Properties of derivatives

Let R be a commutative ring and let p(x) = \sum_{i=0}^n c_ix^i \in R[x]. Recall that the derivative D_x(p) is defined to be D_x(p)(x) = \sum_{i=0}^{n-1} (i+1)c_{i+1}x^i. Prove that for all f,g \in R[x], D_x(f+g) = D_x(f) + D_x(g), D_x(fg) = D_x(f)g + fD_x(g), and (for good measure) D_x(f \circ g) = (D_x(f) \circ g)D_x(g).

Let f(x) = \sum_{i=0}^n c_ix^i and g(x) = \sum_{i=0}^n d_ix^i. (For ease of exposition, pad f or g with zero terms so that they have the same nominal degree.)

Now D_x(f+g)(x) = \sum_{i=0}^{n-1} (i+1)(c_{i+1}+d_{i+1})x^i = \sum_{i=0}^{n-1} (i+1)c_{i+1}x^i + \sum_{i=0}^{n-1} (i+1)d_{i+1}x^i = D_x(f)(x) + D_x(g)(x).

We will now prove the ‘product rule’ in pieces. First, for monic monomials.

Lemma 0: If r \in R, then D_x(rf) = rD_x(f). Proof: D_x(rf) = D_x(\sum_{i=0}^n rc_ix^i) = \sum_{i=0}^{n-1}r(i+1)c_{i+1}x^i = rD_x(f). \square

Lemma 1: For all n,m \in \mathbb{N}, we have D_x(x^nx^m) = D_x(x^n)x^m + x^nD_x(x^m). Proof: If n = 0, then D_x(x^0x^m) = D_x(1 \cdot x^m) = D_x(x^0)x^m + x^0D_x(x^m) as desired. Similarly for m = 0. Now D_x(x^{n+1}x^{m+1}) = D_x(x^{n+m+2}) = (n+m+2)x^{n+m+1} (n+1)x^nx^{m+1} + x^n(m+1)x^m = D_x(x^n)x^m + x^nD_x(x^m). \square

Lemma 2: For all m \in \mathbb{N} and f \in R[x], D_x(x^mf)(x) = D_x(x^m)f + x^mD_x(f). Proof: We have D_x(x^mf) = D_x(\sum_{i=0}^n c_ix^mx^i) = \sum_{i=0}^n c_i D_x(x^mx^i) = \sum_{i=0}^n c_i(D_x(x^m)x^i + x^mD_x(x^m)) = \sum_{i=0}^n c_iD_x(x^i)x^m + \sum_{i=0}^n c_ix^iD_x(x^m) = D_x(x^m)f + x^mD_x(f) as desired. \square

Lemma 3: For all f,g \in R[x], D_x(fg) = D_x(f)g + fD_x(g). Proof: We have D_x(fg) = D_x(\sum_{i=0}^n d_ix^ig) = \sum_{i=0}^n d_i D_x(x^ig) = \sum_{i=0}^n d_i (D_x(x^i)g + x^iD_x(g)) = \sum_{i=0}^n d_i D_x(d^i)g + \sum_{i=0}^n d_ix^iD_x(g) = D_x(f)g + fD_x(g) as desired. \square

Now recall that if f(x) = \sum c_ix^i, then (f \circ g)(x) = \sum c_ig(x)^i.

Lemma 4: For all m \in \mathbb{N}, D_x(x^m \circ f) = (D_x(x^m) \circ f)D_x(f). Proof: We proceed by induction on m. For m = 0 and m = 1, the result is clear. (Note that 1 \circ f = 1.) Suppose the result holds for some m \geq 2. Now D_x(x^{m+2} \circ f) = D_x(ff^{m+1}) = D_x(f)f^{m+1} + fD_x(f^{m+1}) = D_x(f)f^{m+1} + (m+1)f^{m+1}D_x(f) ((m+2)f^{m+1})D_x(f) = (D_x(x^{m+2}) \circ f)D_x(f) as desired. \square

Lemma 5: For all f,g,h \in R[x], (f+g) \circ h = (f \circ h) + (g \circ h). Proof: Let f = \sum c_ix^i, g = \sum d_ix^i, and h = \sum e_ix^i, padding to make the degrees nominally n. Then (f+g) \circ h = \sum(c_i+d_i)x^i \circ h = \sum c_ih^i + \sum d_ih^i = (f \circ h) + (g \circ h). \square

Lemma 6: For all f,g \in R[x], D(f \circ g) = (D(f) \circ g)D(g). Proof: We have D_x(f \circ g) = D_x(\sum c_ix^i \circ g) = \sum c_i D_x(x^i \circ g) = \sum c_i (D_x(x^i) \circ g)D_x(g) = (D(f) \circ g)D(g) as desired. \square

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