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Power Rule

Tags

Calculus

Cegep/1

Word count

225 words

Reading time

2 minutes

Let $f(x)=x^n\text{, }n\in \mathbb{R}$, then

$$\frac{d}{dx}{x}^n=n{x}^{n-1}$$

General Power Rule

Let $f$ be a differentiable function and $h(x)=(f(x){)}^n$ for $n\in \mathbb{R}$, then

$$h^{\prime }(x)=\frac{\mathrm{d}}{\mathrm{d}x}(f(x){)}^n=n(f(x){)}^{n-1}\cdot f^{\prime }(x)$$

Proof

For now, this proof is only for $n \in \mathbb{N}$.

Note the fact that the following equality is true:

$$x^n-a^n=(x-a)\cdot \sum _{i=1}^n{x}^{n-i}{a}^{i-1}$$

Let $f(x)=x^n$, we have to show that $f^{\prime }(a)=n{a}^{n-1}$ for any $a$:

$$\begin{array}{rl}{f}^{\prime }(a)& =\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}\\ & =\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}\\ & =\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}\end{array}$$

Using the fact stated above, we have

$$\begin{array}{rl}& =\underset{x\to a}{lim}\frac{(x-a)\cdot \sum _{i=1}^n{x}^{n-i}{a}^{i-1}}{x-a}\\ & =\underset{x\to a}{lim}\sum _{i=1}^n{x}^{n-i}{a}^{i-1}\\ & =\sum _{i=1}^n{a}^{n-i}{a}^{i-1}\\ & =\sum _{i=1}^n{a}^{n-1}\\ & =n{a}^{n-1}\end{array}$$

$◻$

Contributors

Ajitani Hifumi

Changelog

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