Gradients and First Principles: Exercise 2

by Batool Akmal

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00:00 So, let's look at the next exercise now.

00:02 This one is a little bit easier because we've done an example like this before.

00:06 So, let's just do the algebra and hopefully enjoy this one.

00:10 Remember the definition? I'm going to write that once again, over, and over again.

00:14 So dy by dx as the limit of delta x tends to 0 is f of x + delta x - f of x, all over delta x.

00:27 Okay, we're gonna put that into the numbers or the numerical function.

00:31 Remember what we're doing each x function, we are changing to x + delta x.

00:35 And this in itself is just the original function, which we'll bring in in a second.

00:40 So, let's write this all up.

00:42 So we've got dy by dx, as the limit of delta x tends to 0.

00:46 We have 3, change the x squared, to x + delta x squared - 2, change the x to x + delta x.

00:56 And then we have + 7 at the end.

00:59 Remember to take away the original function, so this is the unchanged function.

01:03 So, we'll have 3 x squared - 2 x + 7 And then it's all being divided by delta x.

01:11 Okay, here's the common mistake that I see students make is this.

01:14 So, when you expand this, please be careful.

01:17 Remember to expand this as two bracket.

01:19 So, this is going to give you a quadratic.

01:21 So, if we just do that here, just to be extra cautious.

01:24 We'll leave it as (x + delta x) (x + delta x) first.

01:28 We can expand this bit.

01:30 So we've got 2 x - 2 delta x + 7.

01:34 And here once again, be careful, because it's multiplying through with a minus the signs inside will change.

01:40 So you have - 3 x squared + 2 x - 7 and it's all being divided by delta x.

01:48 Okay, expand this out.

01:50 Firstly, so we've got 3.

01:52 You're going to times it through.

01:53 So we have x squared + x delta x + another x delta x.

01:58 so that gives you 2 x delta x + delta x squared.

02:03 This is all fine.

02:03 2 x - 2 delta x + 7 - 3 x squared + 2 x - 7 all over delta x times 3.

02:15 So we've got, 3 x squared + 6 x delta x + 3 delta x squared - 2 x - 2 delta x + 7 - 3 x squared + 2 x- 7 we're almost there.

02:31 So all divided by delta x.

02:35 Alright? Here you will see that we can cancel this function out.

02:38 Remember what I said that if we're doing it correctly, this should usually cancel out.

02:42 So, you have positive 3 x squared, and a negative 3 x squared, we have a negative 2 x and a positive 2 x and a - 7 and a + 7.

02:54 If I do this in two steps, so we're now left with 6 x delta x + 3 delta x squared - 2 delta x, all over delta x.

03:05 You now see that you've got a delta x common in each term here, here, here and common with the denominator, so you can cancel delta x out.

03:13 You can cancel this one with this one of them, and this term.

03:18 So, we're now left with 6 x + 3 delta x - 2.

03:25 And lastly, don't forget to apply this, we want our final answer of d y by d x, as the limit of delta x tends to 0, that means that anything that had delta x in it, take it to 0, because it's 3 times 0 will give you 0, giving you a final answer of 6 x - 2.

03:43 So that is your differential.

03:46 So, we'll also do this, the faster, and the shorter way, just because it's easier, and you'll see that you'll prefer it.

03:52 So if I just write this function out again, so we can look at each term and what we're doing to it - 2 x + 7 Remember what we discussed in the previous lecture that if you are differentiating this, you bring the power to the front, and you - 1 from this power.

04:08 So you decrease the power by 1.

04:10 So let's do that one bit at a time.

04:11 So we've got 2 times the 3 stays, that doesn't bother us.

04:15 we've got x to the power of 2 - 1.

04:18 So you have to do this in order, bring the power to the front and decrease the power by 1.

04:22 Again, you do this for the next term.

04:24 So the power of x now is just 1.

04:26 So you can bring 1 to the front, and we then times it with 2 and we have x to the 1 - 1.

04:34 And lastly, this constant here will disappear and it will go to 0.

04:39 And that's because you can imagine writing this as 7 x to the 0.

04:43 So anything to the power of 0 is just 1.

04:45 So, 7 x to the 0 is just 7.

04:48 When you bring the power to the front when you're differentiating it it's just going to be 0 multiplied by 7 x to the power of - 1.

04:57 So your constant will always disappear.

04:59 So any number numbers that you see at the end of your functions that you're differentiating, they just go away, so you can just ignore them.

05:07 Here we have 6 x to the 1 when we tidy this up, and then this gives you 2 x to the 0 is just 1.

05:16 So like we've discussed before, anything to the power of 0 is just 1.

05:19 So this is your final answer, 6 x minus 2, which is exactly the same value that we got when we differentiate it from first principles.

05:28 But you can see that thanks to modern day mathematicians and modern calculus, things have become a lot easier for us.

05:34 And so we encourage that you use these faster methods unless, we actually need to go back to basics, and derive improve using first principles.

The lecture Gradients and First Principles: Exercise 2 by Batool Akmal is from the course Calculus Methods: Gradients and First Principles.

Included Quiz Questions

1. f(x + 1) = x² + (δx)² + 2xδx + x + δx , f'(x) = 2x + 1
2. f(x + 1) = x² + (δx)² + 2xδx + x , f'(x) = 2x
3. f(x + 1) = x² + (δx)² + 2x + x + δx , f'(x) = x + 1
4. f(x + 1) = x² + (δx)² + 2xδx + x + δx , f'(x) = 2x
5. f(x + 1) = x² + 2xδx + x + δx , f'(x) = 2x + 1

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