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Higher-order polynomials

12 November, 2015 - 15:03

So far, we have the following results for polynomials up to order

function

derivative

1 0
t 1
t^2 2t

Interpreting 1 as t^0, we detect what seems to be a general rule, which is that the derivative of t^k is kt^{k-1}.
The proof is straightforward but not very illuminating if carried out with the methods developed in this chapter, so I've relegated it to Derivatives of polynomials. It can be proved much more easily using the methods of To infinity — and beyond!.

Example

If x=2t^7-4t+1, find \dot{x}.

This is similar to Example, the only difference being that we can now handle higher powers of t. The derivative of t^7 is 7t^6, so we have

\begin{align*} \dot{x} &=(2)(7t^6)+(-4)(1)+0 \\ &= 14t^6-4 \end{align*}

Example

Calculate 3^{-1} and 3.01^{-1}. Does this seem consistent with a conjecture that the rule for differentiating t^k holds for k < 0?

We have 3^{-1}\approx 0.33333 and 3^{-1}\approx 0.332223, the difference being -1.1\times 10^{-3}. This suggests that the graph of x=1/t has a tangent line at t=3 with a slope of about

\frac{-1.1\times 10^{-3}}{0.01}=-0.11

If the rule for differentiating t^k were to hold, then we would have \dot{x}=t^{-2}, and evaluating this at x = 3 would give -1/9, which is indeed about -0.11. Yes, the rule does appear to hold for negative k, although this numerical check does not constitute a proof. A proof is given in Example.