AP®︎/College Calculus AB
- Chain rule
- Common chain rule misunderstandings
- Chain rule
- Identifying composite functions
- Identify composite functions
- Worked example: Derivative of cos³(x) using the chain rule
- Worked example: Derivative of √(3x²-x) using the chain rule
- Worked example: Derivative of ln(√x) using the chain rule
- Chain rule intro
f(x)=ln(√x) is a composition of the functions ln(x) and √x, and therefore we can differentiate it using the chain rule.
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- This is a silly thing, but couldn't Sal have solved this without using the chain rule at all? We know from the power law of the natural log is that if the term inside it has an exponent (in this case, 1/2) we can bring it out. So that would leave us with 1/2*ln(x) which is just a million times simpler.(30 votes)
- I am a little lost with this comment and with the response from "SlightlyChaotic". Applying the power rule of the natural log as (1/2*ln(x)) will give the answer to that function. However, to find the derivative of the function we will have to to do so as presented by Sal in his video.
How would you apply the power rule of the natural log to arrive at the derivative of the function <ln(sqrt(x))>?
What am I missing?(2 votes)
- v'(u(x))= 1/u(x)... How did we get to this?
Because.... v'(x)= 1/(x)
So v'(u(x)) = 1/x rootx(3 votes)
- Let u(x)=p. Then we're taking the derivative with respect to p of v(p) is v'(p)=1/p.
Resubstitute p and we get v'=1/(√x).
By chain rule, to get the derivative of v with respect to x, we then multiply by u'(x). So our result is
- why is v'(u(x)) = 1/u(x) ??(3 votes)
- That is just in this case where the derivative of ln(x)* is *1/x , and v is representing *ln(x)*. So, *v'(x)* is just *1/x.*(5 votes)
- Could I write v'(u(x)) as (1/x)*sqrt x, and leave it like that?(3 votes)
- Lets look at the whole equation. It says that,
Using the chain rule, f'(x) is 1/(√x) (1/2√x) (1)(2 votes)
- What if it is f(x)= ln(2x+5)^3, how wpuld the chain rule play into that, with the ^3 on the end?(2 votes)
- Where is the ^3 on?
If it's ln((2x+5)^3), then it equals to 3ln(2x+5), then f'(x)=3(ln(2x+5))'=3/(2x+5)*(2x+5)'=6/(2x+5)
If it's (ln(2x+5))^3, then the derivative equals to f'(x)= 3*(ln(2x+5))^2*(ln(2x+5))'=3*(ln(2x+5))^2*(1/(2x+5))*(2x+5)'=[6(ln(2x+5))^2]/(2x+5).(2 votes)
- At5:25, why √x*√x equals x and not |x|?(1 vote)
- By definition, a square root of a number is any number that when multiplied by itself gives the original number.(2 votes)
- Is that notation correct?
sin' x(sine prime of x)(2 votes)
- You mean the derivative of
It would be more correct to write it as
f'(x), f(x) = sin(x)
I hope this helps!(2 votes)
- Newton's notation seems to make it clear what any given derivative has been taken with respect to, since it sits on the denominator of the 'fraction'. Using Leibniz' notation, is it correct to say that whatever is in the function bracket defines what the derivative has been taken with respect to?(2 votes)
- I believe you have mixed up the notations. According to KA,
Newton's notation is ẏ or ḟ, most commonly used in physics.
This notation does not very clearly show what the derivative is with respect to.
Lagrange's notation is y’ or f’(x), pronounced "f prime".
The "x" in the brackets is what the derivative is wrt.
Leibniz's notation is the most common d/dx, df/dx, or dy/dx.
The "denominator" is the variable the derivative is wrt.
