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## Differential Calculus

### Course: Differential Calculus>Unit 2

Lesson 9: Derivatives of cos(x), sin(x), 𝑒ˣ, and ln(x)

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

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.A (LO)
,
FUN‑3.A.4 (EK)
The derivative of natural log, left parenthesis, x, right parenthesis is start fraction, 1, divided by, x, end fraction:
start fraction, d, divided by, d, x, end fraction, open bracket, natural log, left parenthesis, x, right parenthesis, close bracket, equals, start fraction, 1, divided by, x, end fraction
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 $\ln(x)$natural log, left parenthesis, x, right parenthesis 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 $\ln(x)$natural log, left parenthesis, x, right parenthesis by using the fact that $\dfrac{d}{dx}[e^x]=e^x$start fraction, d, divided by, d, x, end fraction, open bracket, e, start superscript, x, end superscript, close bracket, equals, e, start superscript, x, end superscript 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?

• In the first video at , I don't understand how he was able to move the limit to inside the natural log.
(21 votes)
• 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!
(23 votes)
• 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)
• This would be due to the product log rule being able to separate ln(2x) into ln(x) + ln(2), the latter being a simple constant which does not change the slope
(10 votes)
• in the second video, what does he do in to get to multiply ey to that derivate??
(6 votes)
• It looks like using the chain rule. y is a function of x, so the derivative of e^y with respect should be e^y multiplied by dy/dx.
(2 votes)
• 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.
(3 votes)
• Put simply, 1 / x approaches 0 as x approaches infinity, and vice versa.
If we let u = 1 / x, then
e = lim x -> inf of (1 + u)^x = lim u -> 0 (1 + u)^(1 / u).
(4 votes)
• 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
(2 votes)
• 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.
(5 votes)
• What is the derivative of pi^x or tau^x?
(2 votes)
• the derivative of any constant to the x power is found like this.

start with a^x where a can be any constnat.

Use log rules to rewrite a^x as e^(ln(a^x)) let me know if you don't get how that works.

now you can use the chain rule to derive e^ln(a^x). The chain rule basically lets you solve a composite function f(g(x)). here f(x) is e^x and g(x) is ln(a^x) which can also be simplified to x*ln(a) by log rules.

the chain rule says f(g(x)) gets us f'(g(x))*g'(x) so this gets us e^ln(a^x)*ln(a). You can simplify e^ln(a^x) the same way we got it with log rules so the final simplified answer is a^x*ln(a)

Hopefully that helped but let me know if you don't understand something.

You can simplify
(2 votes)
• At of the first video Sal says that 1/x is unaffected when n approaches to zero. I don't understand this statement, for if n = dx/x then x = dx/n and thus n approaching zero makes x explode towards infinity.

Could someone help me out?
(2 votes)
• dx is infinitely small. If you have an expression like 0.001/1 then it going to be approximately equal 0. And from how we defined n we know n = dx/x. so limit dx->0 => n->0.
(2 votes)
• Why is the derivative of log(u) = u’/u ( u is in place of something polynomial involving x)
(2 votes)
• 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)
• 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 ?
(2 votes)
• 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)