00:00Hi guys, okay, so the
00:02next proof prediction I'm gonna
00:03do is Differentiation, so these
00:06are actually my favorite ones
00:07probably because I like cactus,
00:10but I Think these these
00:14this makes a lot of
00:15sense what it does to
00:16me anyway So we're given
00:17this is actually a past
00:18paper question. We're given some
00:20function f of x And
00:22it says the nth derivative
00:23is f and a vector
00:25the nth derivative and it
00:27says show that proof
00:28with mathematical induction, it has
00:32to say prove by mathematical
00:33induction, but they almost always
00:35do say, prove by mathematical
00:36induction. So that the nth
00:38derivative is given by this.
00:40So that means if I
00:40was to differentiate this thing,
00:4410 times the derivative would
00:46be 2 to the power
00:46of 10 times x plus
00:4910 times 2 to the
00:51power of 9, all times
00:54e to the 2x.
00:56which is interesting in itself
01:00but we need to kind
01:01of prove that. So let's
01:02go through our routine. So
01:05our proposition, proposition is that
01:10the ant derivative, ant derivative
01:13is equal to 2 to
01:17the power of nx plus
01:19n times 2 to the
01:21power of n minus 1,
01:23all times e to the
01:24power of n
01:24part of 2x. Okay, so
01:26we're gonna, we're gonna show
01:27this step one, step one,
01:33prove true for n equals
01:40one. Okay, now to prove
01:44true for n equals one,
01:45we need to get, so
01:47f of x is equal
01:51to
01:52f of x is equal
01:55to x e to the
02:002x. Now to get f
02:01dash of x, to get
02:03f 1 are the first
02:05derivative of x. Well, here
02:08we just need to use
02:08a product rule. So f
02:09dash of x is equal
02:11to, I'm going to do
02:13this quickly guys, but don't
02:14you do this, do it
02:15to the side, but I'm
02:16going to do it in
02:16my head. U times dv
02:19dx.
02:20So u times dv dx,
02:23the derivative of this is
02:242 e to the 2x.
02:26So that's u times dv
02:27dx plus v times dv
02:29dx, which is e to
02:31the 2x. And this is
02:35equal to, um, e to
02:41the 2x times 2x plus
02:461. So this is the
02:47first derivative
02:48this is the first derivative.
02:53And then if we get,
02:56let me write here from
03:01proposition, from proposition f one
03:08of x. This is using
03:11their thing here. f one
03:14of x is the first
03:14derivative of x is equal
03:16to
03:162 to the power of
03:181 x plus 1 times
03:222 to the power of
03:241 minus 1 times, close
03:31this bracket, and I just
03:32want to bring this over,
03:34so I have space times
03:37e to the 2 x,
03:40which is 2 x plus
03:44One times two to the
03:45power of zero is one,
03:46one times one is one.
03:48This is all in the
03:49bracket. E to the two
03:50X and look this. When
03:54I sub in one there,
03:55I do get the first
03:57derivative. So I'm gonna say
03:58therefore, therefore true for n
04:03equals one. Step two. Step
04:07two. Assume, so again, this
04:10is always the same. Assume,
04:12true for n equals k.
04:19So the kth derivative if
04:21you like, the kth derivative
04:22of k of x is
04:25equal to 2 to the
04:27power of kx plus k
04:31times 2 to the power
04:32of k minus 1 times
04:35e to the power of
04:372x. So the kth derivative
04:39is this
04:40We've assumed that. Next step.
04:45Step three. Prove. Prove true
04:52for n equals k plus
04:56one. Right. This again, I
04:59want to say what are
05:00we trying to show? So
05:01trying. I don't get right
05:04in trying. I'm trying to.
05:08prove that f of k
05:16plus 1 is the k
05:17plus 1 derivative. I can
05:19say that is equal to.
05:21So this time I'm going
05:22to put k plus 1
05:24in here instead of k.
05:25So it's 2 to the
05:26power of k plus 1,
05:282 to the power of
05:29k plus 1 x plus
05:31k plus 1 times
05:362 to the power of
05:39k plus 1 minus 1,
05:41which is 2 to the
05:41power of k. Close bracket,
05:46e to the 2x. So
05:48that's what I'm trying to
05:49show. So for this step,
05:56remember, for step three, you
05:58always use step two. You
06:00use the step that you
06:03have your assumption step.
06:04So you assume this. So
06:07f, let's write this out
06:09again. The kth derivative of
06:12x is equal to 2kx
06:16plus k2k minus 1 e
06:20to the 2x. So this
06:24is the kth derivative. So
06:26how do I get, if
06:27this is the kth derivative,
06:28how do I get the
06:29k plus 1th derivative? So
06:31we're assuming this is true,
06:32this is the
06:33So to get the cables
06:35one derivative, the next derivative,
06:37I just need to differentiate
06:39this. So if I differentiate
06:42this, I should get that.
