Proving the product rule for differentiation

[Aion]

Question:

I am working on a problem involving the product of two polynomials and their derivatives, and I need some help understanding a specific part of the derivation.


Suppose we have two polynomials:

$$p(x)= a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$$
$$q(x)= b_0 + b_1x + b_2x^2 + \cdots + b_nx^n$$
We want to prove that:

$$(pq)'(x)= p'(x)q(x) + p(x)q'(x)$$
To be more specific, we have shown that the coefficient of $x^{k-1}$ in the product $pq$ is equal to

$$k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)$$
However, the part I am struggling with involves finding the coefficient of $x^{k-1}$ in $p'(x)q(x)+p(x)q'(x)$. I want to understand why the same result holds for the expression $p'(x)q(x)+p(x)q'(x)$. I have tried expanding both products $p'(x)q(x)$ and $p(x)q'(x)$ and then combining the terms, but I'm not sure how to prove that the sum of these terms also yields the coefficient $k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)$.


My solution attempt is the following:

Suppose
Let $p'(x)=c_0+c_1x+\cdots +c_{n-1}x^{n-1}$ where $c_0=a_1,c_1=2a_2, \cdots, c_{n-1}=na_n$

and $q'(x)=d_0+d_1x+\cdots d_{n-1}x^{n-1}$ for $d_0=b_1, d_1=2b_2,\cdots, d_{n-1}=nb_n$

Now consider the product $p'(x)q(x)$. Its terms will be of the form $c_ib_jx^{i+j}$, and for the exponent of $x$ to equal $x^{k-1}$ we must have $i+j=k-1$. Hence the coefficient is equal to

$$c_0b_{k-1}+c_1b_{k-2}+\cdots+c_{k-1}b_0$$
By substitution, we find that the coefficient for $x^{k-1}$ in the product $p'(x)q(x)$ is equal to

$$a_1b_{k-1}+2a_2b_{k-2}+\cdots+ka_kb_0$$
By a similar method, we find the coefficient for $x^{k-1}$ in the product $p(x)q'(x)$ is equal to

$$ka_0b_k+\cdots+2a_{k-2}b_2+a_{k-1}b_1$$
Here I'm stuck as I fail to see how the addition of these two expressions equals $$k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)$$

 

[Dr.Peterson]

I am working on a problem involving the product of two polynomials and their derivatives, and I need some help understanding a specific part of the derivation.

Why would you want to prove this specifically for polynomials, when it can be proved (more easily, I think) for any differentiable functions? Am I missing something?

But it may help you if you write two terms at each end of those summations:

By substitution, we find that the coefficient for $x^{k-1}$ in the product $p'(x)q(x)$ is equal to

$$a_1b_{k-1}+2a_2b_{k-2}+\cdots+ka_kb_0$$
By a similar method, we find the coefficient for $x^{k-1}$ in the product $p(x)q'(x)$ is equal to

$$ka_0b_k+\cdots+2a_{k-2}b_2+a_{k-1}b_1$$
Here I'm stuck as I fail to see how the addition of these two expressions equals $$k(a_0b_k + a_1b_{k-1} + \cdots + a_kb_0)$$

$$a_1b_{k-1}+2a_2b_{k-2}+\cdots+{\color{red}(k-1)a_{k-1}b_1}+ka_kb_0$$
$$ka_0b_k+{\color{red}(k-1)a_1b_{k-1}}+\cdots+2a_{k-2}b_2+a_{k-1}b_1$$
Add those together, term by term, and factor out k.

 

[Aion]

Why would you want to prove this specifically for polynomials, when it can be proved (more easily, I think) for any differentiable functions? Am I missing something?

But it may help you if you write two terms at each end of those summations:

$$a_1b_{k-1}+2a_2b_{k-2}+\cdots+{\color{red}(k-1)a_{k-1}b_1}+ka_kb_0$$
$$ka_0b_k+{\color{red}(k-1)a_1b_{k-1}}+\cdots+2a_{k-2}b_2+a_{k-1}b_1$$
Add those together, term by term, and factor out k.

Thank you, Dr. Peterson! I completely missed that simplification.