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Determine the value of f'(x) from the first principle if f(x)=2x2-3x-5

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From the first principle, we know that,  $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$, now what we have to do is to plugin $f(x+h)$ and $f(x)$, simplify and find the limit.

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ f'(x) = \lim\limits_{h \to 0} \frac{ 2(x+h)^2 -3(x+h)-5-2x^2 +3x+5}{h} \\ f'(x) =\lim\limits_{h \to 0} \frac{2x^2 +4xh +2h^2-3x-3h-5-2x^2+3x+5}{h}\\ f'(x) = \lim\limits_{h \to 0} \frac{4xh+2h^2-3h}{h} \\ f'(x) = \lim\limits_{h \to 0} \frac{h(4x+2h-3)}{h} \\ f'(x) = \lim\limits_{h \to 0} 4x +2h -3\\ f'(x) = 4x-3$

Hence the final answer is $f'(x) = 4x-3$

by Wooden (882 points)

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