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.
[Math] If a partial derivative exists at a point is then do all directional derivatives exist as well
multivariable-calculus
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Best Answer
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.