Q. 13 - Calculus

Question

Let $f(x,y,z)=\left\{\begin{array}{ll}0& if xyz=0\\ 1& if xyz\ne 0\end{array}\right.$

Use the definition of the partial derivatives to show that $f_x(0,0,0)=f_y(0,0,0)=f_z(0,0,0)=0$. Explain why this example shows existence of partial derivatives at point $(x_0,y_0,z_0)$ for a function f(x,y,z) does not guarantee that f is continuous at that point

Step-by-Step Solution

Answer

The partial derivatives are equal to zero.

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

Step 1: Given information

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

Step 2: Find the partial derivatives

We have

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

Note that the value of the function is zero when one of the x ,y, z is zero and hence the partial derivative with respect to x is zero.

When it is non-zero The value of the function is 1.

Hence, We have,

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

Similarly,

Note that the value of the function is zero when one of the x ,y, z is zero and hence the partial derivative with respect to y, z is zero. And when they are non-zero The value becomes 1

And hence as for x we get the partial derivative of the function with respect to y and partial derivative of the function with respect to z is zero. 

Step 3: Explanation

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