Find the directional derivative of ##f## at the given point

[chwala]

TL;DR

Kindly see attached;

Going through this now: pretty straightforward i just want to check that i have covered all aspects including any other approach...

Ok for 15. I have,

$\nabla f= (yz \cos (xyz), xz \cos (xyz), xy \cos (xyz) )$

so,

$D_v f(1,1,1) = \textbf v ⋅\nabla f(1,1,1)$=$\left(\dfrac {1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3} })\right)⋅( \cos 1, \cos 1 , \cos 1) = \dfrac{\cos 1}{\sqrt{3}} + \dfrac{\cos 1}{\sqrt{3}} + \dfrac{\cos 1}{\sqrt{3}} =\dfrac{3}{\sqrt{3}}\cos 1= \sqrt{3} \cos 1$ Ok for 16. I have,

$\nabla f= (2x e^{yz} , x^2z e^{yz}, x^2y e^{yz} )$

so,

$D_v f(1,1,1) = \textbf v ⋅\nabla f(1,1,1)$=$\left(\dfrac {1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3} })\right)⋅( 2e, e, e) = \dfrac{2e}{\sqrt{3}} + \dfrac{e}{\sqrt{3}} + \dfrac{e}{\sqrt{3}} =\dfrac{4}{\sqrt{3}}e$

 

[haider]

Looks correct to me. You could also evaluate the actual definition to check your answer:

$$ D_{u}f(\mathbf{p}) = \lim_{h \to 0}{\frac{ f(\mathbf{p}+h\mathbf{u}) - f(\mathbf{p}) }{h} } $$

It might be nitpicking for this problem but the gradient formula is only applicable if the function is differentiable (equivalent to saying all partial derivatives exist; they are used in deriving $D_{u}f(\mathbf{p}) = \nabla f(\mathbf{p}) \cdot \mathbf{u}$)

 

[mathwonk]

Don't you need a bit more than existence of partials, to deduce differentiability? I.e. I seem to recall that local existence and continuity of partials is sufficient, and perhaps point wise existence of all directional derivatives, as well as their linear dependence on the direction, is necessary and sufficient.

Pictured geometrically, existence of a directional derivative is existence of a tangent line to the graph in one direction. Existence of all partials is merely existence of tangent lines in each axis direction. Existence of all directional derivatives is existence of tangent lines in every direction, i.e. of a tangent "cone" to the graph, and finally differentiability is existence of that cone plus the fact that it is actually a linear space, i.e. a hyperplane.

 

[haider]

Don't you need a bit more than existence of partials, to deduce differentiability? I.e. I seem to recall that local existence and continuity of partials is sufficient, and perhaps point wise existence of all directional derivatives, as well as their linear dependence on the direction, is necessary and sufficient.

Pictured geometrically, existence of a directional derivative is existence of a tangent line to the graph in one direction. Existence of all partials is merely existence of tangent lines in each axis direction. Existence of all directional derivatives is existence of tangent lines in every direction, i.e. of a tangent "cone" to the graph, and finally differentiability is existence of that cone plus the fact that it is actually a linear space, i.e. a hyperplane.

You're correct. I misremembered the condition. Differentiability implies existence of partials but the inverse is not necessarily true. This looks to be a sign that I should brush up on my multivariable calculus.

According to Apostol (and as you said), the local existence and continuity of partials at a point is the sufficient condition for differentiability at the point.