Problem 1
1. For the function \(f(x,y)\), the point \((a,b)\in\mathbb{R}^2\), define the partial derivative of \(f(x,y)\) with respect to \(y\) at \((a,b)\). (2 points)
Solution. We define the partial derivative of \(f(x,y)\) w.r.t. \(y\) as
\[ \frac{\partial f}{\partial y}(a,b) = \lim_{h\to 0} \frac{f(a,b+h)-f(a,b)}{h}. \]
Problem 2
2. For \( f(x,y) = \begin{cases} \dfrac{3x^2y-y^3}{x^2+y^2}, & \text{if } (x,y)\ne(0,0) \\ 0, & \text{if } (x,y)=(0,0), \end{cases} \) find a formula for \(f_x(x,y)\). (4 points)
Solution. As this is a rational function in \(x\), it is differentiable everywhere except possibly for when \(x^2+y^2=0\). Thus, we first see what the function is doing away from the origin. Use the quotient rule to see
\[ f_x(x,y) = \frac{(x^2+y^2)(6xy)-(3x^2y-y^3)(2x)}{(x^2+y^2)^2} = \frac{8xy^3}{(x^2+y^2)^2}. \]
Next, we use the limit definition of derivative to see what is happening at the origin. We should have \[ f_x(0,0) = \lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0} \frac{\frac{0}{h^2}-0}{h} = \lim_{h\to 0} \frac{0}{h} = \lim_{h\to 0} 0 = 0. \]
Thus, we can say that \[ f_x(x,y) = \begin{cases} \dfrac{8xy^3}{(x^2+y^2)^2}, & \text{if } (x,y)\ne(0,0) \\ 0, & \text{if } (x,y)=(0,0). \end{cases} \]
Problem 3
3. Find the tangent plane to the graph of \(z=f(x,y) = -9x^3-3y^2\) at the point \((2,1,f(2,1))\). (4 points)
Solution. We know that the equation for a tangent plane is \(z = f_x(a,b)(x-a)+f_y(a,b)(y-b)+f(a,b)\). Thus we first find the partial derivatives: \(f_x(x,y) = -27x^2\) and \(f_y(x,y) = -6y\), so we have \(f_x(2,1) = -108\) and \(f_y(2,1) = -6\). We also see that \(f(2,1) = -75\). Putting this together, we have that the equation for the tangent plane to the function \(f(x,y)\) at \((2,1)\) is given by the equation
\[ z = -108(x-2)-6(y-1)-75. \]