = n^2 \cdot \frac{1}{1 + \frac{y^2}{n^2}} \cdot \frac{1}{n} = n^2 \cdot \frac{1}{1 + \frac{y^2}{n^2}} \cdot \frac{1}{n}$$ Simplify denominator: $$1 + \frac{y^2}{n^2} = \frac{n^2 + y^2}{n^2}$$ So $$\frac{\partial f}{\partial y} = n^2 \cdot \frac{n^2}{n^2 + y^2} \cdot \frac{1}{n} = \frac{n^3}{n^2 + y^2}$$ 7. **Find the second partial derivative with respect to $y$: $$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left(\frac{n^3}{n^2 + y^2}\right) = n^3 \cdot \frac{\partial}{\partial y} \left(\frac{1}{n^2 + y^2}\right) = n^3 \cdot \left(- \frac{2y}{(n^2 + y^2)^2}\right) = - \frac{2 n^3 y}{(n^2 + y^2)^2}$$ **Final answers:** $$\boxed{\frac{\partial^2 f}{\partial n^2} = 2 \arctan\left(\frac{y}{n}\right) - \frac{2 n^3 y}{n^2 + y^2} - \frac{2 n y^3}{(n^2 + y^2)^2}}$$ $$\boxed{\frac{\partial^2 f}{\partial y^2} = - \frac{2 n^3 y}{(n^2 + y^2)^2}}$$