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## Class 12 math (India)

Reverse chain rule introduction

What is the “reverse chain rule”, and why it does the same thing as u-substitution.

## Want to join the conversation?

- In what circumstances would it be necessary to resort to u substitution?(20 votes)
- Good Question.

U sub is a method for algebraically simplifying the form of a function so that its anti-derivative can be more easily recognized. This method is intimately related to the chain rule for differentiation, which when applied to anti-derivatives is sometimes called the reverse chain rule. If you can do all the algebraic manipulation and simplification in your head, then you don’t actually need to explicitly write out the the u substitution you performed, though you are actually doing it, that is, recognizing some f and f’ within the integrand. You will find over time, as you do more and more problems, you will be able to do more and more in your head, and that is fine. Sometimes it is helpful to write out the u sub so as just to avoid careless algebraic errors, however, everyone is different with respect to their personal comfort zone and preferences when choosing what to write out when simplifying expressions and what not. U sub is just one simplification technique of many in our toolbox.

On a personal note, I have found that the exercises here at the Khan academy don’t spend as much time converting seemingly intractable anti-derivatives into an integrable form as when I was in university – it made integration much more of an art form than differentiation, which was pretty much cook-book math: recognize the form, apply the appropriate recipe. We used to call the art of integration “Algebra and Innovation.”

I hope this answer was satisfactory.

Keep Studying!(53 votes)

- Good Question.
- I am having a lot of trouble identifying g(x). I understand that sinx is f(x) and cosx is f'(x), so is the ^2 g’ in this case?(9 votes)
- g'(x) would be u^2, while g(x) would be (u^3)/3(13 votes)

- Isn’t this basically just u-substitution? Is there a particular difference or no?(7 votes)
- After a fashion, yes. But the more you can do without resorting to u substitution, the less time you will take. So, learning as many forms as you can will help you solve problems much more efficiently. Why take the extra 5 minutes to do u-sub if you can skip it? On a timed exam, 5 minutes can make a lot of difference.(9 votes)

- integration of e^atanx?(4 votes)
- While there is a way to solve that integral using very advanced calculus, it is impossible to solve that with any method taught in the first few years of studying calculus. The solution itself involves very advanced functions not covered at this level of study.(5 votes)

- Why exactly does the cos x disappear? I understand how he gets ((sin x)^3)/3 but where’d the cosine go?(3 votes)
- When doing the µ-substitution we replace the integrand with new symbols. Symbols that are in the first integrand are sinx, cosx and dx. These all depend on x, so we need to make them all depend on µ. Here is how the substitution was made:

µ=sin(x)

dµ=cos(x)dx <=> [dx=dµ/cos(x)]

Now dx in the first integrand can be replaced with dµ/cos(x), so you get the new integrand:

µ^2*cosx*(dµ/cosx)=µ^2dµ

If we calculate this integrand µ^2dµ we get the same solution of (µ^3)/3+C, which from the substitution equals [sin(x)^3]/3+C. (dont forget the constant at the end!)(3 votes)

- When doing the µ-substitution we replace the integrand with new symbols. Symbols that are in the first integrand are sinx, cosx and dx. These all depend on x, so we need to make them all depend on µ. Here is how the substitution was made:
- Isn’t the integral of g'(f(x)) = (g(f(x))/(f'(x))+C? E.g. ∫e^(2x) = (e^(2x))/2.(3 votes)
- This trick only works when f'(x) is a constant, which means f(x) is of the form ax + b. So it’s true that ∫e^(2x)dx = (e^(2x))/2, and ∫e^(ax + b)dx = (e^(ax + b))/a.

But it is NOT true that ∫e^(x²)dx = (e^(x²))/(2x). (FALSE!) You can see that if you go to check your antiderivative by taking its derivative, if f'(x) is not a constant then you need the quotient rule and it won’t be correct.(2 votes)

- This trick only works when f'(x) is a constant, which means f(x) is of the form ax + b. So it’s true that ∫e^(2x)dx = (e^(2x))/2, and ∫e^(ax + b)dx = (e^(ax + b))/a.
- Wouldn’t Cosx be changed to -Sinx since we’re integrating here?(2 votes)
- If you’re just asking about Integrating cos (x) it would give sin (x) not -sin (x)

but if you’re suggesting changing cos (x) to sin (x) and multiply it with [(sin (x))^3] / 3 then you’re mixing things up between integration by parts & reverse chain rule.

Keep Studying!(3 votes)

- If you’re just asking about Integrating cos (x) it would give sin (x) not -sin (x)
- So in u-substitution the du is the derivative of f(x) so when we take the integral of it we have the original function (according the the Fundamental Theorem) and so the du part doesn’t necessarily disappear but instead it is ‘inside the composite function g(f(x)). Is this right?(2 votes)
- Yes, that’s a good way to think about it. U-substitution is almost just a way to help us see what parts of the integrand could be produced by the chain rule so that we can deal with those parts appropriately when we find the antiderivative (because we know that they’ll appear when we take the derivative).(2 votes)

- when we use the u-subtitution technique we have to calculate the integral in a “new interval” because of the substitution : I=[a.b] “becomes” J=[g(a) . g(b)].

