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I thought that if partial derivatives exist then gradient also exist ,then all direction derivatives should also exist . is this true and if it is not then why am i wrong ? $D_u=▽.u$ where ▽ is the gradient and u is the direction vector along which you need the directional derivative.

rschwieb
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avz2611
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1 Answers1

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The condition is not sufficient. Consider $f(x,y)=|y|^\alpha e^{-\frac{x^2}{y^2}}$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$ with $\alpha\in (0,1)$. The function is continuous at $(0,0)$ as $0\geq f(x,y)\neq |y|^\alpha $ for $(x,y)\neq (0,0)$.

The directional derivative at $(0,0)$ is specified by the versor $v(\theta)=(\cos\theta,\sin\theta)$. Then

$$\frac{f(t\cos\theta, t\sin\theta)}{t}=\frac{|t\sin\theta|^\alpha}{t} e^{-\operatorname{cotg}^2\theta}, $$

if $\theta\neq 0,\pi$ and

$$\frac{f(t,0)}{t}=0. $$

if $\theta = 0,\pi$. It follows that $f_x(0,0)=0$ identically.

However, if $\theta$ is different from $0$ and $\pi$, the limit $\lim_{t\rightarrow 0}\frac{f(t\cos\theta, t\sin\theta)}{t}$ does not converge as $\alpha-1<0$.

A final remark: if a given function is differentiable at a given point, then all directional derivatives- and so also the partial derivatives- at that point exist, however.

Avitus
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  • so the formula that directional derivative is dot product of gradient and vector is wrong ? – avz2611 Sep 08 '14 at 19:25
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    It holds if the function is differentable at that given point – Avitus Sep 08 '14 at 19:26
  • @user142634 did it help? – Avitus Sep 09 '14 at 06:28
  • Thanks, if partial derivatives exist at all points of a function will my statement hold??? – avz2611 Sep 09 '14 at 11:59
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    You are welcome. The existence of partial derivatives is not strong enough to guarantee the existence of all directional derivatives. You need something more, i.e. differentiability of the function at the point under consideration. If at a given point the partial derivatives exist and are continuous, then the function is differentiable at that point. This implies the existence of all directional derivatives, at that given point. For more info, please have a look at http://math.stackexchange.com/questions/521217/partial-derivatives-continuous-implies-differentiability-in-euclidean-space – Avitus Sep 09 '14 at 12:41