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Math Help - Prove that dy/dx = nx^n-1

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

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

    \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} =

     = \lim_{h \to 0}\frac{\sum_{k=0}^{n} (\binom{n}{k}x^kh^{n-k}) - x^n}{h} =  \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} =

    \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...
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  3. #3
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    Quote Originally Posted by sderosa518 View Post
    Can anyone help me out and Prove that dy/dx = nx^n-1.
    let "google" be your friend ...

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

    \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} =

     = \lim_{h \to 0}\frac{\sum_{k=0}^{n} (\binom{n}{k}x^kh^{n-k}) - x^n}{h} =  \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} =

    \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?:

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

    Tonio
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