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Math Help - Need help with derivative

  1. #1
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    Need help with derivative

    Hello, this is my first post. I need help finding the derivative of 1/x^2 at x=2. The answer is supposed to be -2/x^3 and i need to find the derivative by definition. I have worked this out a million times and i keep making mistakes in my algebra i suppose because no matter what i try, i cannot get -2/x^3. any help would be greatly appreciated.

    also, this is the first time ive ever heard of latex so hopefully i got it right, if i didnt i apologize
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  2. #2
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    \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{1}{(x+h)^2}-\frac{1}{x^2}\right)

    \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{1}{x^2+2xh+h^2}-\frac{1}{x^2}\right)

    \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{x^2-(x^2+2xh+h^2)}{x^2(x^2+2xh+h^2)}\right)

    \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{-2xh-h^2}{x^2(x^2+2xh+h^2)}\right)

    \displaystyle \lim_{h\to 0}\frac{1}{h}\left(\frac{h(-2x-h)}{x^2(x^2+2xh+h^2)}\right)

    \displaystyle \lim_{h\to 0}\frac{-2x-h}{x^2(x^2+2xh+h^2)}

    \displaystyle \frac{-2x+0}{x^2(x^2+2x(0)+(0)^2)} = \dots
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  3. #3
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    I always use substitution when it comes to finding derivatives using the limit definition:

    We have \displaystyle f'(x) = \lim_{h \to 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}. Let x+h = a and x = b, then h = b-x = a-b, thus:

    \begin{aligned} f'(x) & = \lim_{a \to b}\frac{\frac{1}{a^2}-\frac{1}{b^2}}{a-b} = \lim_{a \to b}\frac{b^2-a^2}{a^2b^2(a-b)} = \lim_{a\to{b}}\frac{-(a-b)(a+b)}{a^2b^2(a-b)} \\& = -\lim_{a\to{b}} \frac{a+b}{a^2b^2} = -\frac{2b}{b^4} = -\frac{2}{b^3} = -\frac{2}{x^3}   ~ \diamond \end{aligned}<br />



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