# Thread: Prove that dy/dx = nx^n-1

1. ## Prove that dy/dx = nx^n-1

Can anyone help me out and Prove that dy/dx = nx^n-1.

2. Originally Posted by sderosa518
Can anyone help me out and Prove that dy/dx = nx^n-1.
I assume you mean find $\displaystyle \frac{dy}{dx}$, where $\displaystyle y(x) = x^n$ for $\displaystyle n \geq 1$. If so:

$\displaystyle \frac{dy}{dx} = \lim_{h \to 0} \frac{y(x+h) - y(x)}{h} = \lim_{h \to 0}\frac{(x+h)^n - x^n}{h} =$

$\displaystyle = \lim_{h \to 0}\frac{\sum_{k=0}^{n} (\binom{n}{k}x^kh^{n-k}) - x^n}{h} =$ $\displaystyle \lim_{h \to 0} \frac{\binom{n}{0}h^n + \binom{n}{1}xh^{n-1} + ... + \binom{n}{n-1}hx^{n-1} + \overbrace{\binom{n}{n}}^{=1}x^n - x^n}{h} =$

$\displaystyle \lim_{h \to 0} \binom{n}{0}h^{n-1} + \binom{n}{1}xh^{n-2} + ... + \overbrace{\binom{n}{n-1}}^{=n}x^{n-1} = nx^{n-1}$

There is probably a much easier way...

3. Originally Posted by sderosa518
Can anyone help me out and Prove that dy/dx = nx^n-1.

4. Originally Posted by Defunkt
I assume you mean find $\displaystyle \frac{dy}{dx}$, where $\displaystyle y(x) = x^n$ for $\displaystyle n \geq 1$. If so:

$\displaystyle \frac{dy}{dx} = \lim_{h \to 0} \frac{y(x+h) - y(x)}{h} = \lim_{h \to 0}\frac{(x+h)^n - x^n}{h} =$

$\displaystyle = \lim_{h \to 0}\frac{\sum_{k=0}^{n} (\binom{n}{k}x^kh^{n-k}) - x^n}{h} =$ $\displaystyle \lim_{h \to 0} \frac{\binom{n}{0}h^n + \binom{n}{1}xh^{n-1} + ... + \binom{n}{n-1}hx^{n-1} + \overbrace{\binom{n}{n}}^{=1}x^n - x^n}{h} =$

$\displaystyle \lim_{h \to 0} \binom{n}{0}h^{n-1} + \binom{n}{1}xh^{n-2} + ... + \overbrace{\binom{n}{n-1}}^{=n}x^{n-1} = nx^{n-1}$

There is probably a much easier way...

What about the following apparently simpler one?:

$\displaystyle (x^n)'=\lim_{x\rightarrow x_0}\frac{x^n-x_0^n}{x-x_0}=\lim_{x\rightarrow x_0}\frac{(x-x_0)(x^{n-1}+x^{n-2}x_0+...+x_0^{n-1})}{x-x_0}$ $\displaystyle =\lim_{x\rightarrow x_0}(x^{n-1}+x^{n-2}x_0+...+x_0^{n-1})$ $\displaystyle =x_0^{n-1}+x_0^{n-2}x_0+...+x_0^{n-1}=nx_0^{n-1}$

Tonio

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