What is my error in this $\nabla_{\vec{v}} f(x,y,z)$ at $\vec{a} = (-1, -1, 4)$ and $\vec{v} = (\frac{\sqrt 2}{2}, \frac{1}{2}, \frac{1}{2})$ problem I want to find gradient of $f(x,y,z) = \sqrt{xyz}$ in the direction of $\vec{v}$ at a point $\vec{a}$.
That is, $\nabla_{\vec{v}} f(x,y,z)$ at $\vec{a} = (-1, -1, 4)$ and $\vec{v} = (\frac{\sqrt 2}{2}, \frac{1}{2}, \frac{1}{2})$
I computed gradient of $f(x,y,z)$ along $\vec{v}$ to be
$\left(\begin{matrix} \sqrt{\frac{yz}{4x}} \\ \sqrt{\frac{xz}{4y}} \\ \sqrt{\frac{xy}{4z}} \end{matrix}\right)$.
So my answer for the value of gradient at $\vec{a}$ is $\left(\begin{matrix} 1 \\ 1 \\ \frac{1}{4}\end{matrix}\right)$.
But the answer given is
$\left(\begin{matrix} \frac{yz}{\sqrt{4xyz}} \\ \frac{xz}{\sqrt{4xyz}} \\ \frac{xy}{\sqrt{4xyz}} \end{matrix}\right)$.
So the accepted answer for value of gradient at $\vec{a}$ is $\left(\begin{matrix} -1 \\ -1 \\ \frac{1}{4}\end{matrix}\right)$.
Why is my answer wrong?

You need to be careful to differentiate the form $\sqrt{ax}$ when $a$ can be negative (in this case $a=yz$)
Especially $\sqrt{ax}=\sqrt{a}\sqrt{x}$ does not hold when $a$ is negative. I guess this is why you got wrong.
