Question 190304
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(x) = x^2 + 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(x + h) = (x + h)^2 + 2(x + h) = x^2 + 2hx + h^2 + 2x + 2h], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f(x + h) - f(x)}{h} = \frac{x^2 + 2hx + h^2 + 2x + 2h -(x^2 + 2x)}{h} = \frac{h^2 + 2hx + 2h}{h} = h + 2x + 2]


And then, of course,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{df(x)}{dx} = \lim_{x\rightarrow h}\ {\frac{f(x+h)-f(x)}{h}\ =\ \lim_{x\rightarrow h}\ {h + 2x + 2}\ =\ 2x + 2]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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