The same derivative for very different functions Just a curiosity and request for a comment on this. d/(dx)(-2 log (u x-2))=-(2 (x u'))/(u x-2) d/(dx)(-2 log (2-u x))=-(2 (x u'))/(u x-2) What can be said about such results? Clearly there is nothing similar between ux−2 and 2−ux, yet the equations are both the same. Checked with Mathematica and by simple manual computation. Maybe the differentiated functions can be derived one from another?

Diego Barr

Diego Barr

Answered question

2022-10-12

The same derivative for very different functions
Just a curiosity and request for a comment on this.
d d x ( 2 log ( u x 2 ) ) = 2 ( x u ) u x 2
d d x ( 2 log ( 2 u x ) ) = 2 ( x u ) u x 2
What can be said about such results? Clearly there is nothing similar between u x 2 and 2 u x, yet the equations are both the same. Checked with Mathematica and by simple manual computation. Maybe the differentiated functions can be derived one from another?

Answer & Explanation

Sauppypefpg

Sauppypefpg

Beginner2022-10-13Added 23 answers

It boils down to the fact that because:
d d y log y = 1 y
then the chain rule gives us that:
d d y log ( y ) = 1 y = 1 y
However, for real y, at most one of log y and log ( y ) is defined (because log is only defined for positive operands).

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