The differentiation of `log_e x ,i s1/xdot` i.e. `d/(dx)((log_e x)=1/x)`

Explanation

Let `ln x = log_ex`

We know that `d/dx (logₐ x) = 1 / (x ln a)`
Substitute a = e on both sides.
Then we get:


`d/dx (log_e x) = 1 / (x ln e)`


By the properties of natural logarithms, ln e = 1.
So

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