If R from big G to G L V is a representation and X is in little g (this is not necessarily the X in little s l 2 C) we have (R to the nth tensor power) star of X applied to v_1 tensor dot dot dot tensor v_n equals R star X applied to v_1, tensor v_2 tensor dot dot dot tensor v_n, plus v_1 tensor R star X applied to v_2, tensor dot dot dot tensor v_n, plus dot dot dot, plus v_1 tensor dot dot dot tensor R star X applied to v_n In other words, we use the "Leibniz/product rule".
X and Y example
Review
So far we have seen that, given a complex representation R from S U 2 to G L V, we get:

a (real linear) representation R star from little s u 2 to little g l V of the real Lie algebra little s u 2

a (complex linear) representation R star superscript C from little s l 2 C to little g l V of its complexification little s l 2 C equals little s u 2 tensor C.
Define the basis H equals the diagonal matrix 1, 1, X is the 2by2 matrix 0, 1, 0, 0 and Y is the 2by2 matrix 0, 0, 1, 0 of little s u 2. We saw that V splits as a direct sum of weight spaces W_m where W_m is the eigenspace of R star superscript C of H with eigenvalue m and saw that X and Y act on weight spaces as follows:
X takes vectors in W_m to vectors in W_{m+2},
Y takes vectors in W_m to vectors in W_(m minus 2).
We will usually omit the superscript C from R star superscript C and actually, we'll usually omit the R star altogether.
In this video, I want to see how this actually works in practice for the example Sym 2 C 2. But first, we will need to understand how Lie algebras act on symmetric powers and, more generally, tensor products.
Leibniz rule
Recall that R tensor square of g applied to v_1 tensor v_2 equals R of g applied to v_1 tensor R of g applied to v_2. What is the Lie algebra representation corresponding to this? In other words, what is (R tensor squared) star?
For now, we will assume this lemma, work out the example and then justify the lemma.
Example: Sym^{2}(C^{2})
Let V be Sym 2 of C 2. If e_1 and e_2 form a basis for C 2 then e_1 squared (which equals e_1 tensor e_1), e_1 e_2 (which equals e_1 tensor e_2 plus e_2 tensor e_1) and e_2 squared (which equals e_2 tensor e_2) form a basis for Sym 2 C 2. The Lie algebra elements H, X and Y act on e_1 and e_2 as follows:

Since H is diagonal 1, 1, we have H e_1 = e_1, H e_2 equals minus e_2,

Since X equals 0, 1, 0, 0, we have X e_1 = 0, X e_2 = e_1,

Since Y equals 0, 0, 1, 0, we have Y e_1 = e_2, Y e_2 = 0.
Sym 2 H therefore sends e_1 squared equals e_1 tensor e_1 to (H e_1) tensor e_1 plus e_1 tensor (H e_1), which equals 2 e_1 squared by the product rule. Similarly, Sym 2 H of e_1 e_2 equals (H e_1) e_2 plus e_1 (H e_2), which equals e_1 e_2 minus e_1 e_2, which equals 0, and Sym 2 H of e_2 squared equals minus 2 e_2 squared. The weight spaces, that is the eigenspaces of H, are spanned by e_1 squared (weight 2), e_1 e_2 (weight 0) and e_2 squared (weight minus 2).
Let's compute Sym 2 X:
Sym 2 X of e_1 squared equals X e_1 times e_1 plus e_1 times X e_1, which equals 0.
Sym 2 X of e_1 e_2 equals X e_1 times e_2 plus e_1 times X e_2, which equals e_1 squared.
Sym 2 X of e_2 squared equals X e_2 times e_2 plus e_2 times X e_2, which equals 2 e_2 e_2.
Notice that in this last example we used the fact that e_1 e_2 = e_2 e_1 (which holds because it's a symmetric tensor).
In terms of our weight diagram, we can see that Sym 2 X is taking e_2 squared in W_(minus 2) to 2 e_1 e_2 in W_0 (increasing the weight by 2) and e_1 e_2 in W_0 to e_1 squared in W_2 (again, increasing the weight by 2). Finally, it takes e_1 squared in W_2 to 0 in W_4 (necessarily, since W_4 = 0).
It's an exercise to calculate Sym 2 Y.
Proof of lemma
We still need to prove that (R to the nth tensor power) star can be evaluated using the Leibniz rule, as stated in the lemma.
We have R to the nth tensor power of exp t X equals exp of (t (R to the nth tensor power) star X) by our usual formula for Lie algebra representations.
Let's apply both sides to v_1 tensor dot dot dot tensor v_n and expand both sides in powers of t. The righthand side is the identity plus t (R to the nth tensor power) star X plus terms of order t squared) applied to v_1 tensor dot dot dot tensor v_n, which equals v_1 tensor dot dot dot tensor v_n plus t (R to the nth tensor power) star applied to v_1 tensor dot dot dot tensor v_n, plus terms of order t squared. The lefthand side is R of exp t X v_1 tensor dot dot dot tensor R exp t X v_n.
We have R exp t X v_i equals (the identity plus t R star X plus higher order terms) applied to v_i, which equals v_i plus t R star X v_i plus higher order terms, so multiplying out the brackets, the lefthand side becomes: v_1 tensor dot dot dot tensor v_n, plus t times, (R star X v_1) tensor v_2 tensor dot dot dot tensor v_n, plus v_1 tensor R star X v_2 tensor dot dot dot tensor v_n, plus dot dot dot, plus v_1 tensor dot dot dot tensor R star X v_n, plus higher order terms.
By comparing the terms of order t, we get the desired formula for (R to the nth tensor power) star X.
Preclass exercise
Compute the action of \mathrm{Sym}^2(Y) on the basis e_1^2, e_1e_2, e_2^2.