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### Course: AP®︎/College Calculus AB > Unit 2

Lesson 8: Derivatives of cos(x), sin(x), 𝑒ˣ, and ln(x)- Derivatives of sin(x) and cos(x)
- Worked example: Derivatives of sin(x) and cos(x)
- Derivatives of sin(x) and cos(x)
- Proving the derivatives of sin(x) and cos(x)
- Derivative of 𝑒ˣ
- Derivative of ln(x)
- Derivatives of 𝑒ˣ and ln(x)
- Proof: The derivative of 𝑒ˣ is 𝑒ˣ
- Proof: the derivative of ln(x) is 1/x

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

The derivative of $\mathrm{ln}(x)$ is $\frac{1}{x}$ :

The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

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

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

Note: Implicit differentiation is a technique that is taught later in the course.

## Want to join the conversation?

- In the first video at7:26, I don't understand how he was able to move the limit to inside the natural log.(20 votes)
**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!(26 votes)

- 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)
- 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(16 votes)

- in the second video, what does he do in1:15to get to multiply ey to that derivate??(8 votes)
- 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)

- 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(1 vote)
- 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)

- 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.(2 votes)
- 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).(3 votes)

- At4:28of first video ,I dint understand what he said. 0 is not in the domain of ln x; why is that important ? If delta x approaches 0 then n simply approaches 0 . Does it depend on any conditions ?(3 votes)
- Note that
`0 ÷ 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.(1 vote)

- in the 2nd video, at1:21, I thought d/dx [e^y]=e^y? and is d/dx [e^y]=e^y correct?(1 vote)
- 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!(5 votes)

- 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)
- 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)

- Why is the derivative of log(u) = u’/u ( u is in place of something polynomial involving x)(2 votes)
- 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)

If we do that for our ln expression, we get:

d/dx[ln(u)] = d/dx[ln](u) * u' = 1/u * u' = u'/u

Hope this helps.(1 vote)

- at00:40, why is the denominator missing x? if I assume delta is x is same as x, then it feels right isn't it? but it wouldn't be something concrete?

thanks(2 votes)