a unit vector in the direction in which $f$ increases most rapidly $f(x,y,z)=\sqrt {y^2-\sin (x+2z)} $, could anyone tell me how to find a unit vector in the direction in which $f$ increases most rapidly at the point $p=(0,2,0)$? Thanks for helping, I have no idea how to start. 
 A: The direction of steepest ascent will be in the direction of the gradient. So if you find the vector $\nabla f = (\partial_x f, \partial_y f, \partial_z f)$ evaluated at your point and normalize this vector, you should be golden. You may have to worry about points at which your function is not differentiable, for example near parameter at which the argument of square root is zero. However since here $y^2 = 4$ and $\sin$ varies between $-1$ and $1$ you should not have to worry about this.
A: As @DominicReiss said, the steepest ascent is along gradient. Also notice:
$$f=\sqrt g \Longrightarrow f'_t=\frac{g'_t}{2\sqrt g}=\frac{g'_t}{2f} \Longrightarrow \nabla f \ \| \ \nabla g$$
Since we need unit vector only, we can deal with $g$:
$$\nabla g = (g'_x,g'_y,g'_z)=(-\cos(x+2z),2y,-2\cos(x+2z)) \Longrightarrow$$
$$\Longrightarrow \nabla g(p) = \nabla g(0,2,0) = (-1,4,-2)$$
So the unit vector will be as follows:
$$u=\frac{\nabla f}{|\nabla f|}=\frac{\nabla g}{|\nabla g|}=(\frac{-1}{\sqrt{21}},\frac{4}{\sqrt{21}},\frac{-2}{\sqrt{21}})$$
A: Hint: differentiate function by x, y and z, i.e. find partial derivatives
$$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$$
and calculate them in point p=(0,2,0)
Those three values will be the vector with direction you need
A: This is what $$\nabla f=(\partial_xf,\partial_yf,\partial_zf)$$ 
is, just evaluate it at the point $p$ and then normalise it so that it is unit ($\nabla f /\|\nabla f\|$).
