Directional Derivative

**demode** · March 17th 2010, 12:18 AM

Part (c), (d) and (e) of the following problem:

My attempt (please correct me where I am wrong)

(c) $\hat{u} = \| u \| = \sqrt{1^2 + 1^2} = \sqrt{2}$ so

$\hat{u} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$

From part (a) I know that $f_x(0,0)=f_y(0,0)=0$ . Therefore:

$\nabla f(0,0)=f_x(0,0) i + f_y(0,0)j = 0$

Therefore $D_u f(0,0)= \nabla f(0,0). \hat{u} = 0$

(d) Letting $y=x$ we will have $f(x)=\sqrt[3]{2x^3}$ . Therefore

$f'(x)= \frac{2^{(1/3)}}{(x^3)^(2/3) x^2}$

$\nabla f = f(0)i = 0$

Therefore $D_u f(0,0)= \nabla f(0). \hat{u} = 0$

I got 0 again!

(e) I think maybe the method in part (d) is more efficient but neither work here because in part (b) I showed that the function is not differentiable at (0,0), and if a function is not differentiable at a point then the directional derivative does not exist there for all directions $\hat{u}$ .

Is this right? I need to submit this tomorrow so I would appreciate any help or suggestions.

**HallsofIvy** · March 17th 2010, 02:25 AM

Originally Posted by demode

Part (c), (d) and (e) of the following problem:

My attempt (please correct me where I am wrong)

(c) $\hat{u} = \| u \| = \sqrt{1^2 + 1^2} = \sqrt{2}$
so

$\hat{u} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$

From part (a) I know that $f_x(0,0)=f_y(0,0)=0$ . Therefore:

$\nabla f(0,0)=f_x(0,0) i + f_y(0,0)j = 0$

Therefore $D_u f(0,0)= \nabla f(0,0). \hat{u} = 0$

(d) Letting $y=x$ we will have $f(x)=\sqrt[3]{2x^3}$ . Therefore

$f'(x)= \frac{2^{(1/3)}}{(x^3)^(2/3) x^2}$

$\nabla f = f(0)i = 0$

Therefore $D_u f(0,0)= \nabla f(0). \hat{u} = 0$

Then try again! $\sqrt[3]{2x^3}= (\sqrt[3]{2})x$ and its derivative, at x= 0, is NOT 0. You seem to be under the impression that if the numerator of a fraction is 0, then the fraction is equal to 0. That's not true if the denominator is also 0!

I got 0 again!

(e) I think maybe the method in part (d) is more efficient but neither work here because in part (b) I showed that the function is not differentiable at (0,0), and if a function is not differentiable at a point then the directional derivative does not exist there for all directions $\hat{u}$ .

Is this right? I need to submit this tomorrow so I would appreciate any help or suggestions.

No, if a function is not differentiable at a point then the directional derivative may not exist for some or all directions.

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