# Chain, Product and Quotient Rule: Summary

by Batool Akmal

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00:01 So we've looked at the product rule in more than one way.

00:05 We've looked at the product rule numerically, we've looked at the product rule and its definition.

00:11 And we've also done in the exhaustive derivation of the product rule.

00:15 So, let's do one more example before we move on to the next and the final rule.

00:21 Let's have a look at these questions, y equals to x cubed, 3x squared minus 1 to the 4.

00:29 First thing that we need to do, is we need to spot what kind of function this is.

00:35 Firstly, you can see that we have a function x cubed and that is multiplying with another function, so there's definitely a product rule. But the second function you will also notice is a function within a function.

00:47 So we're using more than one rule here to work out the differential.

00:51 Let's just remind ourselves of what the product rule states.

00:55 So the product rule says that dy/dx is vdudx plus udvdx or you could use the dash notation which ever one you prefer. It doesn't matter what order you put them in either, because they're plusing so you could put udvdx first, and then vdudx next, because they're adding, it doesn't really make a difference. So, we're going to split this into two, so we're going to say this is my u function which is x cubed, and this is my v function which is 3x squared minus 1 to the power of 4. We differentiate each one of them separately, so I have du/dx which gives me 3x squared. And then I have dv/dx which gives me, remember that this is now the chain rule. So firstly, the outside function, bring the power down 4, leave everything on the inside as it is, decrease the power by 1 and then don't forget to multiply with the differential of the inside function which is 6x. This gives me a final derivative of 24x, 3x squared minus 1 to the power of 3. All of that is done.

02:10 So now, it's just the matter of putting this into the product rule.

02:14 So the product rule is saying dy/dx, we want vdudx, so v is here.

02:20 So I've got 3x squared minus 1 to the power of 4.

02:24 I now need to multiply this with du/dx which is this function. So it's multiplying with 3x squared, put a plus in the middle, because that's what the rule says.

02:34 We now take our u function which is x cubed, and we multiply this with dv/dx, that's this function here.

02:43 So, if I just put brackets around it's because this has a bit more happening.

02:47 I've got 3x squared minus 1 cubed, and we just tidy this up.

02:52 So I've got 3x squared, 3x squared minus 1 to the power of 4.

02:59 So we multiply this term together, which gives me 24x to the 4, and then 3x squared minus 1 cubed.

03:10 There's one last thing you can do here. You can see here that within this whole expression, you have a common factor of 3x squared minus 1 cubed. So you can take that out of this bracket.

03:21 This is just algebra now, so just working on an algebra skills here. Leaving as now, so, you've taken 3x squared minus 1 cubed out, so that leaves you with 3x squared and 1 bracket of 3x squared minus 1. And in this term here, you've taken the 3x squared minus 1 cubed out, leaving you with just 24x to the 4. So, this is just an extra bit of factorizing, if you would like to do, we can neaten this up, that gives you 3x squared minus 1 cubed, multiply that through.

03:59 We end up with 9x to the 4 minus 3x squared plus 24x to the 4.

04:09 You can now see that you can add x to the 4, x to the 4 terms together.

04:16 So, I got 3x squared minus 1 cubed. You can add 9x to the 4 with 24x to the 4 giving you 33x to the 4 minus 3x squared. So, just the next to a little part of fact of expanding things out and you can see that it's made the solution a little bit easier, a little bit easier to read than what we had here previously. So, we've just tidy this up to make it into a slightly, more simpler, a bit more readable differential. So, we've now used the product rule to find the derivative for x cubed bracket, 3x squared minus 1 to the 4. We used the product rule and the chain rule, and we did some algebraic simplifying to get to our final answer.

The lecture Chain, Product and Quotient Rule: Summary by Batool Akmal is from the course Quotient Rule, Chain Rule and Product Rule.

### Included Quiz Questions

1. f(g(x))' = f'(g(x)) g'(x)
2. f(g(x))' = f(g(x)) g'(x)
3. f(g(x))' = f'(x)g(x) + f(x)g'(x)
4. f(g(x))' = f'(g(x)) g(x)
5. f(g(x))' = f'(g'(x))
1. (f(x)g(x))' = g(x)f'(x) + f(x)g'(x)
2. (f(x)g(x))' = g'(x)f(x) - f'(x)g(x)
3. (f(x)g(x))' = g'(x)f'(x)
4. (f(x)g(x))' = f(x)f'(x) + g(x)g'(x)
5. (f(x)g(x))' = f(x)g(x) + f'(x)g'(x)
1. (u / v)' = ( v u' - u v' ) / v²
2. (u / v)' = ( v u' - u v' ) / v
3. (u / v)' = ( v u' + u v' ) / v²
4. (u / v)' = ( v u' - u v' ) / u²
5. (u / v)' = u' / v + u / v'

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