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Elementary power rule of derivatives/Proof

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\frac{d}{dx}x^n=nx^{n-1}, for n is a natural number.

Proof by Induction


  • The product rule of derivatives, \frac{d}{dx}uv=uv'+vu'


The proof proceeds by mathematical induction. Take the base case k=0. Then:


The induction hypothesis is that the rule is true for n=k:


We must now show that it is true for n=k+1:


Since the power rule is true for k=0 and given k is true, k+1 follows, the power rule is true for any natural number.


Proof by Exponentiation

This proof is validates the power rule for all real numbers 'n' such that the derivative d/dx x^n = nx^(n-1)


  • The exponential rule of derivatives, \frac{d}{dx} e^x = e^x
  • The chain rule of derivatives, \frac{d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)


\frac{d}{dx} x^n= \frac{d}{dx} (e^{\ln(x)})^n
\frac{d}{dx} x^n= \frac{d}{dx} e^{n \cdot\ln(x)}
\frac{d}{dx} x^n= e^{n \cdot\ln(x)} \cdot \frac{d}{dx}(n \cdot \ln(x))
\frac{d}{dx} x^n= x^{n} \cdot n \cdot \frac{1}{x}
\frac{d}{dx} x^n= n\cdot x^{n-1}

Proof by Binomial Expansion

This proof is only valid for positive real integer 'n' exponents, x^n


  • The Binomial Expansion:
(a+b)^n = \sum_{i=0}^n \frac{n!}{i!(n-i)!} \cdot a^{n-i} \cdot b^i
  • The first principle limit definition of the derivative
\frac{d}{dx} f(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}


\frac{d}{dx} x^n = \lim_{h\to 0} \frac{(x+h)^n - x^n}{h}
\frac{d}{dx} x^n = \lim_{h\to 0} \frac{\left(\sum_{i=0}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^i\right) - x^n}{h}
\frac{d}{dx} x^n = \lim_{h\to 0} \frac{\left(x^n + nx^{n-1}h + \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^i\right) - x^n}{h}
\frac{d}{dx} x^n = \lim_{h\to 0} \frac{nx^{n-1}h + \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^i}{h}
\frac{d}{dx} x^n = \lim_{h\to 0} nx^{n-1} + \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^{i-1}
\frac{d}{dx} x^n = nx^{n-1} + \lim_{h\to 0} \sum_{i=2}^n \frac{n!}{i!(n-i)!} \cdot x^{n-i} \cdot h^{i-1}
\frac{d}{dx} x^n = nx^{n-1} + 0
\frac{d}{dx} x^n = nx^{n-1}

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