Question

In Exercises 87 and \(88,\) use the function to prove that (a) \(f_{x}(0,0)\) and \(f_{y}(\mathbf{0}, \mathbf{0})\) exist, and (b) \(f\) is not differentiable at \((\mathbf{0}, \mathbf{0})\). \(f(x, y)=\left\\{\begin{array}{ll}\frac{3 x^{2} y}{x^{4}+y^{2}}, & (x, y) \neq(0,0) \\ 0, & (x, y)=(0,0)\end{array}\right.\)

Short answer

The partial derivatives \(f_x(0,0)\) and \(f_y(0,0)\) can be computed by evaluating the appropriate limits. The differentiability of the function at the point \((0, 0)\) can be ascertained by verifying whether a certain limit (as described in the steps) exists and is finite.

Step-by-step solution

Compute \(f_{x}(0,0)\)

Firstly, calculate the partial derivative of the function \(f(x,y)\) with respect to \(x\) at the point \((0,0)\), \(f_x(0,0)\). As \(f(x, y)=\left\{\begin{array}{ll}\frac{3 x^{2} y}{x^{4}+y^{2}}, & (x, y)\neq(0,0) \ 0, & (x, y)=(0,0)\end{array}\right.\), Using the definition of derivative at a point, we compute the two-sided limit: \[ f_x(0,0) = \lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h} \] Because \(f(0,0)=0\), this simplifies to \[ f_x(0,0) = \lim_{h\to 0} \frac{f(h,0)}{h} = \lim_{h\to 0} \frac{f(h,0)}{h}\] since \((x, y)\neq(0,0)\) for \((0+h,0)\).

Compute \(f_{y}(0,0)\)

Following similar steps, calculate the partial derivative of the function \(f(x,y)\) with respect to \(y\) at the point \((0,0)\), \(f_y(0,0)\). Using the definition of derivative at a point, likewise compute the two-sided limit: \[ f_y(0,0) = \lim_{k\to 0} \frac{f(0,0+k)-f(0,0)}{k} \] This simplifies to: \[ f_y(0,0) = \lim_{k\to 0} \frac{f(0,k)}{k} = \lim_{k\to 0} \frac{f(0,k)}{k}\] since \((x, y)\neq(0,0)\) for \((0, 0+k)\).

Checking the Differentiability

The function \(f\) is said to be differentiable at \((0, 0)\) if the following limit exists and is finite: \[ \lim_{(h,k)\to (0, 0)} \frac{f(h,k)-f(0,0)-f_x(0,0)\cdot h -f_y(0,0)\cdot k}{\sqrt{h^2+k^2}}\] Checking the existance and the finiteness of this limit would determine whether \(f\) is differentiable at \((0, 0)\) or not.