SiRaLeX Posted May 4, 2016 Posted May 4, 2016 d/dx(0.5^x) = -0.693147×0.5^x d/dx(1^x) = 0 d/dx(1.5^x) = 0.405465×1.5^x d/dx(2^x) = 2^x log(2) d/dx(2.71^x) = 0.996949×2.71^x d/dx(e^x) = e^x d/dx(2.72^x) = 1.00063×2.72^x d/dx(3.14^x) = 1.14422×3.14^x d/dx(pi^x) = pi^x log(pi) Explain, anyone?
Jacko Posted May 4, 2016 Posted May 4, 2016 d/dx(0.5^x) = -0.693147×0.5^x d/dx(1^x) = 0 d/dx(1.5^x) = 0.405465×1.5^x d/dx(2^x) = 2^x log(2) d/dx(2.71^x) = 0.996949×2.71^x d/dx(e^x) = e^x d/dx(2.72^x) = 1.00063×2.72^x d/dx(3.14^x) = 1.14422×3.14^x d/dx(pi^x) = pi^x log(pi) Explain, anyone? Magic.
Nyerguds Posted May 4, 2016 Posted May 4, 2016 Man, that shit's been ages. You do your homework yourself, lol.
GiftS Posted May 7, 2016 Posted May 7, 2016 d/dx(0.5^x) = -0.693147×0.5^x d/dx(1^x) = 0 d/dx(1.5^x) = 0.405465×1.5^x d/dx(2^x) = 2^x log(2) d/dx(2.71^x) = 0.996949×2.71^x d/dx(e^x) = e^x d/dx(2.72^x) = 1.00063×2.72^x d/dx(3.14^x) = 1.14422×3.14^x d/dx(pi^x) = pi^x log(pi) Explain, anyone? Do it by definition if it confuses you...
Strategst Posted May 12, 2016 Posted May 12, 2016 The derivative of a function gives its instantaneous rate of change, or slope, at any given point. The exponential function just so happens to have the same value as its slope at every point. The simplest example is exp^0=1, where the slope of exp^x at x=0 is also 1. Continue inductively from there.
SiRaLeX Posted June 15, 2016 Author Posted June 15, 2016 Strategst, I know what a derivative is. Magic. Do it by definition if it confuses you... These are the best answers. Except, I'm not sure I can simplify the limit without L'H and knowing that the derivative of e^x is e^x.
Plokite_Wolf Posted June 15, 2016 Posted June 15, 2016 These are the best answers. Except, I'm not sure I can simplify the limit without L'H and knowing that the derivative of e^x is e^x. There is something called a Google search, which gives the answer (without the L'Hospital rule)... https://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs/e_to_the_x
SiRaLeX Posted June 15, 2016 Author Posted June 15, 2016 Thank you, Plokite_Wolf! It is all over Google, indeed. But your link is exactly what I was looking for!
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