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

#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#

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#