What about the reverse chain rule ? Are we calculating in I or in J ? thx ^^(2 votes) - I am struggling with the integral x(x+6)^1/2. When you attempt to derive the inner function, you get x (outer function) is = 1 (derivative of the inner function)

The actual answer is 2/5(x+6)^5/2-6(2/3)(x-6)3/2

I see a pattern with similar problems, but can you use the reverse chain method?(1 vote)- Ah, the “x” in your function is NOT an outer function. In fact, outer function is something appeared in composite function, ie in the form g(f(x)). Here “x” is just being multiplied with the other thing. So this can’t be solved by reverse chain rule.

So just use u-substitution. Denote u=x+6, and du=dx.

Int ( x(x+6)^(1/2) dx ) = Int ( (u-6)u^(1/2) du ).

Now distribute the u^(1/2) to u-6, which gives Int ( u^(3/2) – 6u^(1/2) du )

And then use Reverse Power Rule, which gives 2/5u^(5/2) – 6(2/3)u^(3/2) +C, the exact same as what you’ve posted.

As Sal mentioned, Reverse Chain Rule is just u-substitution. Indeed it is faster but it is likely to get the wrong answer.

Hope this helps!(2 votes)

- Ah, the “x” in your function is NOT an outer function. In fact, outer function is something appeared in composite function, ie in the form g(f(x)). Here “x” is just being multiplied with the other thing. So this can’t be solved by reverse chain rule.

## Video transcript

– [Voiceover] Hopefully we all remember our good friend the chain rule from differential calculus that tells us that if I were to take the derivative with respect to x of g of f of x, g of, let me write those parentheses a little bit closer, g of f of x, g of f of x, that this is just going to be equal to the derivative of g with respect to f of x, so we can write that as g prime of f of x. G prime of f of x, times the derivative of f with respect to x, times f prime of x. And if you want to see it in the other notation, I guess you could say, it would be, you could write this part right over here as the derivative of g with respect to f times the derivative of f. The derivative of f with respect to x, and that’s going to give you the derivative of g with respect to x. This is just a review, this is the chain rule that you remember from, or hopefully remember, from differential calculus. It’s hard to get, it’s hard to get too far in calculus without really grokking, really understanding the chain rule. So what I want to do here is, well if this is true, then can’t we go the other way around? If I wanted to take the integral of this, if I wanted to take the integral of g prime of f of x, g prime of f of x, times f prime of x, dx, well, this should just be equal to, this should just be equal to g of f of x, g of f of x, and then of course whenever I’m taking an indefinite integral g of, let me make sure they’re the same color, g of f of x, so I just swapped sides, I’m going the other way. So if I’m taking the indefinite integral, wouldn’t it just be equal to this? And of course I can’t forget that I could have a constant here now that might have been introduced, because if I take the derivative, the constant disappears. And so this idea, you could really just call the reverse chain rule. Reverse, reverse chain, the reverse chain rule. Which is essentially, or it’s exactly what we did with u-substitution, we just did it a little bit more methodically with u-substitution. And we’ll see that in a second, but before we see how u-substitution relates to what I just wrote down here, let’s actually apply it and see where it’s useful. And this is really a way of doing u-substitution without having to do u-substitution, or doing u-substitution in your head, or doing u-substitution-like problems a little bit faster. So let me give you an example. So let’s say that we had, and I’m going to color code it so that it jumps out at you a little bit more, let’s say that we had sine of x, and I’m going to write it this way, I could write it, so let’s say sine of x, sine of x squared, and obviously the typical convention, the typical, the sine of x squared, the typical convention would be to put the squared right over here, but I’m going to write it like this, and I think you might be able to guess why. Sine of x squared times cosine of x. Times, actually, I’ll do this in a, let me do this in a different color. Times cosine of x, times cosine of x. So I encourage you to pause this video and think about, does it meet this pattern here, and if so, what is this indefinite integral going to be? Well let’s think about it. If f of x is sine of x, what’s the derivative of that? What’s f prime of x? Well f prime of x in that circumstance is going to be cosine of x, and what is g? Well g is whatever you input into g squared. So what’s this going to be if we just do the reverse chain rule? Well this is going to be, well we take sorry, g prime is taking whatever this thing is, squared, so g is going to be the anti-derivative of that, so it’s going to be taking something to the third power and then dividing it by three, so let’s do that. So if we essentially take the anti-derivative here with respect to sine of x, instead of with respect to x, you’re going to get you’re going to get sine of x, sine of x to the, to the third power over three, and then of course you have the, you have the plus c. And if you don’t believe this, just take the derivative of this, you’ll have to employ the chain rule and you’ll get exactly this. And you say well wait, how does this relate to u-substitution? Well in u-substitution you would have said u equals sine of x, then du would have been cosine of x, dx, and actually let me just do that. That actually might clear things up a little bit. You would set this to be u, and then this, all of this business right over here, would then be du, and then you would have the integral, you would have the integral u squared, u squared, I don’t have to put parentheses around it, u squared, du. U squared, du, well, let me do that in that orange color, u squared, du. Well that’s pretty straightforward, this is going to be equal to u, this is going to be equal to u to the third power over three, plus c, which is equal to what? Well we just said u is equal to sine of x, you reverse substitute, and you’re going to get exactly that right over here. So when we talk about the reverse chain rule, it’s essentially just doing u-substitution in our head. So in the next few examples, I will do exactly that.