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## AP®︎ Calculus BC (2017 edition)

### Course: AP®︎ Calculus BC (2017 edition)>Unit 4

Lesson 10: Optional videos

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

Proving that the derivative of ln(x) is 1/x by using the definition of the derivative as a limit, the properties of logarithms, and the definition of 𝑒 as a limit.

## Want to join the conversation?

• Isn't the definition of e
(1+1/n)^n?
In the video, we used the definition as (1+n)^(1/n). Are those two the same thing?
• You could do a change of variables. If you replace every n with 1/n and every 1/n with n. In the original limit, n went to infinity, so 1/n is going to 0. So the limit is the same as saying the limit as n tends to 0 of (1+n)^(1/n).
• At why can we say that 1/x is not affected as n approaches zero?
• Because there is no 'n' in the expression for 1/x. There's just an x.
• at how do we know that we can not stop there ? I mean if I would substitute Delta X approaching zero, then 1 over Delta X would become infinitely large. Natural log[ of 1 plus (delta x over x) ] would become natural log of 1, since delta x over x would be approaching zero. And ln 1 = 0 . That would give us infinity multiplied by zero and the limit would be zero. Why is that not possible? How do we know that we have to continue on here?
• "zero * ∞" is an indeterminate form – because ∞ is not a real number you can't use the rule that anything times zero is zero.

https://en.wikipedia.org/wiki/Indeterminate_form
• there must be another proof for the definition of e independent of derivatives so this proof wont be circular
• At , "n" is substituted for "delta x" in the limit statement. Nowhere have we let n=delta x. We let n= (delta x)/x. I'm sure there's some magic going on, but why don't we have to state the limit as "nx->0"?
(1 vote)
• It was stated that as ∆x approaches 0, so does n.
It's not so much a substitution as it is choosing to write n instead of ∆x, because they both approach the same thing.
The limit should then affect the function in an equivalent way but the change allows you to manipulate the expression differently.
• At , the limit as n approaches zero and the natural logoritm change place, so that we get ln(lim (1+n)^1/2 instead of the other way around. Why?
• What is the derivative of lnx^lnx
Plz! With solution
(1 vote)
• It is not clear what you mean by lnx^lnx, so I will assume that you mean (lnx)^(lnx).

Using laws of exponents, we rewrite (lnx)^(lnx) as e to some function:
(lnx)^(lnx) = e^{ln[(lnx)^(lnx)]} = e^[lnx * ln(lnx)].

Using the chain and product rules, we find that the derivative of (lnx)^(lnx) is
d/dx of [(lnx)^(lnx)] = e^[lnx * ln(lnx)] * d/dx of [lnx * ln(lnx)]
= (lnx)^(lnx) * [(1/x)*ln(lnx) + (lnx)*(1/x)/(lnx)]
= (1/x) * (lnx)^(lnx) * [ln(lnx) + 1].

Have a blessed, wonderful day!