Derivative of log X = 1 / X Proof in Calculus

Derivative of log x to the base e is equal to 1/x. To find the derivative of log x we use first principle method of derivative calculus math.Derivative of log x is very important in derivative calculus.

[rohanshah]

what about d/dx log |x| ?
Why it is 1/x ? can you give proof

[MediumSizedPizza]

Your explantions are clear and concise! Thumbs up. :)

[stephengarramone]

You have shown a tautology.  You have assumed that (1 + 1/p)^p = e  as p --> ∞ and used that as the basis for your natural logarithms.  This is NOT GOOD, so try this:
let y = log (to any base) x
then y + ∆y = log (x +  ∆x)
then ∆y = log (x + ∆x) - y = log (x + ∆x) - log x = log [(x + ∆x)/x]
∆y = log(1 + ∆x/x)
∆y/∆x = (1/∆x)log(1 + ∆x/x) = log(1 + ∆x/x)^(1/∆x)
consider 1/p = ∆x/x → 0 as p → ∞; 1/∆x = p/x; ∆y/∆x → dy/dx or y’
then
y’ = log(1 + 1/p)^p/x = [log(1 + 1/p)^p]1/x
y’ = (1/x)log(1 + 1/p)^p for ANY base
but lim(1 + 1/p)^p as p → ∞ is e (proven elsewhere) or 2.718281828...
thus y’ = (1/x)log e
If we use the base 10, y’ = (1/x)0.4342944819...)However, if we choose the base e = 2.718281828..., then log e = 1 and y’ = 1/x(1) = 1/x
QED

[DiamondMind]

I throughly enjoy how you say derivative.

[nimmagaddasatyaprasad]

1/x should be written before d/dx before applying log formula