Hope that I helped, and correct me if I'm wrong.(2 votes)
- Why use the chain rule here? Why not simplify to (ln x)/2 and differentiate that? = 1/2 d/dx(ln(x)) =(1/x)/2 = 1/2x. Much quicker.(2 votes)
- Good point. But this isn't always the case. in other cases using chain rule is faster or even it is the only way.(1 vote)
- I'm taking calculus now and am liking it. But I find my algebra isn't as good as I thought when trying to simplify these expressions. Like at4:40Sal simplifies 1/2x^-1/2, I get sort of lost. Is there a spot on Khan Academy to practice and learn just simplifying complex fractions/exponents and such?(2 votes)
- I guess this is going to help you : "https://en.khanacademy.org/math/algebra-home/alg-rational-expr-eq-func".(1 vote)
- [Voiceover] So we have here f of x being equal to the natural log of the square root of x. And what we wanna do in this video is find the derivative of f. And the key here is to recognize that f can actually be viewed as a composition of two functions. And we can diagram that out, what's going on here? Well if you input an x into our function f, what's the first thing that you do? Well, you take the square root of it. So if we start off with some x, you input it, the first thing that you do, you take the square root of it. You are going to take the square root of the input to produce the square root of x, and then what do you do? You take the square root and then you take the natural log of that. So then you take the natural log of that, so you could view that as inputting it into another function that takes the natural log of whatever is inputted in. I'm making these little squares to show what you do with the input. And then what do you produce? Well you produce the natural log of the square root of x. Natural log of the square root of x. Which is equal to f of x. So you could view f of x as this entire set, or this entire, I guess you could say, this combination of functions right over there. That is f of x, which is essentially, a composition of two functions. You're inputting into one function then taking that output and inputting it into another. So you could have a function u here, which takes the square root of whatever its input is, so u of x is equal to the square root of x. And then you take that output, and input it into another function that we could call v, and what does v do? Well it take the natural log of whatever the input is. In this case, in the case of f, or in the case of how I just diagrammed it, v is taking the natural log, the input happens to be square root of x, so it outputs the natural log of the square root of x. If we wanted to write v with x as an input, we would just say well that's the natural log, that is just the natural log of x. And as you can see here, f of x, and I color-coded ahead of time, is equal to, f of x is equal to, the natural log of the square root of x. So that is v of the square root of x, or v of u of x. So it is a composition which tells you that, okay, if I'm trying to find the derivative here, the chain rule is going to be very, very, very, very useful. And the chain rule tells us that f prime of x is going to be equal to the derivative of, you can view it as the outside function, with respect to this inside function, so it's going to be v prime of u of x, v prime of u of x, times the derivative of this inside function with respect to x. So that's just u prime, u prime of x. So how do we evaluate these things? Well, we know how to take the derivative of u of x and v of x, u prime of x here, is going to be equal to, well remember, square root of x is just the same thing as x to 1/2 power, so we can use the power rule, bring the 1/2 out from so it becomes 1/2 x to the, and then take off one out of that exponent, so that's 1/2 minus one is negative 1/2 power. And what is v of x, sorry, what is v prime of x? Well the derivative of the natural log of x is one over x, we show that in other videos. And so we now know what u prime of x is, we know what v prime of x is, but what is v prime of u of x? Well v prime of u of x, wherever we see the x, we replace it, let me write that a little bit neater, we replace that with a u of x, so v prime of u of x is going to be equal to, is going to be equal to one over u of x, one over u of x, which is equal to, which is equal to one over, u of x is just the square root of x. One over the square root of x. This thing right over here, we have figured out, is one over the square root of x, and this thing, u prime of x, we figured out, is 1/2 times x to the negative 1/2, and x to the negative 1/2, I could rewrite that as 1/2 times one over x to the 1/2, which is the same thing as 1/2 times one over the square root of x, or I could write that as one over 2 square roots of x. So what is this thing going to be? Well this is going to be equal to, in green, v prime of u of x is one over the square root of x, times, times, u prime of x is one over two times the square root of x, now what is this going to be equal to? Well, this is going to be equal to, this is just algebra at this point, one over, we have our two and square root of x times square root of x is just x. So it just simplifies to one over two x. So hopefully this made sense, and I intentionally diagrammed it out so that you start to get that muscle in your brain going of recognizing the composite functions, and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class, or in your calculus textbook. But as you get more practice, you'll be able to do it, essentially, without having to write out all of this. You'll say okay, look, I have a composition. This is the natural log of the square root of x, this is v of u of x. So what I wanna do is I wanna take the derivative of this outside function with respect to this inside function. So the derivative of natural log of something, with respect to that something, is one over that something. So it is one over that something, the derivative natural log of something with respect to that something is one over that something, so that's what we just did here. One way to think about it, what would natural log of x be? Well that'd be one over x, but it's not natural log of x. It's one over square root of x, so it's going to be one over the square root of x, so you take the derivative of the outside function with respect to the inside one, and then you multiply that times just the derivative of the inside function with respect to x. And we are done.