1. ## Directional derivative problem

I'd like you to check my answer. This is the most common exercise about directional derivatives...
Let $\displaystyle f(x,y,z)=xyz$, $\displaystyle \bold u =( \cos (\alpha) \sin (\beta) , \sin (\alpha) \sin (\beta) , \cos (\beta) )$ and $\displaystyle \bold x = (1,0,1)$.
Find the directional derivative of $\displaystyle f$ in the direction of $\displaystyle \bold u$, in the point $\displaystyle \bold x$.
Here's my work :
$\displaystyle \frac{\partial f}{\partial \bold u} (\bold x)= \nabla f( \bold x) \cdot \bold u = \sin (\alpha) \sin (\beta)$.

I think it is right, but I'm unsure.

2. It look correct to me.

$\displaystyle \nabla f(x,y.z) = (yz, xz, xy)$

$\displaystyle \nabla f(1,0,1) = (0,1,0)$

$\displaystyle \bold{u}$ is already a unit vector so it doesn't need to be normalized

$\displaystyle D_{\bold{u}}f(1,0,1) = \nabla f(1,0,1) \cdot \bold{u} = \sin \alpha \sin \beta$

3. Originally Posted by Random Variable
It look correct to me.

$\displaystyle \nabla f(x,y.z) = (yz, xz, xy)$

$\displaystyle \nabla f(1,0,1) = (0,1,0)$

$\displaystyle \bold{u}$ is already a unit vector so it doesn't need to be normalized

$\displaystyle D_{\bold{u}}f(1,0,1) = \nabla f(1,0,1) \cdot \bold{u} = \sin \alpha \sin \beta$
Ok thanks. Yeah I've realized I had to check out if $\displaystyle \bold u$ was normalized and I did the calculations after I posted here. Thanks for pointing it out.