The differential of inclusion from S^{2} to R^{3} I've found this question on Loring Tu's "Introduction to Manifolds", numbered 11.4 and titled "Differential of the inclusion map"

On the upper hemisphere of the unit sphere $S^{2}$ we have the coordinate map $\phi = (u,v)$, where $u(a,b,c) = a, \ v(a,b,c) = b$ 
Let $i: S^{2} \to \mathbb{R^3}$ be the inclusion, and $i_*$ it's differential. 
Calculate:
$$
i_*(\frac{\partial}{\partial u}), \ \ 
i_*(\frac{\partial}{\partial v}) 
$$
in terms of  $ \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$, where x,y,z are the standard coordinates

I've fiddled with it a bit, and the closest I've came is something like:
$$i_*(\frac{\partial}{\partial u})(x) = (\frac{\partial}{\partial u})(x \circ i)  = (\frac{\partial}{\partial u}) (u) = 1\\
i_*(\frac{\partial}{\partial u})(y) = (\frac{\partial}{\partial u})(y \circ i) = 0 \\
i_*(\frac{\partial}{\partial u})(z) = (\frac{\partial}{\partial u}) (z \circ i)
$$
I can't really wrap my head around
calculating $(\frac{\partial}{\partial u}) (z \circ i)$.
Some guidance would be very much appreciated.

Here $z\circ i \circ \phi^{-1}=\sqrt{1-u^2-v^2}$ and so
$\frac{\delta}{\delta u} \sqrt{1-u^2-v^2}=$
$=\frac{-u}{\sqrt{1-u^2-v^2}}=-\frac{x}{z}$
so that
$i_*(\frac{\delta}{\delta u })=\frac{\delta}{\delta x}-\frac{x}{z}\frac{\delta}{\delta z}$
Similarly
$i_*(\frac{\delta}{\delta v })=\frac{\delta}{\delta y}-\frac{y}{z}\frac{\delta}{\delta z}$
Merry Christmas 
