WEBVTT
00:00:00.430 --> 00:00:06.620
Find ππ¦ by ππ₯, given that π¦ equals six log base six of six π₯.
00:00:07.510 --> 00:00:12.020
We need to differentiate this expression involving the function that log base six.
00:00:13.140 --> 00:00:19.030
But the only logarithmic function that we know how to differentiate is the logarithm base π β the natural logarithm.
00:00:20.520 --> 00:00:24.360
π by ππ₯ of the natural logarithm of π₯ is one over π₯.
00:00:25.460 --> 00:00:29.400
Itβs not immediately obvious how to differentiate the logarithm base six.
00:00:29.650 --> 00:00:35.130
But if we can somehow rewrite the expression using the natural logarithm, then we know how to differentiate that.
00:00:36.010 --> 00:00:44.750
Luckily, we have the change of base formula: the logarithm base π of π₯ is the logarithm base π of π₯ divided by the logarithm base π of π.
00:00:45.770 --> 00:00:53.590
And choosing π to be π, we can write log base π of π₯ in terms of the logarithm base π β in other words the natural logarithm.
00:00:54.630 --> 00:01:00.110
Instead of writing log base π, we can use ln to remind us that this is the natural logarithm.
00:01:00.830 --> 00:01:08.650
And we can also express the rule using π instead of π₯, itβs the same rule here but all this save us some confusion when we come to substitute.
00:01:09.810 --> 00:01:12.460
We have a log base six of six π₯.
00:01:12.640 --> 00:01:19.850
So comparing this to log base π of π, we see that π is six and π is six π₯.
00:01:20.690 --> 00:01:28.270
And so log base six of six π₯ is the natural logarithm of six π₯ over the natural logarithm of six.
00:01:29.590 --> 00:01:40.000
Now, we can use the fact that the derivative of a number times a function is that number times the derivative of the function to pull out some constants from inside β our π by ππ₯.
00:01:40.720 --> 00:01:46.150
We then just have to find the derivative of the natural logarithm of six π₯ with respect to π₯.
00:01:46.360 --> 00:01:49.180
And this is a straightforward application of the chain rule.
00:01:50.260 --> 00:02:02.020
If we let π§ equals six π₯ and then apply the chain rule, weβll get this π by ππ§ of the natural logarithm of π§ times ππ§ by ππ₯, where π§ is six π₯ as mentioned before.
00:02:02.970 --> 00:02:07.290
You might like to pause the video and check that this is indeed what you get when you apply the chain rule.
00:02:08.430 --> 00:02:11.760
What is π by ππ§ of the natural logarithm of π§?
00:02:12.440 --> 00:02:22.280
The derivative with respect to π₯ of the natural logarithm of π₯ is one over π₯ and so the derivative with respect to π§ of the natural logarithm of π§ is one over π§.
00:02:23.530 --> 00:02:25.200
And π§ is six π₯.
00:02:25.200 --> 00:02:27.570
So we can write this as one over six π₯.
00:02:29.120 --> 00:02:31.020
How about ππ§ by ππ₯?
00:02:31.050 --> 00:02:33.120
Well, π§ is six π₯.
00:02:33.120 --> 00:02:36.180
So differentiating with respect to π₯, we get six.
00:02:36.780 --> 00:02:39.430
Now, we can simplify inside the parentheses.
00:02:40.030 --> 00:02:43.910
One over six π₯ times six is just one over π₯.
00:02:44.670 --> 00:02:50.510
We have shown that the derivative of the natural logarithm of six π₯ with respect to π₯ is one over π₯.
00:02:51.400 --> 00:02:57.170
And of course, this is also the derivative with respect to π₯ of the natural logarithm of just π₯.
00:02:58.140 --> 00:03:04.100
You might like to think about why the derivative of ln six π₯ is the same as the derivative of ln π₯.
00:03:04.900 --> 00:03:07.750
I promise you that this isnβt just an arithmetic error.
00:03:08.050 --> 00:03:08.970
This really is true.
00:03:10.410 --> 00:03:16.470
Anyway, back to the problem at hand, the only thing left to do is to write this productβs fractions as a single fraction.
00:03:17.280 --> 00:03:22.670
Doing so, we find that ππ¦ by ππ₯ is six over π₯ ln six.
00:03:23.590 --> 00:03:38.030
The thing to take away from this video is that if you want to differentiate some expression which involves the logarithm to some base, then first use a change of base formula to rewrite the expression in terms of the natural logarithm because we know how to differentiate that.