Q:

Computing Lie derivative of a sum

I am very well aware of how you compute Lie derivatives or tensor fields and just vector fields via the Lie bracket, but what confuses me is when there are addition operation involved in say the tensor field or the vector fields for example, how would you compute something like this?
$$\mathcal L_X h=L_X \bigg(\frac{x}{x+1}dx\otimes dx+y^2 dy\otimes dy  + \sin z dz\otimes dz\bigg)$$
Where the tensor $h$ is a $2-$tensor defined on some manifold $N =\mathbb{R}\times \mathbb{R}^2: x\in\mathbb{R}$ and $(y,z) \in \mathbb{R}^2.$ I would appreciate a clear example so that I can compute Lie derivatives in a better way. 

A:

The Lie derivative is a derivation of the algebra of tensor fields. That means 

$\mathcal{L}_X(cT) = c\mathcal{L}_XT$ for all $c \in \mathbb{R}$, 
$\mathcal{L}_X(T_1 + T_2) = \mathcal{L}_X(T_1) + \mathcal{L}_X(T_2)$, and
$\mathcal{L}_X(T_1\otimes T_2) = \mathcal{L}_X(T_1)\otimes T_2 + T_1\otimes\mathcal{L}_X(T_2)$.

By the second dot point, we have
\begin{align*}
\mathcal L_X h &= \mathcal{L}_X \left(\frac{x}{x+1}\,dx\otimes dx+y^2\, dy\otimes dy  + \sin z\, dz\otimes dz\right)\\
&= \mathcal{L}_X \left(\frac{x}{x+1}\,dx\otimes dx\right) +\mathcal{L}_X(y^2\, dy\otimes dy)  + \mathcal{L}_X(\sin z\, dz\otimes dz).
\end{align*}
Now each of these three terms can be further expanded using the third dot point. I will only do the first one.
$$\mathcal{L}_X\left(\frac{x}{x+1} dx\otimes dx\right) = \mathcal{L}_X\left(\frac{x}{x+1}\right)dx\otimes dx + \frac{x}{x+1}\mathcal{L}_X(dx)\otimes dx + \frac{x}{x+1}dx\otimes\mathcal{L}_X(dx).$$
It follows from Cartan's magic formula that for any differential form $\omega$, we have $\mathcal{L}_X(d\omega) = d(\mathcal{L}_X\omega)$. So now the computation reduces to calculating the Lie derivative of a function $f$ which is given by $\mathcal{L}_X(f) = X(f)$.
If $X = X^x\partial_x + X^y\partial_y + X^z\partial_z$, then 
$$\mathcal{L}_X\left(\frac{x}{x+1}\right) = X\left(\frac{x}{x+1}\right) = X^x\partial_x\left(\frac{x}{x+1}\right) = \frac{X^x}{(x+1)^2}$$
and 
$$\mathcal{L}_X(dx) = d(\mathcal{L}_X x) = d(X(x)) = d(X^x\partial_xx) = dX^x$$
so 
\begin{align*}
\mathcal{L}_X\left(\frac{x}{x+1} dx\otimes dx\right) &= \mathcal{L}_X\left(\frac{x}{x+1}\right)dx\otimes dx + \frac{x}{x+1}\mathcal{L}_X(dx)\otimes dx + \frac{x}{x+1}dx\otimes\mathcal{L}_X(dx)\\
&= \frac{X^x}{(x+1)^2}\, dx\otimes dx + \frac{x}{x+1}\, dX^x\otimes dx + \frac{x}{x+1}\, dx\otimes dX^x.
\end{align*}

