WEBVTT
00:00:00.560 --> 00:00:08.240
Given that four π₯ squared minus five π₯ plus eight π¦ squared minus three π¦ equals one, find ππ¦ by ππ₯.
00:00:09.420 --> 00:00:15.610
We could try to rearrange this equation so that we had π¦ in terms of π₯ and then differentiate in the normal way.
00:00:16.440 --> 00:00:23.330
This is tricky because not only do we have this minus three π¦ term, but also this eight π¦ squared.
00:00:23.330 --> 00:00:25.140
And so we have a quadratic in π¦.
00:00:26.310 --> 00:00:31.460
Itβs possible to solve this quadratic and find π¦ in terms of π₯ and then differentiate.
00:00:31.730 --> 00:00:32.980
But there is a better way.
00:00:33.900 --> 00:00:40.730
As this equation is a relation between the variables π₯ and π¦, we can differentiate both sides with respect to π₯.
00:00:42.210 --> 00:00:53.890
We can differentiate the left-hand side term by term, using the fact that the derivative with respect to π₯ of π times π₯ to the π is π times π times π₯ to the π minus one.
00:00:54.240 --> 00:00:58.210
We find that our first derivative is eight π₯.
00:01:00.290 --> 00:01:03.510
Similarly, π by ππ₯ of five π₯ is five.
00:01:05.300 --> 00:01:07.940
This term is slightly more difficult to differentiate.
00:01:08.150 --> 00:01:10.820
How do you find π by ππ₯ of a function of π¦?
00:01:11.790 --> 00:01:18.540
Here, we can use the chain rule: ππ by ππ₯ is ππ by ππ¦ times ππ¦ by ππ₯.
00:01:19.500 --> 00:01:32.050
And so just using slightly different notation here, we see that to differentiate a function of π¦ with respect to π₯, we differentiate that function of π¦ with respect to π¦ and then we just multiply by ππ¦ by ππ₯.
00:01:32.900 --> 00:01:35.930
So what is π by ππ₯ of eight π¦ squared?
00:01:36.200 --> 00:01:44.300
We differentiate eight π¦ squared with respect to π¦ and get 16π¦ and then you multiply by ππ¦ by ππ₯.
00:01:45.180 --> 00:01:54.250
Similarly, for π by ππ₯ of three π¦, you differentiate three π¦ with respect to π¦ to get three and then multiply by ππ¦ by ππ₯.
00:01:55.750 --> 00:01:57.750
The right-hand side is more straightforward.
00:01:57.750 --> 00:02:01.510
So the derivative of one with respect to π₯ is just zero.
00:02:02.600 --> 00:02:17.180
We now have an equation in terms of π₯, π¦, and ππ¦ by ππ₯ and we can rearrange this equation to make ππ¦ by ππ₯ the subject β that is we can write ππ¦ by ππ₯ in terms of π₯ and π¦.
00:02:18.210 --> 00:02:23.700
The first thing to do here is to factor out ππ¦ by ππ₯ in the two terms where it appears.
00:02:24.740 --> 00:02:32.260
Then, we can subtract eight π₯ minus five from both sides and then we just have something times ππ¦ by ππ₯ on the left.
00:02:33.260 --> 00:02:42.450
And dividing through by this something, we get that ππ¦ by ππ₯ is negative eight π₯ plus five over 16π¦ minus three.
00:02:43.400 --> 00:02:44.990
This is our final answer.
00:02:44.990 --> 00:02:47.920
Note that it is written in terms of both π₯ and π¦.
00:02:49.150 --> 00:02:55.300
You might feel slightly uneasy about this if youβre accustomed to writing ππ¦ by ππ₯ in terms of π₯ alone.
00:02:56.250 --> 00:03:01.860
But to write ππ¦ by ππ₯ in terms of π₯ alone, weβd have to write π¦ in terms of π₯.
00:03:02.040 --> 00:03:05.780
And as we discussed at the start of the video, thatβs a bit tricky.
00:03:06.980 --> 00:03:11.910
And not only that, weβd probably find that the expression for π¦ in terms of π₯ is quite ugly.
00:03:12.360 --> 00:03:16.050
And so we get quite an ugly expression for ππ¦ by ππ₯.
00:03:16.910 --> 00:03:29.060
So for this question and for questions involving implicit differentiation more generally β that is when you differentiate both sides with respect to a variable β itβs best to leave the answer in terms of both π₯ and π¦.