Q. 11 - Calculus

Question

Let $f(x,y)=\left\{\begin{array}{ll}0& if xy=0\\ 1& if xy\ne 0\end{array}\right.$ use definition of the partial derivatives to show that $f_x(0,0)=f_y(0,0)=0$ Explain why this example shows that the existence of the partial derivatives at a point $(x_0,y_0)$. for a function f(x,y) does not guarantee that f is continuous at $(x_0,y_0)$

Step-by-Step Solution

Answer

We show that both the partial derivatives are zero. Also from the above we know that the function is differentiable at (0,0). but the point is not continuous at (0,0) 

Step 1: Given information

We are given a function $f(x,y)=\left\{\begin{array}{ll}0& if xy=0\\ 1& if xy\ne 0\end{array}\right.$

Step 2: Compute the partial derivatives

By using the definition we have,

$\underset{h\to 0}{\lim }\frac{f(x+h,y)-f(x,y)}{h}$

Note that the value of the function will be zero if any of the x or y will be zero. Hence when one of the x or y is zero the value of partial derivative is zero.

When both are non-zero, Then we have

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{1-1}{k} \\ \underset{h\to 0}{\lim }0 \\ =0 \end{gathered}$$

Similarly the partial derivative with respect to y is 0

Step 3: Explaination

Now from the above we know that the function is differentiable at (0,0). but the point is not continuous at (0,0)