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}