WEBVTT
00:00:00.560 --> 00:00:10.640
In this video, weβll learn how to use implicit differentiation to help us find the derivative of functions expressed implicitly as functions of π₯.
00:00:11.800 --> 00:00:23.560
The majority of differentiation problems you will have come across so far will have involved functions written explicitly as functions of π₯, such as π¦ equals three π₯ squared sin π₯.
00:00:24.160 --> 00:00:40.560
In this video, weβre going to explore how implicit differentiation, which is an extension of the chain rule, allows us to easily differentiate equations such as the Cartesian equation of a circle, π₯ squared plus π¦ squared equals one for example.
00:00:41.360 --> 00:00:46.960
Weβll also explore what this means for the second derivative and higher order derivatives.
00:00:47.920 --> 00:00:51.280
The chain rule allows us to differentiate composite functions.
00:00:51.880 --> 00:01:06.880
It says that for two differentiable functions, π and β, such that π is the composite function π of β of π₯, the derivative of π is the derivative of β of π₯ times the derivative of π evaluated at β of π₯.
00:01:07.840 --> 00:01:19.160
Itβs often more intuitive though to write this as dπ¦ by dπ₯ equals dπ’ by dπ₯ times dπ¦ by dπ’, where π¦ is a function of π’ and π’ is a function of π₯.
00:01:20.080 --> 00:01:25.720
Letβs demonstrate how this chain rule can help us to find the derivative of an implicit function.
00:01:26.560 --> 00:01:30.120
Consider the equation π₯ squared plus π¦ squared equals one.
00:01:30.800 --> 00:01:36.720
Using implicit differentiation, find an expression for dπ¦ by dπ₯ in terms of π₯ and π¦.
00:01:37.640 --> 00:01:39.640
And thereβs a second part to this question.
00:01:40.200 --> 00:01:52.920
For the semicircle where π¦ is greater than or equal to zero, express π¦ explicitly in terms of π₯, then differentiate this expression to get an expression for dπ¦ by dπ₯ in terms of π₯.
00:01:54.080 --> 00:01:55.280
Weβll begin with part one.
00:01:55.840 --> 00:02:01.120
To differentiate the function implicitly, weβre going to begin by differentiating both sides of the equation with respect to π₯.
00:02:02.120 --> 00:02:05.200
This is probably going to look a little strange, but bear with me.
00:02:06.120 --> 00:02:12.600
We say that the derivative of π₯ squared plus π¦ squared with respect to π₯ is equal to the derivative of one with respect to π₯.
00:02:13.240 --> 00:02:16.720
And we absolutely must do this to both sides of the equation.
00:02:17.360 --> 00:02:19.720
Next, we differentiate what we can.
00:02:20.240 --> 00:02:24.280
Itβs quite straightforward to differentiate one with respect to π₯.
00:02:24.920 --> 00:02:26.680
Itβs simply zero.
00:02:27.120 --> 00:02:30.000
But how do we differentiate π₯ squared plus π¦ squared?
00:02:30.960 --> 00:02:33.840
Well, the derivative of π₯ squared is two π₯.
00:02:34.520 --> 00:02:37.680
But the derivative of π¦ squared is a little more strange.
00:02:38.440 --> 00:02:40.480
We know that π¦ squared is a function of π¦.
00:02:41.120 --> 00:02:43.280
And, in turn, π¦ is a function of π₯.
00:02:43.920 --> 00:02:46.440
So here, we can use the chain rule.
00:02:47.400 --> 00:02:55.400
We say that the derivative of π¦ squared with respect to π₯ is equal to the derivative of π¦ squared with respect to π¦ times the derivative of π¦ with respect to π₯.
00:02:56.520 --> 00:03:00.440
Well, the derivative of π¦ squared with respect to π¦ is two π¦.
00:03:00.880 --> 00:03:04.600
And the derivative of π¦ with respect to π₯ is simply dπ¦ by dπ₯.
00:03:05.560 --> 00:03:11.520
So our equation now becomes two π₯ plus two π¦ dπ¦ by dπ₯ equals zero.
00:03:12.360 --> 00:03:14.880
Remember, we want an equation for the derivative.
