In my last post, I talked about how one could guess the equation of the tangent line to a polynomial and then check whether that guess is right by factoring. Let us now turn to the question of computing the derivative without relying on guesswork.
Consider the question we answered in the previous post: what is the tangent line to $y=x^2$ at $x=1$? Suppose that we don’t even have a guess. All we can say is that the tangent line has an equation of $y=mx+b$ for some values of $m$ and $b$. Just like before, the graph of $x^2-(mx+b)$ must be flat against the $x$-axis at $x=1$. Algebraically, that means that $x^2-(mx+b)$ is divisible by $(x-1)^2$. How do we know if that happens?
One approach is to use long division: if we divide $x^2-mx-b$ by $x^2-2x+1$, we get a quotient of $1$ with a remainder of $(2-m)x+(-b-1)$. If $x^2-(mx+b)$ is actually divisible by $(x-1)^2$, then our remainder should have been $0$. That only happens if $m=2$ and $b=-1$. Therefore, the equation of the tangent line to $y=x^2$ at $x=1$ is, indeed, $y=2x-1$.
Using long division, we can now compute the tangent line, and hence the derivative, of a polynomial at a given point. But we can do even better.
Suppose we want to compute the derivative function. We can do that as well. Let us start with a new polynomial, $y=x^3-x^2$. What is the derivative of this polynomial at a general point $x=a$? Well, the tangent line at $x=a$ looks like $y=mx+b$ for some values of $m$ and $b$. Then the function $(x^3-x^2)-(mx+b)$ must be flat against the $x$-axis at $x=a$, so it is divisible by $(x-a)^2$. Long division tells us that $x^3-x^2-mx-b$ divided by $x^2-2ax+a^2$ gives a quotient of $x+2a-1$ and a remainder of $(-m+3a^2-2a)x+(-b-2a^3+a^2)$. Since the remainder needs to be $0$, it must be the case that $m=3a^2-2a$ and $b=-2a^3+a^2$. In particular, the slope at $x=a$ is $3a^2-2a$. In other words, $y'(a)=3a^2-2a$, or after changing the variable $y'(x)=3x^2-2x$. This should look familiar to those who know Power Rule.
Using this technique, we can compute the derivative of any polynomial algebraically, with no need for limits. We can even prove some useful shortcuts like Sum Rule, Constant Multiple Rule, and Power Rule. If polynomials were the only functions in calculus, we would have no need to develop limits. However, other functions do exist, so we will now turn our attention to figuring out what limits are.