derivative of (1+1/x)^x

More derivative examples:

blackpenredpen

[LNRDR]

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).

[Hjerpower]

Can’t wait to be in calculus so I can actually do the problems without explanation

[Sam_on_YouTube]

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.

[gnikola]

I never thought derivatives could be a reason to argue, but now that I've seen this comment section I was clearly wrong

[SatyaVenugopal]

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

[BigDBrian]

Since the methods are nearly identical, I don't have a preference.

[youmilsswagger]

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!

[sergiokorochinsky]

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. :-)

[frbaucop]

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

[CornishMiner]

I like the first way. I like implicit differentiation.

[theSASarethebest]

Yes! Implicit differentiation! It is so powerful

[jensonjoseph]

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.

[TheRedfire]

I prefer the second way, chain rule is love *heart*

[nikitakipriyanov]

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!

[Generalth]

Setting a crazy equation to y and taking the implicit derivative of both sides seems like the much easier approach.

[mathemitnawid]

7:56
Could you do it with the definition of derivative 😬😬?

[ريالمدريد-ضص]

Study the changes of the function |1+1÷x|^x

[moonlightcocktail]

How can people just hate on logarithmic differentiation like this?

[wpbn]

wow usually people argue about different ways to prove integrals, this is the first time ive seen people arguing about derivatives :O

[gianpierrefernandez]

Heeeey! I would like you to solve the integral of this function :(

y=