00:03:15.440 --> 00:03:23.200
So weβre going to make dπ¦ by dπ₯ the subject by first subtracting two π₯ from both sides of our equation.
00:03:23.880 --> 00:03:28.000
That gives us two π¦ dπ¦ by dπ₯ is equal to negative two π₯.
00:03:28.880 --> 00:03:31.480
Next, weβre going to divide through by two π¦.
00:03:31.920 --> 00:03:36.960
And we see that dπ¦ by dπ₯ is equal to negative two π₯ divided by two π¦.
00:03:37.800 --> 00:03:40.000
Well, the twos cancel.
00:03:40.520 --> 00:03:47.160
And we found an expression for dπ¦ by dπ₯ in terms of π₯ and π¦; itβs negative π₯ over π¦.
00:03:48.440 --> 00:03:52.280
For part two of this question, weβre going to go back to our original equation.
00:03:52.720 --> 00:03:56.320
And weβre going to begin by making π¦ the subject of the equation.
00:03:56.960 --> 00:04:00.520
Thatβs expressing π¦ explicitly in terms of π₯.
00:04:01.360 --> 00:04:03.680
We can subtract π₯ squared from both sides.
00:04:04.120 --> 00:04:06.800
And then we take the square root of both sides of the equation.
00:04:07.520 --> 00:04:10.800
Now, usually, we take by the positive and negative square root.
00:04:11.360 --> 00:04:16.600
Here though, weβre told the semicircle is such that π¦ is greater than or equal to zero.
00:04:17.160 --> 00:04:20.240
So weβre just going to consider the positive square root.
00:04:20.840 --> 00:04:23.280
And we have an explicit function in π₯.
00:04:23.960 --> 00:04:33.240
Now, writing it as one minus π₯ squared to the power of a half, we can see that we can use the general power rule to differentiate this.
00:04:34.000 --> 00:04:46.480
This says that if π’ is some function in π₯, the derivative of π’ to the power of π with respect to π₯ is equal to π times π’ to the power of π minus one multiplied by the derivative of π’ with respect to π₯.
00:04:47.160 --> 00:04:50.320
And this is of course when π is a real number.
00:04:51.040 --> 00:04:59.200
So the derivative of one minus π₯ squared to the power of a half is a half times one minus π₯ squared to the power of negative half multiplied by dπ’ by dπ₯.
00:04:59.960 --> 00:05:03.400
But, actually, π’ is equal to one minus π₯ squared.
00:05:03.960 --> 00:05:09.400
So the derivative of π’ with respect to π₯ is just negative two π₯.
00:05:10.360 --> 00:05:13.120
Once again, we have some twos to cancel.
00:05:13.640 --> 00:05:21.560
And our expression for dπ¦ by dπ₯ in terms of π₯ is negative π₯ over the square root of one minus π₯ squared.
00:05:22.400 --> 00:05:33.680
Notice that since we said that π¦ was equal to the square root of one minus π₯ squared, we could have written this as dπ¦ by dπ₯ equals negative π₯ over π¦.
00:05:34.520 --> 00:05:38.280
And thatβs the same answer as we got in part one of this question.
00:05:38.760 --> 00:05:43.760
And, of course, we must remember that π¦ cannot be equal to zero here.
00:05:44.600 --> 00:05:47.200
Now, this example demonstrates some important points.
00:05:47.720 --> 00:05:56.400
Firstly, even though it was easy enough to express this relation as an explicit function, it was simpler to differentiate using implicit differentiation.
00:05:57.240 --> 00:06:03.840
And, in general, when weβre differentiating implicitly, we can use the following version of the chain rule.
00:06:04.720 --> 00:06:14.840
This says the derivative of a function in π¦ with respect to π₯ is equal to the derivative of that function in π¦ with respect to π¦ times dπ¦ by dπ₯.
00:06:15.720 --> 00:06:19.680
Itβs really useful to commit this version of the chain rule to memory.
00:06:20.640 --> 00:06:23.920
Letβs now apply it to a more complicated example.
