10.3 Differentiating polynomials

To differentiate f(x)=xnf(x)=x^{n} some algebra is required77 7 If you’re not confident in conducting algebraic manipulations (i.e. \saydoing algebra) then it really is worth spending time reviewing this; it’s foundational for everything else in mathematics.

limh0f(x+h)f(x)h\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} =limh0(x+h)nxnh\displaystyle=\lim_{h\to 0}\frac{(x+h)^{n}-x^{n}}{h} (10.17)
=limh0xn+(n1)xn1h+(n2)xn2h2++hnxnh\displaystyle=\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} (10.18)
=limh0(n1)xn1h+(n2)xn2h2++hnh\displaystyle=\lim_{h\to 0}\frac{\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^{2}% +...+h^{n}}{h} (10.19)
=limh0(n1)xn1+(n2)xn2h++hn1\displaystyle=\lim_{h\to 0}\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}h+...+h^{n-1} (10.20)
=nxn1\displaystyle=nx^{n-1} (10.21)

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 xn+(xn)=0x^{n}+(-x^{n})=0

  • In Equation 10.20 we divided through by hh.

  • In the final step, we applied the property that hXh\cdot X (where XX is some expression88 8 Where XX\in\mathbb{R}) is 0 as h0h\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 x2+3x8x^{2}+3x-8 (which was the example used above).

ddx(x2+3x8)\displaystyle\frac{d}{dx}(x^{2}+3x-8) =ddx[x2]+ddx[3x]+ddx[8]\displaystyle=\frac{d}{dx}[x^{2}]+\frac{d}{dx}[3x]+\frac{d}{dx}[-8] (10.22)
=2x+3\displaystyle=2x+3 (10.23)

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

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

limh08(8)h\displaystyle\lim_{h\to 0}\frac{-8-(-8)}{h} =limh00h\displaystyle=\lim_{h\to 0}\frac{0}{h} (10.24)
=0\displaystyle=0 (10.25)