Solved IGNOU Assignments and Question papers: IGNOU MTE07 2019 SOLVED ASSIGNMENT 4(a)

$\left| y \right| \le \left| x \right|$

can be written as

$\begin{array}{l} - y \le x\\ y \le x\\ y \le - x \end{array}$

$\left| y \right| > \left| x \right|$

can be written as

$\begin{array}{l} - y > x\\ y > x\\ y > - x \end{array}$

So,

$f\left( {x,y} \right)$ is

$\begin{array}{l} 2xy;{\rm{ - y}} \le {\rm{x, y}} \le {\rm{x, y}} \le {\rm{ - x}}\\ {\rm{( - 1/2)xy; - y > x, y > x, y > - x}} \end{array}$

Now,

$\begin{array}{l} {f_x}\left( {0,0} \right) = {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h,0} \right) - f\left( {0,0} \right)}}{h}\\ = {\lim }\limits_{h \to 0} \frac{{0 - 0}}{h}\\ = 0 \end{array}$

$\begin{array}{l} {f_x}\left( {0,k} \right) = {\lim }\limits_{h \to 0} \frac{{f\left( {h,k} \right) - f\left( {0,0} \right)}}{h}\\ = \frac{{2hk - 0}}{h}\\ = 2k \end{array}$

$\begin{array}{l} {f_{yx}}\left( {0,0} \right) = {\lim }\limits_{k \to 0} \frac{{{f_x}\left( {0,k} \right) - {f_x}\left( {0,0} \right)}}{k}\\ = \frac{{2k - 0}}{k}\\ = 2 \end{array}$

Again,

$\begin{array}{l} {f_y}\left( {0,0} \right) = {\lim }\limits_{k \to 0} \frac{{f\left( {0,0 + k} \right) - f\left( {0,0} \right)}}{k}\\ = {\lim }\limits_{k \to 0} \frac{{0 - 0}}{k}\\ = 0 \end{array}$

$\begin{array}{l} {f_y}\left( {h,0} \right) = {\lim }\limits_{k \to 0} \frac{{f\left( {h,k} \right) - f\left( {0,0} \right)}}{k}\\ = \frac{{( - 1/2)hk - 0}}{k}\\ = ( - 1/2)h \end{array}$

$\begin{array}{l} {f_{xy}}\left( {0,0} \right) = {\lim }\limits_{k \to 0} \frac{{{f_y}\left( {h,0} \right) - {f_y}\left( {0,0} \right)}}{k}\\ = \frac{{( - 1/2)k - 0}}{k}\\ = ( - 1/2) \end{array}$

$\begin{array}{l} {f_{xy}}\left( {0,0} \right) = ( - 1/2)\\ {f_{yx}}\left( {0,0} \right) = 2 \end{array}$

${f_{xy}}\left( {0,0} \right) \ne {f_{yx}}\left( {0,0} \right)$

Solved IGNOU Assignments and Question papers

Saturday, May 25, 2019

IGNOU MTE07 2019 SOLVED ASSIGNMENT 4(a)

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