Q. 12 - Calculus

Question

Consider the function f defined as in exercise 11

a) use the definition of the partial derivatives to show that $f_x(x_0,0)$ every value of $x_0$ but $f_y(x_0,0)$ exists only when $x_0=0$

b)use the definition of the partial derivatives to show that $f_0(0,y_0)$ every value of $y_0$ but $f_x(0,y_0)$ exists only when $y_0=0$.

Step-by-Step Solution

Answer

We proved both the parts and showed the results

Step 1: Given information

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

Part (a) Step 2. The explanation for part (a).

Now consider the partial derivative

we have,

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{f(x+h,y)-f(x,y)}{h} \\ \underset{h\to 0}{\lim }\frac{(x+h)y-xy}{h} \\ \underset{h\to 0}{\lim }\frac{hy}{h} \\ \underset{h\to 0}{\lim }y \end{gathered}$$

This limit is zero at point $(x_0,0)$

Hence $f_x(x_0,0)$ exists for all values of $x_0$

Now consider,

$$\begin{gathered} \underset{k\to 0}{\lim }\frac{f(x,y+k)-f(x,y)}{k} \\ \underset{k\to 0}{\lim }\frac{x(y+k)-xy}{k} \\ \underset{k\to 0}{\lim }\frac{xk}{k} \\ \underset{k\to 0}{\lim }x \end{gathered}$$

At this point, it varies as the value of $x_0$ changes Hence the limit exists only when $x_0=0$

Part (b) Step 1. The explanation for part (b).

Consider the partial derivatives

$$\begin{gathered} \underset{k\to 0}{\lim }\frac{f(x,y+k)-f(x,y)}{k} \\ \underset{k\to 0}{\lim }\frac{x(y+k)-xy}{k} \\ \underset{k\to 0}{\lim }x \end{gathered}$$

At point $(0,y_0)$ this value will be 0.

Hence limit exists for any $y_0$

Now consider,

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

At point, this limit varies hence it will only exist at point $y_0=0$