Chapter 4: Partial Differentiation

Section 4.1: First-Order Partial Derivatives

Example 4.1.5

$$ If $f\=\{\begin{array}{cc}\frac{x y \left(x^2-y^2\right)}{x^2\+y^2}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$ and $\left(a\,b\right)\=\left(0\,0\right)$, obtain $f_x$ and $f_y$ both at $\left(x\,y\right)$ and at $\left(a\,b\right)$. $$ Solution $$ For $\left(x\,y\right)\ne \left(0\,0\right)$, $f_x\left(x\,y\right)\=\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$ and $f_y\left(x\,y\right)\=\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$.   The partial derivatives at $\left(0\,0\right)$ must be calculated by definition, as in Table 4.1.5(a).   $f_x\left(0\,0\right)$ $\=\underset{h\to 0}{lim}\frac{f\left(0\+h\,0\right)-f\left(0\,0\right)}{h}$ $$ $\=\underset{h\to 0}{lim}\frac{\frac{0}{h^2}-0}{h}$ $$ $\=\underset{h\to 0}{lim}0$ $$ $\=0$ $f_y\left(0\,0\right)$ $\=\underset{k\to 0}{lim}\frac{f\left(0\,0\+k\right)-f\left(0\,0\right)}{k}$ $$ $\=\underset{k\to 0}{lim}\frac{0-\frac{0}{k^2}}{k}$ $$ $\=\underset{k\to 0}{lim}0$ $$ $\=0$ Table 4.1.5(a)   Partial derivatives at the origin   Consequently, both first partial derivatives are themselves piecewise defined as in Table 4.1.5(b).   $f_x\=\{\begin{array}{cc}\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$ $f_y\=\{\begin{array}{cc}\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$ Table 4.1.5(b)   First partial derivatives of $f\left(x\,y\right)$. $$

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