Derivative of $e^x$ (i)

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This will be a very basic question, but it will help my thought process to have someone answer it.

$e$ can be defined as the number such that $$\frac{de^x}{dx}=e^x\;\;\;\;\;(1)$$

Taking $x=2$, this implies that $$\frac{de^2}{dx}=e^2\;\;\;\;\;(2)$$

The power rule gives us that $$\frac{de^2}{dx}=2e\;\;\;\;\;(3)$$

Therefore from $(2)$ and $(3)$ we get $$e^2=2e\;\;\;\;\;(4)$$

But a calculator tells me that $(4)$ is false. Where have I gone wrong?

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Equations (2) and (3) are false. $e^2$ is a constant, and the derivative of a constant is zero.

Remember that you evaluate the derivative after differentiating the function. So when you say $\frac{de^x}{dx} =e^x$, you mean that the slope of the line tangent to $y=e^x$ at $x=2$ is $e^2$. In Leibniz notation, the correct form of (2) is $$ \left.\frac{de^x}{dx}\right|_{x=2} = e^2 $$ The correct form of (3) is $$ \left.\frac{dx^2}{dx}\right|_{x=e} = 2e $$ As you can see, these are different functions, differentiated and evaluated at different points, so there's no surprise the derivatives are different.

Dilatory aside. You say that (1) can be taken as the definition of $e$, and you are right, with the insertion of the word positive. Equation (1) is also satisfied with $e$ replaced by $0$, and if $e$ is “defined” to be zero, then (2), (3), and (4) are also all satisfied.

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Your equation (2) is misleading. It should really be written as $$ \frac{d e^x}{d x}|_{x = 2} = e^x|_{x = 2}.$$ Evaluating $\frac{d e^x}{d x}$ at $x = 2$ is not the same as $\frac{d e^2}{d x}.$