Question 1

In the first video at 7:26, I don't understand how he was able to move the limit to inside the natural log.

Accepted Answer

*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!

Question 2

in the second video, what does he do in 1:15 to get to multiply ey to that derivate??

Accepted Answer

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.

Question 3

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.

Accepted Answer

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).

Question 4

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)

Accepted Answer

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

Question 5

in the 2nd video, at 1:21, I thought d/dx [e^y]=e^y? and is d/dx [e^y]=e^y correct?

Accepted Answer

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!

Question 6

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

Accepted Answer

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.

Question 7

At

6:23

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?

Could someone help me out?

Accepted Answer

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.

AP®︎ Calculus AB (2017 edition)

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

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

Here we find the derivative of $\ln (x)$ ‍ directly from the definition of the derivative as a limit.

Here we find the derivative of $\ln (x)$ ‍ by using the fact that $\frac{d}{d x} [e^{x}] = e^{x}$ ‍ and applying implicit differentiation.

Want to join the conversation?

AP®︎ Calculus AB (2017 edition)

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

Here we find the derivative of ln⁡(x)‍ directly from the definition of the derivative as a limit.

Here we find the derivative of ln⁡(x)‍ by using the fact that ddx[ex]=ex‍ and applying implicit differentiation.

Want to join the conversation?

Here we find the derivative of $\ln (x)$ ‍ directly from the definition of the derivative as a limit.

Here we find the derivative of $\ln (x)$ ‍ by using the fact that $\frac{d}{d x} [e^{x}] = e^{x}$ ‍ and applying implicit differentiation.