06:45And then I've, then I
06:46have proven it. So this
06:49I got from sub -in
06:50-k plus one into my
06:52proposition. I sub -in -in
06:54here instead of n. And
06:57this one, I'm going to
06:58get what I have got
06:59from, what this I've
07:01got from the assumption, but
07:03when I differentiate this, I
07:04should get this. And if
07:06I get it, I've proven
07:07it. Okay, so let's do
07:10it. I'm going to say
07:12the K plus one derivative
07:19is equal to U times
07:24dv, dx, this one I
07:25actually am going to write
07:26it out. I'm going to
07:27write it out properly over
07:29using
07:29the product row. So u
07:31is equal to 2 to
07:34the k x plus k
07:37times 2 to the k
07:39minus 1. And v is
07:46equal to e to the
07:472x. d u dx is
07:50equal to 2 to the
07:52k because this is just
07:53a constant, which is k.
07:55And d v dx
07:57is equal to 2e to
07:59the 2x chain rule. So
08:02the k plus 1 derivative
08:03is u times u times
08:07u times u times dv
08:14dx, I think you've actually
08:15let's bring this up, let's
08:20bring this up here and
08:23this down.
08:25Here. So it's u times
08:29dv dx to eat the
08:322x plus v times v
08:36times d u dx to
08:40the k. Now, at this
08:42point I look at this,
08:44right? One more, actually, what
08:45am I trying to show?
08:47So there's an e to
08:48the 2x outside. So yeah,
08:50I'm going to have that.
08:51Let's bring that outside. And
08:53then
08:53Right, look, I think maybe
08:55we can see something, right?
08:58So I'll take out my
08:59e to the 2x, I'll
09:00eat the 2x and inside
09:04this square bracket I will
09:07put, there's a 2 here,
09:10so it's 2 times 2
09:14times 2 to the kx
09:18plus k times 2 to
09:21the k
09:21plus 2 to the k.
09:28So it's not clear I've
09:28taken out this two, I've
09:30taken it out here. This
09:31is just this and this
09:33two k is here and
09:34the e to the 2x
09:34is out there, right? So
09:35maybe, just make sure you
09:39know how I've got from
09:39there to there. Okay here
09:41now I have my e
09:41to the 2x outside. Let's
09:44multiply this out. So I've
09:45two times 2 times 2
09:49to the k
09:49x, right? Two times two
09:53to the k, this is
09:54two to the one if
09:55you like. Two to the
09:57one times two to the
09:58k is actually two to
10:00the k plus one. x.
10:04And even better, even better.
10:11Two times two to the
10:13k minus one is just
10:14going to be two to
10:15the k because one
10:17one plus k minus one
10:19is just k. So this
10:21becomes k times 2k times
10:262 to the k. And
10:30then I have this guy
10:32here, which is plus 2
10:35to the k. Okay, that's
10:39not of k's. Nearly there,
10:43I see this is working
10:44out now. Look, I have
10:45my e to the 2x,
10:48I have my 2 to
10:51the k plus 1x. So
10:54here's my e to the
10:552x. Here's my 2k plus
10:581x. And then here, what
11:01factors is, I can take
11:03out the 2 to the
11:04k, and I'm left with
11:06k plus 1 times 2
11:11to the k, close bracket.
11:13This is what I was
11:15trying to prove. And this
11:17is why, again, why I
11:20like to write this, I
11:21like to write this step
11:23here, this step here, because
11:25when you have that, you
11:27know what you're working towards,
11:28you're working towards that, you're
11:29trying to turn this into
11:30this. If you don't write
11:32that down, I find it
11:33far more difficult. Okay, so
11:36done. That's it. But am
11:39I done? No, I have
11:41to write my step.
11:41step four, you have to
11:44write step four. Now, guess
11:45what I'm gonna do, guys?
11:47Obviously, you don't have the
11:47luxury of doing this, but
11:49I'm gonna go to my
11:51previous example. I'm going to
11:55copy it. I'm gonna paste
11:58it. So, I'm doing this
12:00for two reasons. One, I
12:03don't have to write it
12:04out again. That's quite quick.
12:05But two, to show you
12:07that it's literally the exact
12:09same thing.
12:09I don't need to change
12:14it for the different types
12:21of proof -brand -duction. It's
12:24literally the exact same thing.
12:25The exact same statement. True
12:26-friend equals 1. If true
12:28-friend equals k, then true
12:29-friend equals k equals 1.
12:31Therefore, by the principle, my
12:32data -modern -conduction, true for
12:33all n is an element
12:34of the positive integers.
12:37That's differentiation. I need to
12:40do divisibility. And actually there
12:42was one other one I
12:43think that I didn't mention
12:44that I'm going to do,
12:45which is inequality. So see
12:47you in those next two
12:48lessons.