## keonlo 3 years ago Is e^x the only function that has a same derivative of itself? if yes, how to proof it?

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1. shandelman

f(x) = 0 has a derivative of 0.

2. JingleBells

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!

3. FoolAroundMath

$\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)

4. JingleBells

In this case the constant k is just a translation, and the exponential still has to have e as it's base, right?

5. FoolAroundMath

Yes the exponential has to have $$e$$ as its base, and k can be any real number..

6. JingleBells

Thanks!

7. shandelman

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.

8. JingleBells

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?

9. shandelman

f(x) = 0 is a horizontal line overlapping the x-axis. x=0 is a vertical line overlapping the y-axis.

10. JingleBells

Does it qualify as a function was what I was wondering.

11. FoolAroundMath

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.

12. alexaniandavid

h