# Gradients and First Principles: Exercise 1

by Batool Akmal

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00:01 Welcome back, I hope you've attempted the questions that we just previously set and you've enjoyed the two challenges that we're in there.

00:08 Let's get you out of your misery now and just go through the solution.

00:13 We were looking at differentiate square root of x.

00:16 Now remember I said, attempt to differentiate it using first principles and also using the faster method of differentiation, so we'll see both now.

00:27 So we have a function y equals to square root of x or f of x equals to square root of x.

00:34 I'm going to remind you of the first principles formula 'cause that's where we'll start with.

00:39 So remember the first principle states that the gradient dy/dx, as the limit of delta x tends to zero is f of x plus delta x minus f of x all over delta x.

00:51 Right, let's now use this function or this definition on to our function that has been given to us.

01:01 So in this case when we use dy/dx of the limit, delta x tends to zero.

01:07 Our original function is now square root of x, so we need to add delta x to the value of x.

01:13 So what we'll have here is x plus delta x, take away the original function and divide it by delta x.

01:21 You can re-write this function as a power, so if you recall that the square root has a power of a half so we can write it as x plus delta x minus x to the power of a half, that's also to the power of a half divided by delta x.

01:38 Okay, now usually this stage isn't a problem because all we have to do is expand it.

01:44 However, expanding something like this isn't as straightforward as the questions that we have done in the examples.

01:50 If you really think hard, you will remember that in order to differentiate a function like this, you need to use the binomial theorem.

01:58 For the ones you've forgotten we'll quickly just have a recap on the side of binomial theorem.

02:04 So if we just do that here on the side, if I remind you of binomial theorem in general, so we have binomial theorem here.

02:12 Looking at a function or expanding something like a plus x to the power of n.

02:22 In order to expand this, you start with the highest power of a and the lowest power of x and you use nCr which is the factorial method.

02:32 So if I just remind you how that works it goes a to the n, x to the power of zero over zero factorial.

02:39 Our next value will be n so you bring the power to the front.

02:43 We now reduce the power of a and minus 1 and you increase the power of x over 1 factorial.

02:50 And then you can see that you were starting to form a pattern, let me just go on the next line.

02:55 Our next value will be n, n minus 1, a, n minus 2 to x to the power of 2 over 2 factorial and so on and you keep going. So the next term will be n, n minus 1, n minus 2, a to the n minus 3 and then x to the power of 3 over 2 factorial.

03:17 So in order to expand something like this we will have to use the binomial theorem on here.

03:23 Now we only really need to use the first two terms of the binomial theorem in this case.

03:28 And I'll explain why that is shortly when we actually get to the final result.

03:32 Okay, so now that we've derived the binomial theorem or we have it here on the side, we're now going to apply it to this expansion and we'll expand it binomially.

03:42 Let's have a look at how we do this, so we've got x plus delta x to the power of a half.

03:48 If you compare each term with the binomial theorem you can see that this is now a.

03:53 This is your x and this term here is your n.

03:56 So let's start to put each one of them in place of the binomial theorem here.

04:02 So if we start with the first term, we've got x to the power of a half and then we have delta x to the power of zero.

04:09 Our next term becomes a half 'cause we bring the power down, that's what n is.

04:14 We have x to the power of a half minus 1 and then we have delta x to the power of 1 all over 1 factorial.

04:24 Now there are more terms but we're going to ignore them for now and I'll explain why that is shortly. It makes life a lot easier to know or to do some forward thinking in this case because it'll save us a little bit of work.

04:37 If we tidy this up now, this term here, gives you x to the half because anything to the power of zero will always just be 1 so we end up with x to the half plus, here we have a half, we have x to the half minus 1 which gives me x to the minus a half and then this is multiply by delta x. One factorial is just 1, so we can just leave that as it is.

05:03 And remember we're not just letting go of the other terms, we're just ignoring them for now and you'll see why that is.

05:09 Right, so we have now expanded this term or this bracket.

05:15 So we can put it all back into differentiation from first principles such as come back to dy/dx. I'm still applying that the limit tends to zero but I'll leave that out in this section just to save some more space.

05:27 I have x to the half plus a half, x to the minus a half, delta x and then remember that we take this away from the original function which is there so I still have x to the half and then you divide it by delta x.

05:42 Right, good things start to happen, you'll see that x to the half positive and x to the half negative cancels out, leaving you with a half x to the minus a half, delta x divided by delta x. The delta x's also cancel out.

05:59 And we're now concluding that dy/dx as the limit of delta x tends to zero is a half, x to the minus a half. Now, this doesn't really apply in this case because we have no delta x terms left in this expansion.

06:16 However, if we had kept all of these terms here that I said we can just ignore.

06:21 They will all have a delta x term within them. So they'll have some form of delta x within them.

06:26 Which when it tends to zero, this entire thing will become zero and that is the reason why we ignore it and it just becomes easier because any terms with delta x or more than delta x squared will always have delta x in it, so that will become zero. In conclusion then, we are saying that root, square root of x differentiates the half x to the minus a half, if we just quickly try that on the side now with our faster method, so if we just try that out here, if I have y equals to square root of x which we know has the same power as x to the half, we can now say that dy/dx, remember what we said, bring the power to the front, so bring the power down as it is and then you decrease the power by 1, so you'll have a half minus 1 which gives you a half x to the minus a half.

07:16 Which you'll see is a lot simpler and a lot faster, so there is light at the end of the tunnel.

07:22 We don't always have to do things by first principles but it's a skill, it's an appreciation of where it comes from and you'll see that we'll use it in a lot of other functions which need to be, which can't be differentiated in this straightforward method.

The lecture Gradients and First Principles: Exercise 1 by Batool Akmal is from the course Calculus Methods: Gradients and First Principles.

### Included Quiz Questions

1. f'(x) = 5/[2√(x+3)]
2. f'(x) = 1/[2√(x+3)]
3. f'(x) = 5/√(x+3)
4. f'(x) = 10[√(x+3)]³/3
5. f'(x) = 5√(x+3) / 2

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