$How do you prove that if $y = x^n$ then the derivative is $\frac{dy}{dx} = nx^{n-1}$$

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$How do you prove that if $y = x^n$ then the derivative is $\frac{dy}{dx} = nx^{n-1}$$

The title basically says it all. How do you prove that for any value of n this will always be the case.

Precalculus

Mr Smurf

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We have
\[(x^n)'=\lim_{h \to 0} {(x+h)^n-x^n\over h}\]
\[=\lim_{h \to 0} {x^n+nx^{n-1}h+{n(n-1)\over 2}x^{n-2}h^2+\cdots+h^n-x^n\over h} \]
\[=\lim_{h \to 0} nx^{n-1}+{\frac {n(n-1)}{2}}x^{n-2}h+\cdots+h^{n-1}\]

\[=nx^{n-1}, \]

since all terms exept the first one have a factor of $h$ and go to zero as $h$ goes to zero.

We have used the Binomial formula:

https://en.m.wikipedia.org/wiki/Binomial_theorem

Erdos

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Erdos

+1

Let me know if you need any clarifications.
- Mr Smurf
  
  +1
  
  Thanks for the solution :)
Erdos

+1

My pleasure :)

$How do you prove that if $y = x^n$ then the derivative is $\frac{dy}{dx} = nx^{n-1}$$

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