Problem 1
1. For the function \( f(x,y) = \begin{cases} \dfrac{2x^2y^2}{\sqrt{x^2+y^2}}, & \text{if } (x,y)\ne(0,0) \\ 0, & \text{if } (x,y)=(0,0), \end{cases} \) verify that the partial derivative with respect to \(x\) is continuous at \((0,0)\). Then use the limit definition to show that \(f(x,y)\) is differentiable at \((0,0)\).
Solution. First, we calculate \(f_x\) away from the origin. This is
\[ f_x = \frac{4xy^2\sqrt{x^2+y^2} - 2x^3y^2(x^2+y^2)^{-1/2}}{x^2+y^2}. \]
Then, to check the behavior of this function at \((0,0)\) we take the limit as it approaches the origin. We use a polar substitution to do so. We have \[ \begin{align*} \lim_{(x,y)\to(0,0)} f_x(x,y) &= \lim_{r\to 0,\, \theta\in\mathbb{R}} \frac{4r^4(\cos(\theta)\sin^2(\theta)) - 2r^5\cos^3(\theta)\sin^2(\theta)r^{-1}}{r^2} \\ &= \lim_{r\to 0,\, \theta\in\mathbb{R}} r\left(4\cos(\theta)\sin^2(\theta) - 2\cos^3(\theta)\sin^2(\theta)\right) \\ &= 0. \end{align*} \]
On the other hand, we can take the limit definition of derivative to find the actual value of \(f_x(0,0)\). We have that \[ f_x(0,0) = \lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0} \frac{\frac{2h^2\cdot 0}{h}-0}{h} = \lim_{h\to 0} 0 = 0. \]
As the limit of the function is equal to the value of the function at the origin, we say that \(f_x\) is continuous at \((0,0)\).Problem 2
2. Find \(D\mathbf{F}(2,3,1)\) for the function \(\mathbf{F}(x,y,z) = (x^2y^3z,\, e^{xy^2z^3},\, \cos(xyz))\). (5 points)
Solution. By taking the correct derivatives of the component functions with respect to the variables, we get that
\[ D\mathbf{F}(x,y,z) = \begin{pmatrix} 2xy^3z & 3x^2y^2z & x^2y^3 \\ y^2z^3e^{xy^2z^3} & 2xyz^3e^{xy^2z^3} & 3xy^2z^2e^{xy^2z^3} \\ -yz\sin(xyz) & -xz\sin(xyz) & -xy\sin(xyz) \end{pmatrix} \]
which leads us to \[ D\mathbf{F}(2,3,1) = \begin{pmatrix} 108 & 108 & 108 \\ 9e^{18} & 12e^{18} & 54e^{18} \\ -3\sin(6) & -2\sin(6) & -6\sin(6) \end{pmatrix}. \]