derivative of (1+1/x)^x
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- čas přidán 5. 08. 2024
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derivative of x^x, two ways: • derivative of x^x
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I like the second way better, because everything stays in terms of x, so you don't have to think about dealing with y. It's also similar to the way I like to solve indeterminate powers with L'Hospital's Rule, which I find much easier than the way that is commonly taught since you don't have to remember about ln(L).
funnily i like the first way more cause it´s straightforward. get rid of the x in the exponent and apply chainrule. okay the other one too but it is a bit more prepwork and more difficult in the computation. first one just flies on the paper for me
Can’t wait to be in calculus so I can actually do the problems without explanation
Hjerpower You need to show your process in the the exams
@@ianmoseley9910 he means do the problems without explanation (from the teacher)
@@ano.g.6217 Thank God this was cleared up 2 years later lol
Are you lgbtq
You can use either of these methods to derive a more general formula:
If y(x)=f(x)^g(x) then
y'=y[f'(g/f) + g'ln(f)]
You could substitute in for y and simplify slightly, but being able to plug your original equation in to that y function is more useful than the slightly simpler looking formula. The simpler looking formula is nice because it looks similar to the product rule, which might be easier to memorize, if that helps you:
y'=[f'×g + g'×f×ln(f)]×f^(g-1)
Up until the natural log part, it looks like the product rule.
I'm using y'=y*[ln(f)*g]'
When i was 15, our math teacher in the school presented a problem to us: we had to find a derivative of x^x. I solved that less than in a minute. I was always impressed by the fact that ln(f(x))' = f'(x)/f(x). So once I noticed that I can easily find ln(x^x)' = (x ln(x))' = ln(x)+1, I was confident this solves. The easy part was to combine both ideas, to multiply to x^x and that's the answer. What was a happy part is that I invented this trick by myself, nobody taught me how to find such derivatives, and it helped me to be the first who told the correct answer.
Here the same idea works very well, the only difference we are solving (ln(1+1/x)^x)' = (x ln(1+1/x))' = ln(1+1/x) + x*(-1/x^2)/(1+1/x) = ln(1+1/x) - 1/(1+x). And again now multiply to the original function and that's the answer.
The fun and systematic contitnuation is to take a limit x→∞ of the derivative. The original function under this limit is e^x, the derivative should be the same, but there should be the proof!
I never thought derivatives could be a reason to argue, but now that I've seen this comment section I was clearly wrong
Challenge: Do this same question as a limit problem. Find the derivative of (1 + 1/x)^x from the definition of the derivative, and evaluate that limit. :D
Impossibru
Maybe take the natural log of the expression and/or factor out something like (1+1/x)^x or (1+1/(x+h))^x, should be doable
This was SO helpful!! Thank you!!!!!
First method is implicit differentiation. Second method is explicit differentiation. My opinion is that the second method is slightly "easier", but both of them work effectively here!
Since the methods are nearly identical, I don't have a preference.
mrBorkD People should practice both for exams - exam question may require one method and if you have always tended to use the other they may come unstuck.
Well explained...!
Is the limit of that derivative the same as when you derive e^x?
Is it possible to anyhow eliminate logarithmic and express it only in terms of x and y?
I prefer to just use the derivation rule for this case (analogous to multiplication rule). The derivative is just the derivative of the power keeping the exponent constant, plus the derivative of the exponential keeping the base constant.
In other words:
d/dx[f(x)^g(x)]=
=d/dx[f(x)^g]+
+d/dx[f^g(x)]
...which makes sense, if you think in the physical meaning of a derivative. :-)
thank you so much !
Setting a crazy equation to y and taking the implicit derivative of both sides seems like the much easier approach.
I like the first way. I like implicit differentiation.
Where is your video about the limit of (1+1/x)^x?
Yes! Implicit differentiation! It is so powerful
implicit function theorem implies inverse function theorem
What happens if you take the limit for this new one as x goes to infinity?
You're awesome!
Bonjour
One quick question : what is the limit of (1+1/x)^x (-1/(x+1) + ln(1 + 1/x)) when x -> +Inf ?
Regards, Philippe
The first factor converges to e, the second one to 0, so the whole thing converges to e * 0 = 0.
