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Let F (x) = 2^x. Which of the following represents the difference quotient with h = 0.001 which can be used to estimate F '(3)?
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Maybe: \(\dfrac{2^{3.001} + 2^3}{0.001}\)?
Whoops! Minus, not Plus. \(\dfrac{2^{3.001} - 2^{3}}{0.001}\). Sorry about that.
later on you will know it is \(2^3\times \ln(2)\)
note that \[\frac{2^{3.001}-2^3}{.001}=2^3\left(\frac{2^.001-1}{.001}\right)\] by the laws of exponents
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