AP®︎/College Calculus AB
- 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
The derivative of is :
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 directly from the definition of the derivative as a limit.
Here we find the derivative of by using the fact that and applying implicit differentiation.
Note: Implicit differentiation is a technique that is taught later in the course.
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- In the first video at7:26, I don't understand how he was able to move the limit to inside the natural log.(22 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!(25 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.(3 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(10 votes)
- in the 2nd video, at1:21, I thought d/dx [e^y]=e^y? and is d/dx [e^y]=e^y correct?(3 votes)
- 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!(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.(3 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).(4 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(2 votes)
- 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.(5 votes)
- What is the derivative of pi^x or tau^x?(2 votes)
- the derivative of any constant to the x power is found like this.
start with a^x where a can be any constnat.
Use log rules to rewrite a^x as e^(ln(a^x)) let me know if you don't get how that works.
now you can use the chain rule to derive e^ln(a^x). The chain rule basically lets you solve a composite function f(g(x)). here f(x) is e^x and g(x) is ln(a^x) which can also be simplified to x*ln(a) by log rules.
the chain rule says f(g(x)) gets us f'(g(x))*g'(x) so this gets us e^ln(a^x)*ln(a). You can simplify e^ln(a^x) the same way we got it with log rules so the final simplified answer is a^x*ln(a)
Hopefully that helped but let me know if you don't understand something.
You can simplify(2 votes)
- At6:23of 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?(2 votes)
- 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)
- Why is e=(1+n)^(1/n)?(1 vote)
- e = "lim" (n->infinity) (1+n)^(1/n)
That is the definition. Take a look at the proof theorem for e^x d/dx = e^x.(3 votes)
- I am bit confused at the below explanation2:07What is e to the natural log of x going to be?2:12The natural log of x is the power2:14I need to raise e to, to get to x.2:16So if I actually raise e to that power,2:18to that exponent, I'm going to get x.
e^(ln(x)) = x^(ln(e)) = x^1 = x. I could get to the x in the denominator but I was unable to understand2:16through to 2.18.(2 votes)