WEBVTT
00:00:00.440 --> 00:00:07.810
If π¦ equals eight cot π₯ plus five sec π₯, find ππ¦ by ππ₯ at π₯ equals π by six.
00:00:08.820 --> 00:00:11.390
Weβre looking for the derivative of ππ¦ by ππ₯.
00:00:11.980 --> 00:00:16.980
Well, actually weβre looking for the value of ππ¦ by ππ₯ at π₯ equals π by six.
00:00:17.710 --> 00:00:21.830
But to find this, we first need to find ππ¦ by ππ₯ in terms of π₯.
00:00:22.040 --> 00:00:23.600
So letβs differentiate.
00:00:24.120 --> 00:00:28.320
Weβre looking for the derivative of eight cot π₯ plus five sec π₯.
00:00:28.950 --> 00:00:47.650
And we use the fact that the derivative of a sum of functions is the sum of their derivatives and the fact that the derivative of a number times a function is that number times the derivative of the function to write this derivative ππ¦ by ππ₯ in terms of the derivatives of cot π₯ and sec π₯.
00:00:48.470 --> 00:00:55.140
The main part of this question is finding the derivative of these two trigonometric functions: cot and sec.
00:00:55.990 --> 00:01:02.840
There are six trigonometric functions that we tend to use: sin, cos, tan, csc, sec, and cot.
00:01:03.470 --> 00:01:10.400
And certainly we could memorize all six of their derivatives if we wanted to, and this isnβt a terrible idea.
00:01:10.520 --> 00:01:15.050
It would allow us to differentiate any trigonometric function very quickly.
00:01:16.050 --> 00:01:21.980
Certainly if we memorized the derivatives of sec and cot, then weβd pretty much be done with this question.
00:01:22.860 --> 00:01:30.380
But if you have forgotten the derivatives of sec and cot or if youβve never memorized them in the first place, all is not lost.
00:01:30.880 --> 00:01:43.700
The key thing to do is to remember the first two derivatives: the derivative of sin π₯ with respect to π₯ is cos π₯, and the derivative of cos π₯ with respect to π₯ is minus sin π₯.
00:01:44.310 --> 00:01:47.900
From these two derivatives, we can derive the other four.
00:01:48.820 --> 00:01:54.440
Letβs see how we can find the derivative of cot when we know the derivatives of sin and cos.
00:01:55.140 --> 00:01:59.260
Okay, so weβre looking for the derivative of cot π₯ with respect to π₯.
00:02:00.120 --> 00:02:02.020
Now what is cot π₯?
00:02:02.600 --> 00:02:09.810
Well, you might know that itβs one over tan π₯, and so weβre looking for the derivative of one over tan π₯ with respect to π₯.
00:02:10.560 --> 00:02:19.570
And as tan π₯ is sin π₯ over cos π₯, cot π₯ being the reciprocal of this must be cos π₯ over sin π₯.
00:02:20.610 --> 00:02:25.090
Now weβve succeeded in writing cot π₯ in terms of sin and cos.
00:02:25.720 --> 00:02:30.610
In fact, weβve shown that cot π₯ is the quotient of cos π₯ and sin π₯.
00:02:31.190 --> 00:02:34.220
And so, we can differentiate it using the quotient rule.
00:02:35.110 --> 00:02:43.070
So hereβs the quotient rule, and now we can apply it setting π of π₯ to be cos π₯ and π of π₯ to be sin π₯.
00:02:43.830 --> 00:02:49.660
Now we have the derivative weβre looking for written in terms of sin π₯ cos π₯ and their derivatives.
00:02:50.580 --> 00:02:55.320
And we know the derivatives of sin π₯ and cos π₯, so we can substitute them in.
00:02:56.280 --> 00:03:06.830
The derivative of cos π₯ is minus sin π₯ and so the first term in the numerator becomes sin π₯ times minus sin π₯ or minus sin π₯ squared.
00:03:07.570 --> 00:03:09.970
How about the second term in the numerator?
00:03:09.970 --> 00:03:14.690
Well, the derivative of sin π₯ with respect to π₯ is cos π₯.
00:03:15.320 --> 00:03:21.010
And so, the second term becomes cos π₯ times cos π₯ or cos π₯ squared.
00:03:21.810 --> 00:03:23.850
Thereβs not much we can do to the denominator.
00:03:23.850 --> 00:03:29.780
It stays as sin π₯ squared, which we can eventually write as sin squared π₯.
00:03:30.550 --> 00:03:38.800
But notice the numerator is minus sin squared π₯ minus cos squared π₯, which is minus sin squared π₯ plus cos squared π₯.
00:03:39.340 --> 00:03:48.890
And hopefully we recognize that we can apply the identity sin squared π₯ plus cos squared π₯ equals one, which means our numerator is negative one.
00:03:49.070 --> 00:03:55.220
And so the derivative of cot π₯ with respect to π₯ is negative one over sin squared π₯.
00:03:56.030 --> 00:04:02.650
And taking the minus sign outside the fraction, we get negative one over sin squared π₯.
