using the derivative of e^x,show that if y=ln(x),then y'=1/x?

e^y = x
e^y dy/dx = 1
dy/dx = 1 / e^y
dy/dx = 1 / x

y = ln(x)
e^y = e^ln(x)
e^y = x
Derive implicitly

e^(y) * dy = dx

Solve for dy/dx

dy/dx = 1 / e^(y)

We know that e^(y) = x, so

dy/dx = 1/x

If y = ln(x) then by definition
e^y = x, then differentiate wrt y
e^(y) = dx/dy = x
dy/dx = 1/x

y=ln(x)
=>
e^y=x
=>
(e^y)y'=1
=>
y'=1/e^y
=>
y'=1/e^(ln(x)
=>
y'=1/x