Question

Why is `d/dxe^x=e^x`?

Answer 1

This follows from the definition of natural logarythm and its inverse.

Explanation:

`ln e^x = x`

`d/dx ln e^x = d/dx x`

`1/e^x d/dx e^x = 1`

`d/dx e^x = e^x`

Answer 2

The "why" depends on how you've defined `e^x`.

Explanation:

Define `lnx` first

One approach is to define `lnx = int_1^x 1/t dt` and
then to define `exp(x)` to be the inverse of `lnx` and,
finally, define `e^x = exp(x)`.

In this case `y=e^x` if and only if `lny=x`.

Differentiating implicitly gets us `1/y dy/dx = 1`.

So, `dy/dx = y = e^x`

Define `e^x` independently of `lnx`

Definition 1

For positive `a`, define

`a^x = lim_(rrarrx)a^x` with `r` in the rational numbers
(We owe you a proof that this is well-defined.)

Then, using the definition of derivative:

`d/dx(a^x) = lim_(hrarr0)(a^(x+h)-a^x)/h = lim(hrarr0)(a^x(a^h-1)/h)`

We then define `e` to be the number that satisfies `lim_(hrarr0)(e^h-1)/h = 1`

With this definition we get
`d/dx(e^x) = lim_(hrarr0)(e^(x+h)-e^x)/h = lim_(hrarr0)(e^x(e^h-1)/h)`

`= e^xlim_(hrarr0)((e^h-1)/h) = e^x`

Definition 2

`e^x = 1+x/1+x^2/(2*1)+x^3/(3*2*1) + x^4/(4!)+x^5/(5!)+ * * *`

(`n! = n(n-1)(n-2)* * * (1)`)
(We owe you a proof that this is well defined.)

Differentiating term by term (we owe you a proof that this is possible), we get

`d/dx(e^x) = 0+1+(cancel(2)x)/(cancel(2)*1)+(cancel(3)x^2)/(cancel(3)*2*1) + x^3/(3!)+x^4/(4!)+* * *`

Which simplifies to `1+x/1+x^2/(2*1)+x^3/(3*2*1) + x^4/(4!)+x^5/(5!)+ * * *` which is again `e^x`