# The directional derivative of a function

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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I'm getting 1/2 now. – Wouter Zeldenthuis Nov 6 '12 at 12:55
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. – Martin Sleziak Nov 6 '12 at 13:06

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.