# Why is the derivative of the exponent 0?

Greetings, dear Readers! I am sure that many of you from your school mathematics course remember perfectly a wonderful function – an exponent, the derivative of which, no matter how much you take it, is equal to the original function.

However, how many of you know why this is happening? Today I want to tell this in the simplest possible language. Go! Consider two exponential functions:

Let us now recall the classical definition of the derivative of a function as the limit of the ratio of the increment of a function to the increment of its argument when the increment of the argument tends to zero.

In simple words: we analyze the rate of change of the function f (x) with an infinitesimal change in its argument, which we will denote by ∆x.

In the formulas for the first function, it looks like this:

Let’s calculate something on the calculator, namely the expression under the limit sign. For example, let the change in the function ∆x = 0.001. Then:

*However, this will not give us anything …* Until we calculate a similar expression for a function based on 3:

But this is already interesting. If we recall a little mathematical analysis, then it pops up in my head **second Bolzano-Cauchy theorem** or intermediate value theorem…

In our case, it allows us to assert that the function under consideration (meaning the fraction (x ^ ∆х-1) / ∆х) for some x equals one! *If we find such an x, then by definition we obtain the equality of the function of its derivative!* Begin! We equate our function to one:

*The second remarkable limit is the ratio known from the school course, which invariably leads to the Euler number.* So the proof is over!

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