00:04:03.510 --> 00:04:06.440
We can apply this result to the derivative we wanted to find.
00:04:07.240 --> 00:04:13.760
The first term of ππ¦ by ππ₯ is, therefore, eight times negative one over sin squared π₯.
00:04:14.450 --> 00:04:27.710
Before we move on, finding the derivative of sec π₯ with respect to π₯ and hence the second term of ππ¦ by ππ₯, letβs first recap how we found the derivative of cot π₯ with respect to π₯.
00:04:28.360 --> 00:04:32.210
We first wrote cot π₯ in terms of sin π₯ and cos π₯.
00:04:32.650 --> 00:04:41.730
And then seeing that we had a quotient, we applied the quotient rule using the derivatives of sin π₯ and cos π₯ with respect to π₯, which we memorized.
00:04:43.090 --> 00:04:53.740
And simplifying by applying the identity sin squared π₯ plus cos squared π₯ equals one gave us the simple expression negative one over sin squared π₯.
00:04:54.560 --> 00:04:59.850
So now letβs follow the same steps to find the derivative of sec π₯ with respect to π₯.
00:05:00.390 --> 00:05:03.700
We can write sec π₯ in terms of sin π₯ and cos π₯.
00:05:03.800 --> 00:05:05.470
In fact, we only need cos π₯.
00:05:05.850 --> 00:05:07.620
Sec π₯ is one over cos π₯.
00:05:08.400 --> 00:05:15.260
And we differentiate this by applying the quotient rule with π of π₯ equal to one and π of π₯ equal to cos π₯.
00:05:16.220 --> 00:05:22.750
The derivative of one with respect to π₯ is just zero, and so the first term in the numerator vanishes.
00:05:23.350 --> 00:05:30.960
And so weβre left with just the second term: minus one times the derivative of cos π₯ with respect to π₯.
00:05:31.390 --> 00:05:34.810
And we know what the derivative of cos π₯ is with respect to π₯.
00:05:34.810 --> 00:05:36.240
Itβs minus sin π₯.
00:05:36.650 --> 00:05:46.740
So combining this with the minus one, we get sin π₯ in the numerator and of course cos squared π₯ in the denominator.
00:05:47.870 --> 00:05:52.670
And now that we found the derivative of sec π₯ with respect to π₯, we can apply this to our problem.
00:05:53.640 --> 00:05:59.580
The second term of ππ¦ by ππ₯ becomes five sin π₯ over cos squared π₯.
00:06:00.640 --> 00:06:02.540
Okay, now we can clear away our working.
00:06:03.270 --> 00:06:06.240
Now weβve found ππ¦ by ππ₯ in terms of π₯.
00:06:06.850 --> 00:06:10.390
We now just need to evaluate this when π₯ is π by six.
00:06:10.730 --> 00:06:18.110
Now we replace π₯ by π by six and we get some expression involving sin π by six and cos π by six.
00:06:18.910 --> 00:06:25.670
And as π by six is a special angle, hopefully we know the values of sin π by six and cos π by six.
00:06:25.900 --> 00:06:30.470
Sin π by six is a half and cos π by six is root three over two.
00:06:31.280 --> 00:06:35.370
We can substitute these values and simplify.
00:06:35.370 --> 00:06:37.380
For example, a half squared is a fourth.
00:06:38.130 --> 00:06:41.310
And root three over two squared is three-fourths.
00:06:42.630 --> 00:06:49.110
Negative one over a fourth is negative four, and a half over three-fourths is two-thirds.
00:06:49.320 --> 00:06:53.350
And so we get eight times negative four plus five times two-thirds.
00:06:54.200 --> 00:07:02.530
Simplifying, we find that the value of ππ¦ by ππ₯ at π₯ equals π by six is negative 86 over three.
00:07:03.450 --> 00:07:06.930
This problem required us to find the derivatives of cot and sec.
00:07:08.070 --> 00:07:15.220
We found that we could work out what these derivatives were using the known derivatives of sin and cos and also the quotient rule.
00:07:16.030 --> 00:07:24.390
The derivative of cot π₯ with respect to π₯ is [minus] one over sin squared π₯, which can also be written as minus csc squared π₯.
00:07:25.040 --> 00:07:35.490
And the derivative of sec π₯ with respect to π₯ we found to be sin π₯ over cos squared π₯, which can also be written as sec π₯ times tan π₯.
00:07:36.300 --> 00:07:43.280
However, we left these derivatives in terms of sin and cos to make it easy to evaluate them at π₯ equals π by six.
00:07:44.050 --> 00:07:54.590
We donβt need to memorize the derivatives of all six trigonometric functions, although it might be a good idea to do so as long as we remember the derivatives of sine and cos.
00:07:55.340 --> 00:08:00.400
To be fair, even the derivatives of sin and cos could be worked out from other more basic facts.
00:08:01.640 --> 00:08:06.160
But itβs definitely worth memorizing the derivatives of sin and cos at the very least.