A calculus of the absurd

10.3 Differentiating polynomials

To differentiate \(f(x) = x^n\) some algebra is required6969 If you’re not confident in conducting algebraic manipulations (i.e. “doing algebra”) then it really is worth spending time reviewing this; it’s foundational for everything else in mathematics.

\begin{align} \lim _{h \to 0} \frac {f(x+h) - f(x)}{h} &= \lim _{h \to 0} \frac {(x+h)^n - x^n}{h} \\ &= \lim _{h \to 0} \frac { x^n + \binom {n}{1} x^{n-1}h + \binom {n}{2} x^{n-2}h^2 + ... + h^n - x^n}{h} \label {applying binomial theorem} \\ &= \lim _{h \to 0} \frac {\binom {n}{1} x^{n-1}h + \binom {n}{2} x^{n-2}h^2 + ... + h^n}{h} \label {cancellation} \\ &= \lim _{h \to 0} \binom {n}{1} x^{n-1} + \binom {n}{2} x^{n-2}h + ... + h^{n-1} \label {div by h} \\ &= nx^{n-1} \end{align}

Note that in the process of carrying out the expansion

  • • In Equation 10.18 we used the binomial theorem (as in Equation 3.64).

  • • In Equation 10.19 we used the fact that \(x^n + (-x^n) = 0\)

  • • In Equation 10.20 we divided through by \(h\).

  • • In the final step, we applied the property that \(h \cdot X\) (where \(X\) is some expression7070 Where \(X \in \mathbb {R}\)) is \(0\) as \(h \to 0\).

We can then combine this with the rule for the derivatives of sums from above to find the derivatives of any polynomial.

For example, we can find the derivative of \(x^2+3x-8\) (which was the example used above).

\begin{align} \frac {d}{dx}(x^2+3x-8) &= \frac {d}{dx}[x^2] + \frac {d}{dx}[3x] + \frac {d}{dx}[-8] \\ &= 2x + 3 \end{align}

Why is \(\frac {d}{dx}(-8) = 0\)?

Let’s suppose we have a function \(f(x)=c\), then the derivative of \(f(x)\) is just

\begin{align} \lim _{h \to 0} \frac {-8 - (-8)}{h} &= \lim _{h \to 0} \frac {0}{h} \\ &= 0 \end{align}