Question 1005097
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ +\ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\right 0}\ \frac{f(x\ +\ h)\ -\ f(x)}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\right 0}\ \frac{(x\ +\ h)^2\ +\ (x\ +\ h)\ -\ (x^2\ +\ x)}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\right 0}\ \frac{(x^2\ +\ 2xh\ + h^2)\ +\ (x\ +\ h)\ -\ (x^2\ +\ x)}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\right 0}\ \frac{2xh\ +\ h^2\ +\  h)}{h}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\right 0}\ 2x\ +\ h\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ 2x\ +\ 1]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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