Derivatives of polynomials

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Some ideas in this proof are due to Tom Goodwillie. Theorem: For n= 0, 1, 2, . . . ,the derivative of the function xdefined by x(t) = $t^{n}$ is x˙ = $nt^{n-1}$ .

The results for n= 0 and 1 hold by direct application of the definition of the derivative.

For n>1, it suffices to prove $\dot{x}$ (0) = 0 and $\dot{x}$ (1) = n, since the result for other nonzero values of tthen follows by the kind of scaling argument used on page 13 for the n= 2 case.

We use the following properties of the derivative, all of which follow immediately from its definition as the slope of the tangent line:

Shift. Shifting a function x(t) horizontally to form a new function x(t+c) gives a derivative at any newly shifted point that is the same as the derivative at the corresponding point on the unshifted graph.

Flip. Flipping the function x(t) to form a new function x(−t) negates its derivative at t= 0.

Add. The derivative of the sum or difference of two functions is the sum or difference of their derivatives.

For even n, $\dot{x}$ (0) = 0 follows from the flip property, since x(−t) is the same function as x(t). For n= 3, 5, . . . , we apply the definition of the derivative in the same manner as was done in the preceding section for

n= 3.

We now need to show that $\dot{x}$ (1) = n. Define the function uas

u(t) = x(t+ 1) −x(t)

= 1 + nt+ ... 　 ,

where the second line follows from the binomial theorem, and. . . represents terms involving $t^{2}$ and higher powers. Since we’ve already established the results for n= 0 and 1, differentiation gives