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Course: Class 12 math (India)>Unit 9

Lesson 7: Reverse chain rule

Reverse chain rule example

Using the reverse chain for (x/2)sin(2x^2 + 2).

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• I don't see how using this method makes it any simpler to calculate antiderivatives compared to u-substitution. Instead of calling it u, you say it's f(x), but ultimately you're still giving names to certain parts of the expression.
(24 votes)
• Yes, it is the same deal - but u-sub was born from reversing the chain rule, not the other way around. Be aware of both - use what works for you.
(40 votes)
• I have been integrating functions and differentiating functions.....but what is their purpose and why do I need to know them for my future (hopefully) field of physics. I know it's used a lot, but how?
(2 votes)
• Physics and engineering are chock full of reasons why you need calculus. Integration and differentiation are used in physics when calculating distance, speed (derivative of distance) and acceleration (derivative of speed), jerk, joust. Calculations on movement within and interactions between either electrical or magnetic fields also require a good bit of calculus. Finding volume (integral of surface area), surface area (derivative of volume), linear distance, 3-D rotational volume (triple integrals), maximum/minimum container volume use calculus, and even the Laws of Thermodynamics are actually built on calculus. Moreover, forensics, astronomy, astrophysics, computer science and engineering as well as literally dozens of scientific fields all use aspects of physics and, as a result, also use calculus.
(6 votes)
• why did he not integrate f'(x) at ?
(4 votes)
• Because when/if you take the integral of the derivative you just get the original function back. So if you take the derivative of 4x you get 2x back...
(1 vote)
• But the usual equation/formula I see is the integrate sign f'(g(x))(g'(x)) ... In the example of this video, which would be f(x) and which would be g(x)? I tried to apply it to the questions, but sometimes I get frustrated because I just can't "see" which part of the equation makes good f(x) or g(x). Any tips? Thank you for helping me out.
(1 vote)
• Sal has the f and g switched from what you have. So with your notation:
g(x) = 2x^2 + 2
f(x) = -cos(x)
(3 votes)
• I paused the video and solved it with no problem, but i didn't know that what i was doing was the chain rule reversed...
It's sad how teachers teach you how to solve a problem but not how to understand it !
(2 votes)
• I still prefer u-substitution to this.
(2 votes)
• Yes, u-sub is easier to think of. In this video Sal set f(x) = 2x^2+2, but in u-substitution you set u = 2x^2+2. The key is to check whether there is a derivative of the function multiplied outside.
(1 vote)
• When you write the final answer why don't you put another 1/2 out the front because usually when you integrate sin(ax) you would get -1/acos(ax)?
(1 vote)
• This is basically U-Substitution without technically doing. It's a bit more in your head. Am I right?
(1 vote)
• at aren't the subtitles suppose to say "We could have used u-substitution, but hopefully..." instead of "We could have used substitution, but hopefully...". I know this is a minor detail, but I did get a bit confused while only looking at the Transcript.
(1 vote)
• wait im confused why is f'(x) not taken into account?
(1 vote)

Video transcript

- [Voiceover] Let's see if we can evaluate the indefinite integral x over two times sine of two x squared plus two, dx. And try to pause the video and see if you can work through it on your own. So, let's see what is going on here. So, I have this x over two, and then I have sine of two x squared plus two. Now, if I were just taking the indefinite integral of sine of x, that is pretty straightforward. The indefinite integral of sine of x. Well, we know that the derivative of cosine of x is equal to negative sine of x. So if I were to take the derivative of negative cosine of x, that's going to be positive sine of x. So this is just going to be negative cosine of x. And I could have made that even clearer. I could have put a negative here and then a negative here. And you see, well look, the anti-derivative of negative sine of x is just cosine of x, and then I have this negative out here, negative cosine of x. But that's not what I have here. I don't have sine of x. I have sine of two x squared plus two. But then I have this other thing with an x here, and so what your brain might be doing, or it's good once you get enough practice when your brain will start doing this, say okay, this is interesting. This kind of looks like the derivative of this. And this is, if we were to call this... I'm tired of that orange. If we were to call a... Why am I doing this, color changing this... If we were to call this f of x. If two x squared plus two is f of x, Two x squared plus two is f of x. What is f prime of x? Well, then f prime of x, f prime of x is going to be four x. And this thing right over here isn't exactly four x, but we can make it, we can do a little rearranging, multiplying and dividing by a constant, so this becomes four x. What if, what if we were to... What if we were to multiply and divide by four, so we multiply by four there and then we divide by four, and then we take it out of the integral sign. And even better let's take this two out so let's just take. So, let's take the one half out of here, so this is going to be one half. And so I could have rewritten the original integral as one half times one fourth, so it's one eighth times the integral, times the integral of four x times sine of two x squared plus two, dx. Well, now this is interesting, because if this is f of x, if... If this business right over here if f of x, so we're essentially taking sine of f of x, then this business right over here is f prime of x, which is a good signal to us that, hey, the reverse chain rule is applicable over here. We can rewrite this, we can also rewrite this as, this is going to be equal to one. We can rewrite this as, the same things as one eighth... They're the same colors. I'm using a new art program, and sometimes the color changing isn't as obvious as it should be. So one eighth times the integral of f prime of x, f prime of x times sine, sine of f of x, sine of f of x, dx, throw that f of x in there. So, sine of f of x. And so when you view it this way, you say, hey, by the reverse chain rule, I have... I have a function, and I have its derivative here, so I can really just take the antiderivative with respect to this. This is essentially what we're doing in u-substitution. You could do u-substitution here, you could set u equalling this, and then du is going to be four x dx. But now we're getting a little practice, starting to do a little bit more in our heads. So, what would this interval integrate out to be? Well, this would be one eighth times... Well, if you take the antiderivative of sine of f of x with respect to f of x, well, we already saw that that's negative cosine of x, so this is going to be times negative cosine, negative cosine of f of x. Negative cosine of f of x, negative cosine of f of x. Woops, I was going for the blue there. And then of course you have your plus c. So what is this going to be? Well, instead of just saying f pri.. Instead of saying in terms of f of x, we just say it in terms of two x squared. This is going to be... Or two x squared plus two is going to be one eighth. Times, I have a function and I have its derivative, so I can really just integrate with respect to that function, so it's going to be times negative negative cosine of two x squared plus two, two x squared plus two, and then, of course, I have my... I keep switching to that color. I have my plus c, and of course, I could just take the negative out, it would be negative one eighth cosine of this business and then plus c. And we're done. We have just employed the reverse chain rule. We could have used substitution, but hopefully we're getting a little bit of practice here. Hey, I'm seeing something here, and I'm seeing it's derivative, so let me just integrate with respect to this thing, which is really what you would set u to be equal to here, integrating with respect to the u, and you have your du here. This times this is du, so you're, like, integrating sine of u, du. I encourage you to try to use u-substitution here, and you'll see it's the exact same thing that we just did.