00:06:25.160 --> 00:06:34.600
Find dπ¦ by dπ₯ by implicit differentiation if negative π of to the π¦ sin π₯ is equal to four π₯π¦ plus two π₯.
00:06:35.800 --> 00:06:41.440
To differentiate this function implicitly, we begin by differentiating both sides of the equation with respect to π₯.
00:06:42.240 --> 00:06:52.040
We begin by writing this as d by dπ₯ of negative π to the π¦ times sin π₯ equals d by dπ₯ of four π₯π¦ plus two π₯.
00:06:52.880 --> 00:06:55.080
Letβs differentiate each bit with respect to π₯.
00:06:55.680 --> 00:06:59.800
Well, the derivative of two π₯ is straightforward; itβs simply two.
00:07:00.560 --> 00:07:08.280
Weβre going to have to use the product rule combined with the chain rule to differentiate four π₯π¦ and negative π to the π¦ sin π₯.
00:07:09.200 --> 00:07:20.600
The special version of the chain rule that we need says the derivative of π of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π¦ times dπ¦ by dπ₯.
00:07:21.400 --> 00:07:30.040
And the product rule says that the derivative of π’ times π£ is equal to π’ times the derivative of π£ plus π£ times the derivative of π’.
00:07:30.920 --> 00:07:32.800
Weβll begin by differentiating four π₯π¦.
00:07:33.280 --> 00:07:36.680
Weβre going to let π’ be equal to four π₯ and π£ be equal to π¦.
00:07:37.440 --> 00:07:40.840
Then the derivative of π’ with respect to π₯ is simply four.
00:07:41.600 --> 00:07:52.080
The derivative of π£ with respect to π₯ is equal to the derivative of π¦ with respect to π¦ which is one times dπ¦ by dπ₯ which is simply dπ¦ by dπ₯.
00:07:52.920 --> 00:07:58.840
π’ times dπ£ by dπ₯ is then four π₯ multiplied by dπ¦ by dπ₯.
00:07:59.640 --> 00:08:03.680
And π£ multiplied by dπ’ by dπ₯ is equal to π¦ times four.
00:08:04.480 --> 00:08:12.640
And, therefore, the right-hand side of our equation is four π₯ dπ¦ by dπ₯ plus four π¦ plus two.
00:08:13.720 --> 00:08:17.920
Letβs now repeat this process for negative π to the power of π¦ times sin π₯.
00:08:18.680 --> 00:08:23.680
This time, letβs say π’ is equal to negative π to the power of π¦ and π£ is equal to sin π₯.
00:08:24.400 --> 00:08:26.400
The derivative of sin π₯ is cos π₯.
00:08:26.880 --> 00:08:37.000
And then the derivative of π’ with respect to π₯ is equal to the derivative of negative π to the π¦ with respect to π¦, thatβs negative π to the π¦, times dπ¦ by dπ₯.
00:08:37.800 --> 00:08:48.320
And then when we use the product rule, we see that dπ¦ by dπ₯ is equal to negative π to the π¦ cos π₯ minus π to the π¦ sin π₯ dπ¦ by dπ₯.
00:08:49.000 --> 00:08:52.040
So our equation is now as shown.
00:08:52.760 --> 00:08:56.400
Remember, weβre trying to find an equation for dπ¦ by dπ₯.
00:08:57.120 --> 00:09:02.800
So weβre going to rearrange and make dπ¦ by dπ₯ the subject.
00:09:03.280 --> 00:09:15.720
When we do, we see that negative π to the π¦ cos π₯ minus four π¦ minus two is equal to four π₯ plus π to the π¦ sin π₯ all multiplied by dπ¦ by dπ₯.
00:09:16.800 --> 00:09:20.560
Now, we can actually factor negative one on the left-hand side.
00:09:21.160 --> 00:09:24.960
And then, we divide through by four π₯ plus π to the π¦ sin π₯.
00:09:25.640 --> 00:09:35.720
And we see that dπ¦ by dπ₯ is equal to negative π to the π¦ cos π₯ plus four π¦ plus two over four π₯ plus π to the π¦ sin π₯.