How to change subject
y=(1+1/x)^x
how do we know that differentiating both sides is okay?
When I tried to do it I sometimes got weird (wrong) results?
(esp. with implicitly defined functions)
Arthur Marcuss,
show us the example, and we will show you the error.
Arthur Marcuss If you think of differentiation as a function, it makes sense that you can take the derivative of both sides and everything will check out. Just make sure when you’re finding dy/dx, if you take the derivative of a variable other than x, you have to apply the chain rule and generate a d(that variable)/dx there.
great very much helpfull
I think if I was to do this question, I would probably do the first way, because it would be a lot easier to write down the answer with no workings. The second one wouldn’t really be feasible to do in your head, unless you are really good at doing things in your head or something.
Sir what r u holding in ur hands?
I just by hearted d (x^x)/dx = x^x (1+ln (x))
Then do simple chain rule, how you differentiate normally and get answer fast
Especially useful if you want to be fast and you deal with such powers a lot.
One can see f(x) as u(x)^v(x)
Where u(x) = (1+1/x) and v(x) = x
Since f(x) is a function composition the derivative of f with respect to x is
df/dx = df/du du/dx + df/dv dv/dx
Since u(x) = (1+1/x) and v(x) = x it comes du/dx = -1/x² and dv/dx = 1
On the oter hand
df/du = derivative of f with respect to u(x) when v(x) is a constant (let's say v)
df/du = derivative of (1+1/x)^v with respect to (1+1/x) = v(1+1/x)^v-1 = x(1+1/x)^(x-1)
df/dv = derivative of f with respect to v(x) when u(x) is a constant (e.g. u)
df/dv = derivative of u^x with respect to x = u^x ln(u) = (1+1/x)^x ln(1+1/x)
df/dx = df/du du/dx + df/dv dv/dx
df/dx = x(1+1/x)^(x-1) (-1/x²) + (1+1/x)^x ln(1+1/x) 1
df/dx = (1+1/x)^(x-1) (-1/x) + (1+1/x)^x ln(1+1/x)
df/dx = (1+1/x)^x (ln(1+1/x) -1/(x+1))
Regards, Philippe
Nice! I prefer the second way
1° Love it!
wow usually people argue about different ways to prove integrals, this is the first time ive seen people arguing about derivatives :O
Awesome
The two methods require almost exactly the same work. If I had a preference, it would have to be based on whether I am more familiar with the small amount of algebra involved. However, it turns out that my algebra skills are well up to the task. No strong preference, but the first method occurred to me first.
I prefer the first way because I learned implicit differentiation before the trick of e^ln(f(x)).
Did with the second way by myself, before he start :)
I used multivariate chain rule on u^x, then set u(x) = 1+1/x.
What is the derivative of (1/x)^x
7:56
Could you do it with the definition of derivative 😬😬?
감사링~
Study the changes of the function |1+1÷x|^x
PLEASE do the indefinite integral of f(x)=1/(1+sqrt(tanx))
HHH xokran mn l3irfan
I prefer the second way, chain rule is love *heart*
what is implicit differentiation? a fancy name for the chain rule. seriously it´s all there is too it. and at least to me it looks more aesthetic to actually have an equation through all the steps. Vanity is not the sin that will get me to hell, but i think if you can write it neatly just do it. (and well i personally think it is easier but difficulty is the most subjective thing there is, isn´t it)
Chen lu*
@@Metalhammer1993 can you find the derivative of ln(x) without knowing it's 1/x
@@aldobernaltvbernal8745 with a trick yes.
Let's rename it first.
y=lnx
Now let's raise e to both sides of the equation
e^y=e^(lnx)
e^y=x
Now we take the derivative with respect to x. Careful now: y is a function of x. Meaning we have to apply the chain rule and multiply by its derivative! Remember we want to know what dy/dx is so we somehow needed to introduce it, and via the chain rule or rather implicit differentiation we did that
We obtain
e^y dy/dx=1
Now we can do two things in interchangeable order. Either substitute e^y for x or divide it over. Whatever you prefer. I prefer resubstitution as the last step so I divide over now
dy/dx=1/e^y
Remember e^y=x
So dy/dx=1/x
This imho is the star example of implicit differentiation
Tell us how to differentiate it second time
I vot smth weird like
(-1/x )*(1 + x)^(x-1)
O man really practice makes a man perfect and also by thinking logically
Heeeey! I would like you to solve the integral of this function :(
y= (1)/((x^(4)-5x^(3)+x^(2)-5x+1))
The coefficients of the integral look symmetric. I wonder if that lets you do something special?
