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I have to find the derivative of $\frac{2}{x^2}$ using only the definition. I have tried to do this several times but I always get the wrong answer - could you tell me where I made an error?

$$\frac{d}{dx}[\frac{2}{x^2}] = \lim_{h \to 0} \frac{\frac{2}{(x+h)^2}- \frac{2}{x^2}}{h} = \lim_{h \to 0} \frac{2x^2-2(x+h)^2}{x^2(x+h)^2h} = \lim_{h \to 0} \frac{-4xh -2h^2}{x^4h + 2x^3h^2 + x^2h^3} = \lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4}{x^4}$$ As you see, I keep getting $x^4$ instead of $x^3$ but I simply cannot see where I made a mistake - any suggestions?

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    $\begingroup$ You have $-4x$ within the limit before cancelling $h$, and haven't carried the $x$ through. $\endgroup$ – Mark Bennet Jan 9 '18 at 19:54
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    $\begingroup$ Your final expression should be $-4x/x^4$. $\endgroup$ – Yves Daoust Jan 9 '18 at 19:55
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You missed an $x$$$\lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4}{x^4}$$ It should have been $$\lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4x}{x^4}= \frac{-4}{x^3}$$

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From first principles we have: $$\frac{2}{(x + h)^2} - \frac{2}{x^2} = hf'(x)$$ $$\frac{2x^2 - 2x^2 - 4hx - 2h^2}{x^4 + 2hx^3 + h^2 x^2} = hf'(x)$$ $$-4hx - 2h^2 = (x^4 + 2hx^3 + h^2 x^2)hf'(x)$$ $$-4x - 2h = (x^4 + 2hx^3 + h^2 x^2)f'(x)$$ If we set $h → 0$ this becomes: $$-4x = x^4 f'(x)$$ $$f'(x) = \frac{-4}{x^3}$$

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