00:09:36.360 --> 00:09:46.800
And, of course, this derivative stands when the denominator is not equal to zero, when four π₯ plus π to the π¦ sin π₯ is not equal to zero.
00:09:47.440 --> 00:09:54.560
Itβs fairly common to use implicit differentiation to find the equation of a tangent to a curve defined implicitly.
00:09:55.560 --> 00:10:02.040
In our next example, weβll look at how we can use implicit differentiation to address a problem of this type.
00:10:02.920 --> 00:10:11.120
The equation π¦ squared minus 24π₯ cubed plus 24π₯ equals zero describes a curve in the plane.
00:10:11.760 --> 00:10:17.040
1) Find the coordinates of two points on this curve, where π₯ equals negative one-half.
00:10:17.880 --> 00:10:25.120
2) Determine the equation of the tangent at the points where π₯ equals negative one-half and the π¦-coordinate is positive.
00:10:25.880 --> 00:10:32.200
3) Find the coordinates of another point, if it exists, at which the tangent meets the curve.
00:10:33.040 --> 00:10:42.120
For part one, to find the points where π₯ is equal to negative one-half, weβre going to substitute this value of π₯ into our equation and solve for π¦.
00:10:42.920 --> 00:10:50.080
Thatβs π¦ squared minus 24 times negative a half cubed plus 24 times negative a half.
00:10:50.720 --> 00:10:51.960
And thatβs equal to zero.
00:10:52.760 --> 00:11:01.200
When we evaluate this, we get π¦ squared plus three minus 12 equals zero or π¦ squared minus nine equals zero.
00:11:02.080 --> 00:11:05.800
Weβre going to solve this equation by adding nine to both sides.
00:11:06.960 --> 00:11:15.800
And our last step is to find the square root of both sides of this equation, remembering to take by the positive and negative square roots of nine.
00:11:16.600 --> 00:11:19.080
The square root of nine is three.
00:11:19.640 --> 00:11:25.320
So we see that π¦ is equal to three and negative three, when π₯ is equal to negative one-half.
00:11:26.320 --> 00:11:34.160
In coordinate form, thatβs negative a half, three and negative a half, negative three.
00:11:35.080 --> 00:11:36.440
We now consider part two.
00:11:37.280 --> 00:11:43.040
We need to find the equation of the tangent at a point where π₯ equals negative a half and the π¦-coordinate is positive.
00:11:43.880 --> 00:11:47.840
Thatβs the coordinate negative a half, three.
00:11:48.560 --> 00:11:52.320
Weβll first, however, need to find the gradient of the tangent to the curve.
00:11:52.920 --> 00:11:59.240
This will be the derivative of the equation for the curve evaluated at π₯ equals negative a half and π¦ equals three.
00:11:59.880 --> 00:12:02.000
So weβre going to differentiate our equation implicitly.
00:12:02.480 --> 00:12:12.320
Thatβs d by dπ₯ of π¦ squared minus 24π₯ cubed plus 24π₯ equals d by dπ₯ of zero.
00:12:12.960 --> 00:12:19.480
The derivative of π¦ squared is the derivative of π¦ squared with respect to π¦ times the derivative of π¦ with respect to π₯.
00:12:20.360 --> 00:12:23.040
Itβs two π¦ dπ¦ by dπ₯.
00:12:23.760 --> 00:12:29.720
The derivative of negative 24π₯ cubed is three times negative 24π₯ squared.
00:12:30.400 --> 00:12:32.920
Thatβs negative 72π₯ squared.
00:12:33.400 --> 00:12:36.800
The derivative of 24π₯ is 24.
00:12:37.360 --> 00:12:39.440
And the derivative of zero is zero.
00:12:40.240 --> 00:12:48.280
So we see that two π¦ dπ¦ by dπ₯ minus 72π₯ squared plus 24 is equal to zero.
00:12:49.280 --> 00:12:51.720
We need an equation for dπ¦ by dπ₯.
00:12:52.520 --> 00:12:57.320
So weβre going to rearrange to make dπ¦ by dπ₯ the subject.
00:12:57.920 --> 00:13:03.000
We add 72π₯ squared to both sides of this equation and subtract 24.
