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RE: LeoThread 2025-03-10 23:28

in LeoFinance7 months ago

Part 2/5:

This formula is simple and applies to the natural logarithm of positive numbers only, making it limited when considering negative values or scenarios where ( x ) might approach zero.

Challenges with Negative Values

When examining functions like ( y = x^{1.3} ), one can quickly run into issues when substituting negative values. For instance, if ( x = -1 ) is input into the function, it leads to ( \ln(-1^{1.3}) ), which becomes undefined in the realm of real numbers. Consequently, one cannot straightforwardly apply ordinary logarithmic differentiation methods in these cases.

Utilizing Absolute Values