Q. 86 - Calculus

Question

Prove, in two ways, that the power rule holds for negative integer powers

a) by using the $z\to x$ definition of the derivative

b) by using the $h\to 0$ definition of the derivative

Step-by-Step Solution

Answer

We prove the power rule for negative powers.

Step 1: Given information

We are given a function $f\left(x\right)=x^{-n}$

Step 2: Find the derivative using z → x

We get,

$$\begin{gathered} \underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x} \\ \underset{z\to x}{\lim }\frac{z^{-n}-x^{-n}}{z-x} \\ \underset{z\to x}{\lim }\frac{z^k-x^k}{z-x} [Replace -n by k] \\ \underset{z\to x}{\lim }\frac{(z-k)(z^{k-1}+z^{k-2}x+....+x^{k-1})}{z-x} \\ \underset{z\to x}{\lim }(z^{k-1}+z^{k-2}x+....+x^{k-1}) \\ =x^{k-1}+x^{k-1}+.....+x^{k-1} \\ =k{x}^{k-1} \\ Now replace k by -n \\ =-n{x}^{-n-1} \end{gathered}$$

Step 3: Using h → 0

We get,

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ \underset{h\to 0}{\lim }\frac{{(x+h)}^{-n}-x^{-n}}{h} \\ Replace -n by k \\ \underset{h\to 0}{\lim }\frac{{(x+h)}^k-x^k}{h} \\ \underset{h\to 0}{\lim }\frac{x^k+k{x}^{k-1}h+....+h^k-x^k}{h} \\ \underset{h\to 0}{\lim }\frac{k{x}^{k-1}h+....+h^k}{h} \\ =k{x}^{k-1} \\ Now replace k by -n \\ =-n{x}^{-n-1} \end{gathered}$$