proof of lnx=1/x?

Question

proof of lnx=1/x

Answer 1

The natural log of x does not equal 1/x, however the derivative of ln(x) does:

The derivative of log(x) is given as:

d/dx[ log-a(x) ] = 1 / (x * ln(a))

where "log-a" is the logarithm of base a. However, when a = e (natural exponent), then log-a(x) becomes ln(x) and ln(e) = 1:

d/dx[ log-e(x) ] = 1 / (x * ln(e))

d/dx[ ln(x) ] = 1 / (x * ln(e))

d/dx[ ln(x) ] = 1 / (x * 1)

d/dx[ ln(x) ] = 1 / x

Answer 2

Derivative Of Lnx Proof

Answer 3

ln x does not equal 1/x!

But D(ln x) = 1/x

Proof- e^(ln x) = x by definition. Take the derivative of both sides.
D(e^(ln x)) = D(x) = 1
But D(e^(ln x)) = e^(ln x) D( ln x) by the chain rule.
We already said e^(ln x) = x, so xD(ln x) = 1, or D(ln x)= 1/x

Answer 4

Don't you mean, d/dx ln x = 1/x?

Go to this site for the proof:

http://www.math.com/tables/derivatives/more/ln.htm

Answer 5

d(lnx)/dx =
lim as h--->0 1/(h)*(ln(x+h)-ln(x)) =
lim as a--->0 1/(a*x)*(ln(x+a*x)-ln(x)) (because we can choose h=a*x and then let a go to 0) =
lim as a--->0 1/x*ln(((x+a*x)/x))^(1/a)) =
lim as a--->0 1/x*ln((1+a)^(1/a)) =
lim as y--->infinity 1/x*ln((1+1/y)^y) (by taking y=1/a) =
lim as y--->infinity 1/x*ln(e) =
lim as y--->infinity 1/x*1 =
1/x.

Answer 6

ln(x) = lim(d->0) [ ln(x+d) - ln(x) ] / d = lim ln((x+d)/x) / d
= lim (1/d) ln(1 + d/x) = lim [ ln (1 + d/x)^(1/d) ].
Set u=d/x and substitute:

lim(u->0) [ ln (1 + u)^(1/(ux)) ] = 1/x ln [ lim(u->0) (1 + u)^(1/u) ]
= 1/x ln (e)
= 1/x.

Answer 7

use this formula to prove them ∫[f'(x)/ f(x)] dx = ln f(x) + c where f'(x) = derivative of f(x) secx =secx (secx+tanx)/ (secx+tanx) =>secx =sec^2 x + secx tanx)/(secx+tanx) ∫secx dx = ∫sec^2 x + secx tanx) dx/(secx+tanx) now it appears like that formula, so write = ln(secx + tanx) + C do try cscx by ur self.... ok..

Answer 8

do you mean d/dx (ln(x)) = 1/x ?

Answer 9

try this website.

Hope it helps =)
http://math2.org/math/derivatives/more/ln.htm