Is it actually correct to put entire equation in parenthesis and then write something outside?
RUSapache Yes. It’s just a shorthand to do something to each side.
How can people just hate on logarithmic differentiation like this?
I love e^x more than ln(x) so I'll say I like the second method better as it had e^x.
I used the second way before actually watching the video. So I think its better xd
I saw that you have ran 7 marathons, have you ever used calculus to run faster and do you run year round?
Andrew Kaye How would calculus help him run faster?
Good
Please solve this d/dx(1+1/x)^x!
I like the first way better, because I like substitution.
huuh?! that is a little weird. I mean sure that is the derivative, but normally we see that with the limit as x-> infinity. The case of which we end up with the number e, which the derivative of a constant is zero. I expected to see everything cancel down to zero. To answer the question though of which way: both. implicit differentiation is powerful, and using properties is always good as it shows knowledge of the subject. So this graph of the function would have a horizontal asymptote at e, and the derivative is showing the slope as we approach e. interesting that we never actually covered that. Most textbooks I have seen stop with defining e and that is it. I do not know what we can gain from see that graph and understanding the slope, but since e is related to natural growth and decay in many fields, then maybe just the pure knowledge is worth it.
🙌
Remember Chen Lu!
OK I think you're doing great work but how can dy/dx have value by its own
Cool
I don't understand why you can't just use the chain rule on the original expression. Why do you have to change it at all? I'd just think it's (1+1/x)^x*ln(1+1/x)*(-1/x^2).
Nice that u grew a beard! Keep it.
Acabo de ser el like número 1000
So, you are saying the long answer you got, in both ways of doing the problem, can't be simplified at all? My brain would like to simplify the answer to having one or at most, two x's. Having so many x's... it kind of has the feel like it could be simplified, but you just didn't want to do it for some reason.
bprp how about d/dx(n^x) ?
somerandomdudeable first you a have a looong name,than how ? how prove thatd/dx n^x is n^x*ln x ?
Mihai Ciorobitca I don't know how to prove it but it makes sense because if you let n be e then you still get e^x because ln(e)=1
George Pennington It's pretty straight forward actually - n^x = exp(log(n) * x, so
(n^x)'=exp(log(n)*x)*log(n)=n^x * log(n)
May I ask, at what point in American middle or highschools you do calculus?
Flewn thanks, that makes sense now. I don't know when Americans do calculus because I'm from England
If given n is a constant,
d/dx (n^x) = n^x * ln(n)
However if n is nonconstant,
let y = n^x
lny = ln(n^x)
lny = xln(n)
(1/y)(dy/dx) = (1)ln(n) + (x)(0) as per product rule
(1/y)(dy/dx) = ln(n)
dy/dx = (ln(n)) * (1/y)
however, we established y = n^x, so 1/y = (1)/(n^x) or n^-x
dy/dx = (ln(n)) * (1/y)
dy/dx = (ln(n)) * n^-x
or you could say dy/dx = ln(n) / n^x
I do not know implicit derivation but the second way is great (:
Carl Fels It’s differentiation. And you just take a derivative of y to generate a dy/dx.
second one much better
I would prefer the first way .
Second way is more beautiful method
Second way is better
Hi
The first one.
NOOOOO
If only you could have talked about the sign of the derivative ...
That's what is problematic to me so I can deduce the variation of y
Both ways are really the same way 😂🤣😂🤣
The first way is better
I like second
Second way is most important
Please 😭
nie pomolges stary
Why don’t you add closed captioning where people not in your ethnic group could understand what you are saying.
Jerry Easley Well that’s offensive. Consider no one other than you has ever made this request in the comments, I’m going to conclude that the issue is close mindedness. I’m white and have no difficulty understanding.
@@cpotisch He cannot understand because he is stupid