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What is the directional derivative of $f(x,y)=\frac{\sin(xy^2)}{x^2+y^2}$ (defined to be 0 at origin), at (0,0) in the direction (1,1). Part of my question is whether or not there are multiple ways to do this problem. I am familiar with taking the limit $\lim_{h\to 0} \frac{f((0,0)+h(1,1)-f((0,0))}{h}$. It's not clear to me how to solve this limit. Is there a better way?

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  • $\begingroup$ I'm getting 1/2 now. $\endgroup$ – Wouter Zeldenthuis Nov 6 '12 at 12:55
  • $\begingroup$ You probably wanted to write $f((0,0)+h(1,1))-f((0,0)$ and not $f((0,0)+h(1,1)-f((0,0))$ in the numerator. $\endgroup$ – Martin Sleziak Nov 6 '12 at 13:06
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When you know for a fact that the function to which you are taking a partial derivative is "sufficiently nice" (that certain numbers of its derivatives are continuous), then there are different algebraically equivalent ways of expression the same directional derivative.

But when you do not know that this is the case a priori (as in the case under consideration), you have to go back to the definitions. Observe that $$ (0,0) + h(1,1) = (h,h) $$ so $$ f((0,0) + h(1,1)) = f(h,h) = \frac{\sin h^3}{2h^2} $$ Hence $$ D_{(1,1)} f(0,0) = \lim_{h\to 0} \frac{\frac{\sin h^3}{2h^2} - 0}{h} = \lim_{h\to 0}\frac{\sin h^3}{2h^3} = \frac12 $$ as you noted in your comment.

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