Question

What function do you know from calculus is such that its first derivative is itself? (Do not use the function f(x) = 0.)

Answer 1

`f(x)=e^x`

Explanation:

Proof:

Let
`e^x=z`

take logs in both sides
`ln(e^x)=ln(z)`

using log properties we have
`xln(e)=ln(z)`. We know that
`ln(e)=1` then
`x=ln(z)`

Taking derivatives with respect to x, we have:
`1=(1)/(z)(dz)/(dx)`

Next we can move things around
`z=(dz)/(dx)`

Replacing z for
`e^x` as defined in the first line, we get

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