## 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. $\square$