November 12, 1997

y = ln x . We could re-write the equation in exponential form: e y = e ln x . e y = x . Next, we can find the derivative using implicit differentiation: e y dy dx = 1 . Then, we can solve for . dy dx : dy dx = 1 e y . Furthermore, we can substitute for ‘y’ since y = ln x : dy dx = 1 e ln x . Finally, this simplifies: dy dx = 1 x ................
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