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Is e^x the only function that has a same derivative of itself? if yes, how to proof it?

OCW Scholar - Single Variable Calculus
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f(x) = 0 has a derivative of 0.
Yes, but I think 0 is just a constant not a function, no? I believe the value of e is unique just like pi. But correct me if I'm wrong!
\[\frac{df}{dx} = f\]\[\Rightarrow \frac{df}{f}=dx\]\[\Rightarrow\int \frac{df}{f} = \int dx\]\[\Rightarrow ln(f) = x+c \]\[\Rightarrow f(x) = e^{x+c}\]\[\Rightarrow f(x) = k.e^{x}\] Here, \(k\) is any real number. So, any function of the form \(f(x) = k.e^{x} \) has the same derivative as itself. (k=0 implies that f(x) = 0 also is a solution)

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In this case the constant k is just a translation, and the exponential still has to have e as it's base, right?
Yes the exponential has to have \(e\) as its base, and k can be any real number..
Thanks!
I would argue that f(x) = 0 is completely a function despite the fact that it is constant. It's graph passes the vertical line test, does it not? I will grant you it's not particularly the most interesting function but it is a function nonetheless.
Well by definition it maps all points on to 0. So it is a many-to-one function. but I always wondered what x=0 on the x-y coordinates would be?
f(x) = 0 is a horizontal line overlapping the x-axis. x=0 is a vertical line overlapping the y-axis.
Does it qualify as a function was what I was wondering.
a function is defined by \(f: D \rightarrow R \) D is the domain R is the range such that all elements in D have a map to R and for \(x_{1}\) in D, \(f(x_{1})\) can have only one value in R. i.e. it must be one->one or many->one. f(x) = 0 satisfies this definition.
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