00:13:03.800 --> 00:13:06.480
Next, weβre going to divide through by two π¦.
00:13:07.240 --> 00:13:19.560
And we see that dπ¦ by dπ₯ is equal to 72π₯ squared minus 24 over two π¦, which simplifies to 36π₯ squared minus 12 over π¦.
00:13:20.400 --> 00:13:24.600
Remember, we want to find the gradient of the tangent of the curve at negative a half, three.
00:13:25.080 --> 00:13:31.040
So weβre going to substitute π₯ equals negative a half and π¦ equals three into the equation for the derivative.
00:13:31.640 --> 00:13:41.600
When we do, we see that the gradient of our tangent is 36 times negative a half squared minus 12 all over three, which is negative one.
00:13:42.480 --> 00:13:46.960
Finally, we substitute what we know about our tangent into the equation of a straight line.
00:13:47.720 --> 00:13:52.600
And we get π¦ minus three equals negative one times π₯ minus negative a half.
00:13:53.320 --> 00:14:01.520
Distributing the parentheses and simplifying then rearranging for π¦, we get π¦ equals five over two minus π₯.
00:14:02.440 --> 00:14:03.960
And now we consider part three.
00:14:04.600 --> 00:14:09.920
We need to find a point, if it exists, where this tangent meets the curve again.
00:14:10.400 --> 00:14:23.480
We therefore want to solve simultaneously the equations π¦ squared minus 24π₯ cubed plus 24π₯ equals zero and π¦ equals five over two minus π₯.
00:14:24.480 --> 00:14:31.480
We can achieve this by substituting π¦ equals five over two minus π₯ into the original equation for the curve.
00:14:32.320 --> 00:14:40.680
That gives us five over two minus π₯ all squared minus 24π₯ cubed plus 24π₯ equals zero.
00:14:41.400 --> 00:14:46.280
Expanding the parentheses and multiplying through by negative one, and we end up with the equation shown.
00:14:47.600 --> 00:14:51.120
Next, we could solve this using a scientific calculator.
00:14:51.800 --> 00:14:56.920
Alternatively, we know that π₯ equals negative a half is a root to this equation.
00:14:57.480 --> 00:15:00.560
And we can therefore factor out π₯ plus a half.
00:15:01.280 --> 00:15:06.240
And we can do this by using long division or matching coefficients.
00:15:06.960 --> 00:15:11.560
The first step towards equating coefficients would be to write the equation shown.
00:15:12.400 --> 00:15:26.600
And whilst itβs outside the scope of this video to spend a lot of time performing this, we should find that π is equal to 24, π is negative 13, and π is negative 25 over two.
00:15:27.640 --> 00:15:33.400
And you can pause the video here if you like and see if you can complete that step yourself.
00:15:34.400 --> 00:15:43.320
Our last step is to solve the quadratic equation 24π₯ squared minus 13π₯ minus 25 over two.
00:15:44.120 --> 00:15:49.640
And we could do that by using the quadratic formula or completing the square.
00:15:50.360 --> 00:15:53.200
When we solve this, we see that π₯ is equal to negative a half.
00:15:53.720 --> 00:15:55.240
So thatβs a repeated root.
00:15:55.960 --> 00:15:58.880
And π₯ is equal to 25 over 24.
00:15:59.840 --> 00:16:09.360
Weβre finding the coordinates, so we substitute π₯ equals 25 over 24 into the equation five over two minus π₯.
00:16:10.280 --> 00:16:14.800
And that gives us a π¦-value of 35 over 24.
00:16:15.120 --> 00:16:29.720
And so the tangent with equation π¦ equals five over two minus π₯ meets the curve at negative a half, three and twenty-five twenty-fourths, thirty-five twenty-fourths.
00:16:31.040 --> 00:16:36.800
In our very final example, weβll look at how implicit differentiation can help us to find higher order derivatives.
00:16:38.040 --> 00:16:46.840
Given that two sin π¦ minus five cos π₯ equals negative four, determine the second derivative of π¦ by implicit differentiation.
00:16:48.120 --> 00:16:55.360
To find to the second derivative of π¦, sometimes called π¦ double prime, weβre going to need to differentiate our function twice.
