If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Proof: the derivative of ln(x) is 1/x

The derivative of ln(x) is 1x:
ddx[ln(x)]=1x
The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

Here we find the derivative of ln(x) directly from the definition of the derivative as a limit.

Khan Academy video wrapper
Proof: the derivative of ln(x) is 1/xSee video transcript

Here we find the derivative of ln(x) by using the fact that ddx[ex]=ex and applying implicit differentiation.

Note: Implicit differentiation is a technique that is taught later in the course.
Khan Academy video wrapper
Derivative of ln(x) from derivative of 𝑒ˣ and implicit differentiationSee video transcript

Want to join the conversation?

  • blobby green style avatar for user Melinda Yuan
    In the first video at , I don't understand how he was able to move the limit to inside the natural log.
    (20 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user bhvima
      As all the n values were inside the natural logarithm, he was able to move the limit inside and arrive at the correct answer.

      Here is another proof that may interest you:
      y = lnx
      x = e^y
      The derivative of x with respect to y is just e^y
      Then the derivative of y with respect to x is equal to 1/(e^y)
      As y = lnx,
      1/(e^y) = 1/(e^lnx) = 1/x


      Hope this helped!
      (26 votes)
  • aqualine ultimate style avatar for user Philip Chen
    Why is the derivative of ln2x = 1/x ? Is there a proof? I don't understand how 2 different functions (lnx, ln2x) can have the same gradient (1/x)
    (2 votes)
    Default Khan Academy avatar avatar for user
  • male robot donald style avatar for user Ismael Jiménez
    in the second video, what does he do in to get to multiply ey to that derivate??
    (8 votes)
    Default Khan Academy avatar avatar for user
  • primosaur ultimate style avatar for user Yeezbear
    please explain number "e" to me.. Who found it? Who made it? Who calculated it? It seems like it is a number created by "god" him(or her)self
    (1 vote)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user David Staver
      Note that numbers like Pi and e aren't defined like regular numbers (like 1 or 2). For example, Pi can be defined as the circumference of a circle whose diameter is 1. The fact that Pi happens to be exactly 3.14159... is because of our base-10 number system.

      When Sal gets to the end of the proof and has the expression lim n->0 (1 + n)^(1 / n), this expression actually is the number e, because e is defined as the result of that expression.
      (7 votes)
  • blobby green style avatar for user Kathleen Oday
    Sal has presented two alternate expressions defining the number e: one set up and explained like a compound interest calculation i.e. e=lim of (1+1/x)^x as x approaches infinity and the other as e=lim of (1+x)^(1/x) as x approaches 0. The first definition is readily understood and there are Sal’s and many other demonstrations of truth of that first definition out there in web land. But if that first definition is supposed to lead inexorably to the second, I am missing something. Does Sal or anyone else have a video demonstration or other type “proof” as to why the latter is true? The graphs of (1+1/x)^(x) and (1+x)^(1/x) are both weird, undefined at x=0 and so on but they do not look similar. At very large x values the first does appear to approach a horizontal asymptote at the value f(x)=e (which is satisfying), but the second just kind goes nuts around x=zero (although it does approach e from x>0). This has been driving me crazy so any help would be most appreciated. Sal uses this second definition a lot but I can’t find any place he explains it.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Moly
    At of first video ,I dint understand what he said. 0 is not in the domain of ln x; why is that important ? If delta x approaches 0 then n simply approaches 0 . Does it depend on any conditions ?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Chuck B
      Note that 0 ÷ a = 0 for all a ≠ 0. That condition is important, because 0 ÷ 0 is undefined. So when Sal takes lim{Δx→0} Δx/x, it's important that x is never 0, because then the limit would be undefined. Thus, he notes that because 0 is not in the domain of the ln function, x will not be 0, so the limit exists.
      (1 vote)
  • aqualine ultimate style avatar for user Leon
    in the 2nd video, at , I thought d/dx [e^y]=e^y? and is d/dx [e^y]=e^y correct?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • male robot donald style avatar for user Venkata
      Not quite. Observe our variable. If you did d/dx of (e^x), it would be e^x. But, as we are doing d/dx (e^y), using the chain rule, it would be e^y (dy/dx). You'll understand this better once you learn about implicit differentiation, which should be coming up ahead!
      (5 votes)
  • leaf orange style avatar for user Justin Hogue
    I'm confused as to why the derivative of "lnx" is equal to a function that is defined outside the original function's domain (1/x). I understand that "y=1/x" is the slope of everything to the right of the y axis in "y=lnx," but why does the given solution for the derivative have values defined for x values outside of the domain of "lnx?" I'm trying to refer to the values of "1/x" that are left of the y axis.
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf orange style avatar for user Luke de Wet
      It can seem paradoxical that the 2 functions have different domains yet are related in the differential context. Though we basically "ignore" the whole domain for x<0 which then results in it being valid. We only use ln(x) when x>0 so we only use 1/x when x>0.

      Basically, think of it as we don't ever use ln(x) when x<0 so we won't ever use the derivative of ln(x) (1/x) for when x<0.
      (2 votes)
  • duskpin ultimate style avatar for user sunshine123u
    Why is the derivative of log(u) = u’/u ( u is in place of something polynomial involving x)
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Iron Programming
      Howdy sunshine123u,

      Did you mean the derivative of ln(u)? Because the derivative of ln(x) is 1/x, if we have the derivative of ln(u), where u is some polynomial, then we must use u-substitution, which says that d/dx[f(g(x))] = f'(g(x))*g'(x)
      If we do that for our ln expression, we get:
      d/dx[ln(u)] = d/dx[ln](u) * u' = 1/u * u' = u'/u

      Hope this helps.
      (1 vote)
  • blobby green style avatar for user gunank312
    at , why is the denominator missing x? if I assume delta is x is same as x, then it feels right isn't it? but it wouldn't be something concrete?

    thanks
    (2 votes)
    Default Khan Academy avatar avatar for user