Log in Melinda Yuan 6 years agoPosted 6 years ago. Direct link to Melinda Yuan's post “In the first video at 7:2...” In the first video at • (21 votes) bhvima 6 years agoPosted 6 years ago. Direct link to bhvima's post “*As all the n values were...” 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: Hope this helped! (26 votes) Philip Chen 3 years agoPosted 3 years ago. Direct link to Philip Chen's post “Why is the derivative of ...” 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) Tim 3 years agoPosted 3 years ago. Direct link to Tim's post “This would be due to the ...” 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 (14 votes) Ismael Jiménez 5 years agoPosted 5 years ago. Direct link to Ismael Jiménez's post “in the second video, what...” in the second video, what does he do in • (7 votes) RowanH 4 years agoPosted 4 years ago. Direct link to RowanH's post “It looks like using the c...” 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) Yeezbear 7 years agoPosted 7 years ago. Direct link to Yeezbear's post “please explain number "e"...” 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) David Staver 7 years agoPosted 7 years ago. Direct link to David Staver's post “Note that numbers like Pi...” 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) Kathleen Oday 5 years agoPosted 5 years ago. Direct link to Kathleen Oday's post “Sal has presented two alt...” 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) Alex 5 years agoPosted 5 years ago. Direct link to Alex's post “Put simply, 1 / x approac...” Put simply, 1 / x approaches 0 as x approaches infinity, and vice versa. (4 votes) Liang a year agoPosted a year ago. Direct link to Liang's post “in the 2nd video, at 1:21...” in the 2nd video, at • (2 votes) Venkata a year agoPosted a year ago. Direct link to Venkata's post “Not quite. Observe our va...” 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! (6 votes) Moly 5 years agoPosted 5 years ago. Direct link to Moly's post “At 4:28 of first video ,I...” At • (3 votes) Chuck B 3 months agoPosted 3 months ago. Direct link to Chuck B's post “Note that `0 ÷ a = 0` for...” Note that (1 vote) Michele Franzoni 4 years agoPosted 4 years ago. Direct link to Michele Franzoni's post “At 6:23 of the first vide...” At Could someone help me out? • (2 votes) cossine 4 years agoPosted 4 years ago. Direct link to cossine's post “dx is infinitely small. I...” 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) Justin Hogue 16 days agoPosted 16 days ago. Direct link to Justin Hogue's post “I'm confused as to why th...” 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) Luke de Wet 16 days agoPosted 16 days ago. Direct link to Luke de Wet's post “It can seem paradoxical t...” 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) sunshine123u 2 years agoPosted 2 years ago. Direct link to sunshine123u's post “Why is the derivative of ...” Why is the derivative of log(u) = u’/u ( u is in place of something polynomial involving x) • (2 votes) Iron Programming 2 years agoPosted 2 years ago. Direct link to Iron Programming's post “Howdy sunshine123u,Did ...” 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) Hope this helps. (1 vote)Want to join the conversation?
7:26
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 1:15
If we let u = 1 / x, then
e = lim x -> inf of (1 + u)^x = lim u -> 0 (1 + u)^(1 / u). 1:21 4:280 ÷ 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. 6:23
If we do that for our ln expression, we get:
d/dx[ln(u)] = d/dx[ln](u) * u' = 1/u * u' = u'/u