00:16:56.200 --> 00:17:00.440
Notice how the function itself is expressed implicitly as functions of π₯.
00:17:01.160 --> 00:17:05.760
So weβre going to use implicit differentiation and begin by finding the derivative of both sides.
00:17:06.800 --> 00:17:11.040
Now the derivative of negative four with respect to π₯ is fairly easy to work out.
00:17:11.640 --> 00:17:13.720
Itβs just zero.
00:17:14.520 --> 00:17:19.560
Similarly, we can find the derivative of negative five cos π₯ with respect to π₯.
00:17:20.360 --> 00:17:21.720
Thatβs five sin π₯.
00:17:22.480 --> 00:17:25.680
But what about the derivative of two sin π¦ with respect to π₯?
00:17:26.600 --> 00:17:30.520
Well, here we use a special version of the chain rule.
00:17:31.160 --> 00:17:40.480
This says that the derivative of a function in π¦ with respect to π₯ is equal to the derivative of that function with respect to π¦ times dπ¦ by dπ₯.
00:17:41.360 --> 00:17:45.640
Well, the derivative of two sin π¦ with respect to π¦ is two cos π¦.
00:17:46.560 --> 00:17:52.240
So our equation is two cos π¦ dπ¦ by dπ₯ plus five sin π₯.
00:17:52.960 --> 00:17:54.040
And thatβs equal to zero.
00:17:54.480 --> 00:18:01.960
We can find the first derivative then by subtracting five sin π₯ from both sides of the equation and dividing through by two cos π¦.
00:18:02.880 --> 00:18:05.760
Remember, though, we were looking to find the second derivative.
00:18:06.280 --> 00:18:13.560
So here weβre going to need to use the quotient rule to differentiate negative five sin π₯ over two cos π¦.
00:18:14.400 --> 00:18:22.400
According to our notation, we can let π’ be equal to negative five sin π₯ and π£ equal two cos π¦.
00:18:23.120 --> 00:18:26.360
dπ’ by dπ₯ is negative five cos π₯.
00:18:27.000 --> 00:18:37.840
Then since the derivative of π£ with respect to π¦ is negative two sin π¦, we can see that dπ£ by dπ₯ is negative two sin π¦ dπ¦ by dπ₯.
00:18:38.720 --> 00:18:41.800
We substitute each of these into the formula for the quotient rule.
00:18:42.200 --> 00:18:49.600
We then simplify and spot that we said that dπ¦ by dπ₯ is negative five sin π₯ over two cos π¦.
00:18:50.280 --> 00:18:55.280
So we can substitute this into the formula for the second derivative.
00:18:56.120 --> 00:19:01.040
To simplify this somewhat, we multiply both the numerator and denominator of our fraction by two cos π¦.
00:19:01.720 --> 00:19:04.840
And weβre going to look to simplify just a little more.
00:19:05.360 --> 00:19:18.640
Distributing this negative one, we can see that the second derivative of our function is 25 sin squared π₯ sin π¦ minus 10 cos π₯ cos squared π¦ all over cos cubed π¦.
00:19:19.520 --> 00:19:23.360
And this formula is valid as long as cos π¦ is not equal to zero.
00:19:24.600 --> 00:19:34.040
In this video, weβve seen that when given a function defined implicitly, we can use a special version of the chain rule to differentiate it.
00:19:34.640 --> 00:19:45.280
The derivative of a function in π¦ with respect to π₯ is equal to the derivative of that function π¦ with respect to π¦ times dπ¦ by dπ₯.
00:19:46.360 --> 00:19:57.440
We also saw that if it was possible to rewrite a relation as an explicit function, it was sometimes simpler to use implicit differentiation.
00:19:58.200 --> 00:20:03.920
We also saw that when we differentiate implicitly, weβll get an expression for dπ¦ by dπ₯ in terms of both π₯ and π¦.
00:20:04.640 --> 00:20:10.240
We also saw that we can find higher order derivatives by using implicit differentiation.
00:20:11.040 --> 00:20:18.520
And in these cases, weβll need to substitute expressions for lower order derivatives to help us simplify the expressions.