WEBVTT
00:00:00.000 --> 00:00:07.480
Determine dπ¦ by dπ₯, given that π¦ is equal to eight times the sin of four π₯ all to the power two π₯.
00:00:08.120 --> 00:00:12.760
Weβre asked to find the derivative of π¦ with respect to π₯.
00:00:13.760 --> 00:00:21.200
But if we look at our function π¦, we have an exponent two π₯, which is an expression in π₯.
00:00:22.240 --> 00:00:32.240
And this means that we canβt apply the usual rules of differentiation, the product, quotient, or chain rules, directly.
00:00:33.160 --> 00:00:36.600
What we can do, however, is use logarithmic differentiation.
00:00:37.600 --> 00:01:03.720
This is a four-step process for a function π¦ is π of π₯, where step one is to apply the natural logarithm to both sides, recalling that the natural logarithm is the log to the base π where π is Eulerβs number, which is approximately 2.71828 and so on.
00:01:04.480 --> 00:01:14.280
In our case, we have π¦ is equal to eight times the sin of four π₯ all to the power two π₯, and so we take the natural logarithm on both sides.
00:01:15.120 --> 00:01:19.720
At this point, we are to specify that π¦ has to be greater than zero.
00:01:20.640 --> 00:01:26.080
Thatβs because the log of zero is undefined and the log doesnβt exist for negative values.
00:01:26.760 --> 00:01:39.160
If we do want to include negative values, we must include also the absolute value signs around π¦ and π of π₯ and, in this case, specify that π¦ is nonzero.
00:01:39.840 --> 00:01:44.880
For our problem, however, weβll simply specify that π¦ is greater than zero.
00:01:45.800 --> 00:01:52.560
Now, our second step in logarithmic differentiation is where we start to see how taking logs can help us.
00:01:53.360 --> 00:01:57.160
We use the laws of logarithms to expand or simplify our right-hand side.
00:01:58.000 --> 00:02:04.920
And since in our argument for the logarithm on the right-hand side we have an exponent, weβre going to use the power rule for logarithms.
00:02:05.680 --> 00:02:13.520
And this says that log to the base π of π raised to the power π is π times log to the base π of π.
00:02:14.320 --> 00:02:19.560
That is, we bring the exponent π down in front of the log and multiply by it.
00:02:20.160 --> 00:02:28.800
In our case, this translates to the natural logarithm of π¦ is two π₯ times the natural logarithm of eight sin four π₯.
00:02:29.880 --> 00:02:34.800
And now we see that on our right-hand side, we have the product of two expressions in π₯.
00:02:35.640 --> 00:02:43.360
And this brings us to our third step in logarithmic differentiation, thatβs differentiate both sides with respect to π₯.
00:02:44.120 --> 00:02:50.840
And since we have a product on our right-hand side, we can use the product rule for differentiation.
00:02:51.720 --> 00:03:09.480
This says that if π’ and π£ are differentiable functions of π₯, d by dπ₯ of π’π£, the product, is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.
00:03:10.160 --> 00:03:25.880
So if we now let π’ equal two π₯ and π£ be the natural algorithm of eight times the sin of four π₯, then to use the product rule, we need to find dπ£ by dπ₯ and dπ’ by dπ₯.
00:03:26.480 --> 00:03:30.720
dπ’ by dπ₯ is two by sight.
00:03:31.720 --> 00:03:46.720
And to differentiate π£, we can use the result that d by dπ₯ of the natural logarithm of a differentiable function π of π₯ is one over π of π₯ times dπ by dπ₯.
00:03:47.560 --> 00:03:56.440
So dπ£ by dπ₯ is one over eight sin four π₯ times d by dπ₯ of eight sin four π₯.
00:03:57.440 --> 00:04:20.840
To differentiate eight sin four π₯, we use the fact that d by dπ₯ of sin π’, where π’ is a differentiable function of π₯, is dπ’ by dπ₯ times cos π’ so that dπ£ by dπ₯ is one over eight times the sin of four π₯ all times four times eight cos four π₯.
00:04:21.560 --> 00:04:35.320
We have a common factor of eight on the denominator and the numerator which cancel each other out so that dπ£ by dπ₯ is four cos four π₯ over sin four π₯.
00:04:35.960 --> 00:04:51.880
So we have π’ is two π₯ and dπ’ by dπ₯ is two, π£ is the natural algorithm of eight sin four π₯, and dπ£ by dπ₯ is four cos four π₯ over sin four π₯.
00:04:52.800 --> 00:05:03.880
And to simplify this even further since cos over sin is the cot, we can say dπ£ by dπ₯ is four times the cot of four π₯.
00:05:04.840 --> 00:05:25.040
So now applying the product rule for differentiation to our right-hand side, we have two π₯, which is π’, times four times the cot of four π₯, which is dπ£ by dπ₯, plus the natural logarithm of eight sin four π₯, which is π£, times two, which is dπ’ by dπ₯.
00:05:25.720 --> 00:05:31.120
And now making some room, we can rewrite this and rearrange it.
00:05:32.040 --> 00:05:43.920
We have d by dπ₯ of the natural logarithm of π¦ is equal to two times the natural logarithm of eight sin four π₯ plus eight π₯ times the cot of four π₯.
00:05:44.640 --> 00:05:51.400
Weβre not quite finished with our third step since we still need to differentiate the natural logarithm of π¦.
00:05:51.960 --> 00:06:05.760
And again, we can use our known result that d by dπ₯ of the natural logarithm of a function π of π₯ is one over π of π₯ times dπ by dπ₯.
00:06:06.360 --> 00:06:15.440
And since our π¦ is indeed a function of π₯, our left-hand side is simply one over π¦ dπ¦ by dπ₯.
00:06:16.280 --> 00:06:23.200
And this brings us to our final step in logarithmic differentiation which is to solve for dπ¦ by dπ₯.
00:06:24.000 --> 00:06:35.360
We can do this by multiplying both sides by π¦ so that on the left-hand side, the π¦βs cancel each other out, and weβre left with dπ¦ by dπ₯.
00:06:35.920 --> 00:06:47.040
On our right-hand side, we can reintroduce our function π¦ so that weβre multiplying by eight sin four π₯ to the power two π₯ and taking out a common factor of two.
00:06:47.760 --> 00:07:10.840
Using logarithmic differentiation to differentiate the function π¦ is eight sin four π₯ all to the power two π₯, we have dπ¦ by dπ₯ is two times eight sin four π₯ to the power two π₯ multiplied by the natural logarithm of eight sin four π₯ plus four π₯ times the cot of four π₯.