Calculus, all content (2017 edition)
Course: Calculus, all content (2017 edition) > Unit 2Lesson 27: Exponential functions differentiation
- Derivatives of sin(x), cos(x), tan(x), eˣ & ln(x)
- Derivative of aˣ (for any positive base a)
- Derivatives of aˣ and logₐx
- Worked example: Derivative of 7^(x²-x) using the chain rule
- Differentiate exponential functions
- Derivative of 2ˣ (old)
- Differentiating exponential functions review
Derivative of 2ˣ (old)
An older video where Sal finds the derivative of 2ˣ using the derivative of eˣ and the chain rule. Created by Sal Khan.
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- Hm, why did e get brought into this? Why wouldn't the answer be x2^x-1?(41 votes)
- Doesn't work if the variable is in the exponent, you have to use logarithms.(34 votes)
- Why is the derivative of (ln2)x equal to ln2? Shouldn't it be 1/(2x)? Thanks for your help!(31 votes)
- Basically, the "ln2" in (ln2)x acts as the coefficient in front of x. It would be like taking the derivative of 1x, which would be 1, but instead replace "1" with "ln2".(52 votes)
- At0:26Sal wrote 2 as e^(ln 2) . Please explain .(39 votes)
- it actually a property of logarithm if you remember, just like this :
a^(a log b) = b, but now we got e^(e log 2) = 2 (remember ln is just log with the base of e)
i hope that help :)(9 votes)
- Will this work with any constant?(8 votes)
- Yes, you could do
A = e^(ln(A))
d/dx [(e^(ln(A)))^x] = d/dx[e^(x * ln(A))]
= e^(x * ln(A)) * ln(A)
= A^x * ln(A)(22 votes)
- At1:00, why did he convert 2^x to (e^ln 2)^x? Why should we write it in that form?(6 votes)
- He did it that way to avoid using implicit differentiation. But, you don't have to do it like that. For example:
log₂ y = log₂(2^x)
log₂ y =x
d(log₂ y) = dx
dy/(y ln 2)= dx
dy/dx = y ln 2
dy/dx = (2^x )(ln 2)
Although in this simple case there was not much to be gained by using the fact that f(x) = e^(ln (f(x)), there are problems where trying to solve the derivative any other way is nightmarishly difficult. So, it is a good method to learn.(13 votes)
- what is e and where can i find a video about it?(7 votes)
- Around2:35I get confused when Sal starts using the Lebowitz notation in a way that is not '''d/dx". Can someone explain what is happening or refer me to some material that will allow me to master the Lebowitz notation, because that seems to be significantly hampering my understanding of Derivates/Calculus.(9 votes)
- Watch the two preceding videos on the chain rule (Chain rule intro & Chain rule definition) to understand a less confusing example of it's application. Here is what is going on at2:35in a notation you might be more comfortable with:
f(x)=2^x and 2^x=(e^ln2)^x
d f(x)/d((ln2)x)=e^((ln2)x)) In plain language this is saying "the derivative of the function f(x) with respect to the natural log of 2 times x"(1 vote)
- Are logarithms, the number e and trigonometric functions related to one another in some way?
If so then how?(4 votes)
- Logarithms, e, and trigonometric functions are all related to some extent with e and trigonometric functions having the strongest relationship. Using Euler's Formula that e^ix = cos x + i(sin x) (or cis x pronounced like an osculation) we can see that cos x = (e^ix + e^-ix)/2 and that sin x = (e^ix - e^-ix)/2i. A weaker link between logarithms and e is that it is common to use logs with base e, i.e., ln, i.e., the natural log. I hope that this has been informative and if you don't know what osculate means you should look it up because it is a fun word.(7 votes)
- I'm confused about why the chain rule has to be used at all after the expression is rewritten at about0:40into the form e^x. I thought the derivative of e^x is exactly just e^x? Yet at about3:30, Sal goes on to multiply e^x by the derivative of the x. Is it that we have to do that if the x on the e^x isn't just a simple x, to put it crudely, but is written as a specific operation like (ln 2)x? Thanks in advance.(5 votes)
- Actually, the chain rule applies to all derivatives. However, we skip stating it explicitly when it gives trivial results.
