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Can someone help me compute the following directional derivative using the formal definition: $ D_u = \lim_{t\to 0 } \frac{f(x_0 +tu)-f(x_0) }{t}$ ?

the function is: $ f(x,y)= \arctan(x^2 + y^2 ) $ , at the point $ x_0 = (1,1)$, in the direction $u = (1,-2) $ .

The answer should be $ -\frac{2}{5 \sqrt5 } $ .

After putting it all together I received the limit of the following expression: $$\frac{\arctan ((1+\frac{t}{\sqrt{5} })^2 +(1-\frac{2t}{\sqrt{5}} ) ^2 ) - \arctan(2)}{t}$$ and I have no idea how to compute it.

Thanks in advance !

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HINT. The function

$f(x,y)= arctan(x^2 + y^2 )$

is differentiable (for sufficient condition for differentiability of a function of two variables: "a function is differentiable at a point if the partial derivatives exist and are continuous in some neighborhood of the point"). Then we can apply

$D_v=\nabla f(x_0,y_0)\cdot v$

(gradient formula)

where $v=(1,-2)$ and $(x_0,y_0)=(1,1)$.

Instead, if you want to use the definition with the limit:

$D_u = \lim_{t\to 0 } \frac{\arctan ((1+\frac{t}{\sqrt{5} })^2 +(1-\frac{2t}{\sqrt{5}} ) ^2 ) - \arctan(2)}{t}=\frac{0}{0}$. Then we apply L'Hôpital's rule:

$\lim_{t\to 0 } \frac{\arctan ((1+\frac{t}{\sqrt{5} })^2 +(1-\frac{2t}{\sqrt{5}} ) ^2 ) - \arctan(2)}{t}= \lim_{t\to 0 } \frac{\frac{2}{\sqrt{5}}(1+\frac{t}{\sqrt{5}})-\frac{4}{\sqrt{5}}(1-\frac{2t}{\sqrt{5}})}{\left(\left(1+\frac{t}{\sqrt{5} }\right)^2 +\left(1-\frac{2t}{\sqrt{5}} \right) ^2\right)^2+1}$

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@Mark: I wanted to use the limit definition, and not this theorem. Thanks anyway ! – joshua Jul 7 '12 at 15:05
Thanks a lot !!!! – joshua Jul 7 '12 at 15:18

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