How can i prove that the differential of log x is equal to 1/x?

Answer 1

If you are saying the derivative of log x is 1/x, you are talking about the natural log x often written ln x.
By the definition of if log x = y then e^y = x
Taking the derivative in terms of x:
d(e^y)/dx = 1
e^y* dy/dx = 1
dy/dx = 1/e^y
But looking at the definition e^y = x, substituting:
dy/dx = 1/x

Answer 2

the derivative of log x is 1/x
if log x = y
then e^y = x
Taking the derivative in terms of x:
d(e^y)/dx = 1
e^y* dy/dx = 1
dy/dx = 1/e^y
But looking at the definition e^y = x
substituting in dy/dx = 1/e^y
dy/dx = 1/x

Answer 3

The hard way is to use the definition of a derivative:

d(ln(x))/dx = lim (e->0){ln(x+e)-ln(x)}/e

write ln(x+e) = ln(x*(1+e/x)) =ln(x)+ln(1+e/x)

SInce e/x << 1, can expand ln(1+e/x)

Ln(1+e/x) ~ e/x +O((e/x)^2)+... where O((e/x)^2 means terms of that go like (e/x)^2 or higher powers

Then ln(x+e)-ln(x) ~ ln(x) +e/x +O((e/x^2)) +...-ln(x) = e/x +O((e/x)^2)+...

Finally d(ln(x))/dx = lim[e->0]{e/x +O((e/x)^2)+...}/e
= lim[e->0](1/x +O(e/x^2)+O(e^2/x^3)+...}
= 1/x

Answer 4

f(x)=lnx

f'(x)= lim_{h-->0} [f(x+h)-f(x)]/h = lim_{h-->0} [ln(x+h)-lnx]/h =
lim_{h-->0} ln[(x+h)/x]/ h = lim_{h-->0} ln[(1+h/x)^(1/h)] =
lim_{h-->0} ln [ 1+ 1/ (x/h)]^[(x/h) · (1/x)] =

lim_{h-->0} ln {[ 1+ 1/ (x/h)]^(x/h) }^ (1/x) = ln [e^(1/x)] =1/x

when x>0

saludos.