Partial Derivatives — Question 9

Question 9 – Solution

Strategy. At the exceptional point, use one-variable difference quotients along the coordinate axes.

Step 1: $x$-partial at the origin $$u_x(0,0)=\lim_{h\to 0}\frac{u(h,0)-u(0,0)}h=\lim_{h\to 0}\frac{0}{h}=\boxed 0.$$

Step 2: $y$-partial at the origin $$u_y(0,0)=\lim_{k\to 0}\frac{u(0,k)-u(0,0)}k=\lim_{k\to 0}\frac{0}{k}=\boxed 0.$$

Step 3: Away from the origin Quotient differentiation gives $$\boxed{u_x=\frac{2xy^3}{(x^2+y^2)^2}},\qquad \boxed{u_y=\frac{x^2(x^2-y^2)}{(x^2+y^2)^2}}\quad ((x,y)\ne(0,0)).$$ These formulas were derived where the denominator is nonzero, so substituting $(0,0)$ would be invalid. The definition supplies the missing partial values.