For example, the derivative of x² is 2x dx, and "dx" may be thought of as the derivative of x, an application of the chain rule. But since the derivative of x is just 1 dx, we don't usually need to focus on the fact that the chain rule actually applies in such trivial cases.
So, the derivative of e^x is e^x dx, where dx can be considered the derivative of x, an application of the chain rule. Likewise, e^[f(x)] = e^[f(x)} f'(x), the same type of application of the chain rule -- although, in this case, the results are not trivial.(4 votes)
- i understand the chain rule for f(g(x)) but it seems in this case that it also applies for g(x)^f(x) and i do not recall he said that in previous videos. is it true? and if so, how do we know it?(3 votes)
Let's see if we can take the derivative with respect to x of 2 to the x power. And you might say, hold on a second. We know how to take the derivative of e to the x. But what about a base like 2? We don't know what to do with 2. And the key here is to rewrite 2 to the x so that we essentially have it as e to some power. And the key there is to rewrite 2. So how can we rewrite 2 so it is e to some power? Well, let's think about what e to the natural log of 2 power is. The natural log of 2 is the power that I would have to raise e to to get to 2. So if we actually raise e to that power, we are going to get to 2. So what we could do, instead of writing 2 to the x, we could rewrite this as e. We could rewrite 2 as e to the natural log of 2, and then raise that to the x power. So this is the x power in yellow. And so let's do that right over here. So instead of taking the derivative with respect to x of 2 to the x, let's say, let's just take the derivative with respect to x of the exact same expression rewritten, of e to the natural log of 2 raised to the x power. Let me put this x in that same color, dx. Now we know from our exponent properties if we raise something to some power, and then raise that to another power, we can take the product of the two powers. Let me rewrite this just to remember. If I have a to b, and then I raise that to the c power, this is the exact same thing as a to the b times c power. So we can utilize that exponent property right here to rewrite this as being equal to the derivative with respect to x of e to the natural log of 2 times x. And what's neat about this is now we've got this into a form of e to the something. So we can essentially use the chain rule to evaluate this. So this derivative is going to be equal to the derivative of e to the something with respect to that something. Well, the derivative e to the something with respect to that something is just e to that something. So it's going to be equal to e to the natural log of 2 times x. So let me make it clear what I just did here. This right over here is the derivative of e to the natural log of 2 times x with respect to the natural log of 2-- let me make it a little bit clearer-- with respect to the natural log of 2 times x. So we took the derivative of e to the something with respect to that something-- that's this right here, it's just e to that something. And then we're going to multiply that by, this is just an application of the chain rule, of the derivative of that something with respect to x. So the derivative of natural log of 2 times x with respect to x is just going to be natural log of 2. This is just going to be natural log of 2. The derivative of a times x is just going to be equal to a. This is just the coefficient on the x. And just to be clear, this is the derivative of natural log of 2 times x with respect to x. So we're essentially done. But we can simplify this even further. This thing right over here can be rewritten. And let me draw a line here just to make it clear that this equals sign is a continuation from what we did up there. But this e to the natural log of 2x, we can rewrite that, using this exact same exponent property, as e to the natural log of 2, and then all of that raised to the x power. And of course, we're multiplying it times the natural log of 2, so times the natural log of 2. Well, what is e to the natural log of 2? Well, we already figured that out. That is exactly equal to 2. This right over here is equal to 2. And so now we can simplify. This whole thing, the derivative of 2 to the x, is equal to-- and I'll switch the order a little bit-- it is the natural log of 2, that's this part right over here, times 2 to the x. Or we could write it as 2 to the x times the natural log of 2.