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haiha
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How can can prove that the derivative of y=(1+1/x)^x is always positive?
Thank you
Thank you
haiha said:y' = (1+1/x)^x [ln(1+1/x)-1/(x+1)]
The problem here is that I can not prove y' always positive
it's not. at x = 0 it's undefined, for instancehaiha said:How can can prove that the derivative of y=(1+1/x)^x is always positive?
Thank you
kesh said:it's not. at x = 0 it's undefined, for instance
because the question asked about the function without any constraints on x's domain.radou said:So?
kesh said:because the question asked about the function without any constraints on x's domain.
part of the problem people have while tackling questions like this is they don't emphasise the domain, as you pointed out yourself
also a graph may make it "pretty obvious", but this isn't a proof
depends how far into analysis you get. undefined points are crucial and interesting at some levelsradou said:Well, constraints on domains are self understood. It wouldn't make any sense to talk about values of the function on an interval where it isn't defined, would it?
being really pedantic: the domain is any set you (or the questioner) chooses so long as the rule is defined on that setHallsofIvy said:The domain of this function is [itex](-\infty, -1) U (0, \infty)[/itex], not just x> 0.
The formula for the derivative of y=(1+1/x)^x is y' = (1+1/x)^x * ln(1+1/x) * (1/x^2) - (1+1/x)^(x-1).
To simplify the derivative of y=(1+1/x)^x, first rewrite the equation as y=e^(x*ln(1+1/x)). Then, apply the chain rule by multiplying the derivative of the exponent (which is ln(1+1/x)) by the derivative of the inner function (which is x). Finally, use the power rule to find the derivative of e^(x*ln(1+1/x)).
The domain of y=(1+1/x)^x is all real numbers except for x=0.
The graph of y=(1+1/x)^x is a smooth curve that approaches the x-axis as x approaches infinity. It has a minimum point at x=e-1 and a vertical asymptote at x=0.
Yes, the derivative of y=(1+1/x)^x can be negative. It is negative when x is between 0 and e-1. This means that the slope of the graph is decreasing at those points.