WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com and welcome back to AP Calculus.
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Today, we are going to talk about this technique called implicit differentiation.
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Let us jump right on in.
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Suppose, we are given the following function.
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Suppose, we are given x³ + y = 7x, what is dy dx?
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What is dy dx or y‘, whichever symbol you want to use.
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For this one, we can solve explicitly for y and just differentiate with we have been doing all this time.
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By explicitly, I mean just y on one side of the equal sign, and then, everything else in x on the other side of the equal sign.
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Here we can solve explicitly for y, then differentiate as before.
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Basically, I just move everything over and I end up with y is equal to 7x -3x³.
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And then, y’ is 7 -, it is just x³ not 3x³, my apologies.
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This just becomes -3x².
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This one is easy to handle.
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In other words, your function is always going to be given to you in terms of y = this.
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Sometimes, you have to put it in that form, before you actually take the derivative.
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What about this, what about x³ + y⁴ = -9xy.
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How do we handle something like this?
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Here we cannot solve explicitly for y.
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We cannot rearrange this equation or manipulate it mathematically such that y = something.
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Even if we could, the expression might be so complicated.
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By taking the derivative of it is going to be intractable.
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It is just not something you want to do.
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Fortunately, there is a way around this.
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This equation, it still defines, if a relation between x and y.
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It is saying that if I x³, I had the y⁴, it is equal to -9xy.
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There is a relation here between x and y, but the relation is implicit.
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Explicit means you have one variable on one side of the equality sign.
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And then, the function whatever it is on the other side, only that variable.
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Y is a function of x, y = something in x.
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Here it is implicit, it is implied in this.
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The idea is that theoretically, you might be able to but how do we find the derivative of it, if we do not have y = something.
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There is a way of doing so and that is called implicit differentiation, when you write all of these out.
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Here, we cannot or do not want to express y explicitly, in terms of x.
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But, the equation, it still defines a relation between x and y, an implicit relation.
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In other words, the x and y are sort of mixed up in the equation, that is an implicit relation.
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To find dy dx or y’ with respect to x, we use implicit differentiation.
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Very important.
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Here is how you do it.
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Let me go to blue here.
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Treat y as a function of x.
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When you differentiate y, when you take the derivative of y, use the chain rule.
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It does not really mean much as written.
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Let us see what actually happens.
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Use the chain rule then isolate the symbol dy dx.
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Let us see what this means.
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Let us start with letters that we had.
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We had x³ + y⁴ = -9xy.
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We are going to implicitly differentiate this entire thing, left side and right side.
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We will start with differentiate everything.
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You are differentiating with respect to x because that is what we are looking for.
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We are looking for dy dx.
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The variable that we are differentiating with respect to is x.
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It is going to look like this.
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We are going to take the ddx of x³ + the ddx of y⁴ = the ddx of -9xy.
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That is all we are saying. We are saying, here we have this equation.
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What we do to the left side, we do to the right side.
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If we differentiate the left side, we differentiate the right side.
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It retains the equality.
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That is all we are doing.
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You differentiate everything.
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The derivative with respect to x³, that is going to be 3x².
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The derivative with respect to x of y⁴, we treat y as a function of x.
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Chain rules says, it becomes 4y³ dy dx.
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That is what the chain rule is.
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It says this is a global function, there are two things going on.
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There is the y⁴, there is the power function, and then there is y itself which is a function of x.
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We are presuming that it is a function of x.
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It is implicit that it is a function of x.
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Therefore, we have to write that dy dx, the derivative of -9xy.
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This -9 is a constant, xy, we use the product rule.
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It is going to be -9 × this × the derivative of that.
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It is going to be x × dy dx + y × the derivative of that.
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Y × 1, the derivative of x with respect to x is 1.
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We have 3x² + 4y³ dy dx =, I’m going to distribute this, the 9 over this.
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- 9x dy dx - 9y.
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I’m going to put all the terms involving dy dx together on one side.
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It is going to be 4y³ dy dx.
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I’m going to bring this one over, + 9x dy dx.
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Over on the other side, I’m going to put everything else.
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I’m going to move this over.
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It is going to be -3x² – 9y.
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I have 4y³ dy dx + 9x × dy dx = this, factor out the dy dx.
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I get dy dx × 4y³ + 9x = - 3x² -9y, now divide by 4y³ + 9x.
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In other words, isolate the dy dx.
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Dy dx = -3x² -9y/ 4y³ + 9x.
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I have my derivative.
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The only issue with this derivative is not really an issue.
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It is just something that you have not seen, is that these derivative actually involves both variables x and y.
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When a function is given explicitly like y = sin (x), the derivative is only going to involve the variable x.
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Implicit derivatives, implicit differentiation expresses the derivative in terms of both variables x and y.
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That is not a problem because in general, we are going to know what x and y are.
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In the sense that we are going to pick some point, 5 3, -6 9, to put them in to find the derivative.
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Again, the derivative is two things, it is a slope and it is a rate of change.
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That is all it is.
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Having it expressed with a variable itself, the y itself does not really matter.
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For any value of x and y, this gives the slope at that point.
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That is it, no big deal.
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We are just adding that y in there.
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Notice, dy dx is expressed in terms of both x and y.
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It is generally going to be true.
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Again, this is not a problem.
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We want an expression for dy dx for y’.
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This implicit differentiation gives us that expression.
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Nice and straightforward.
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The best thing to do is just do examples.
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Once you see the examples, get used to this idea of differentiating y, treating it as of function of x using chain rule on it.
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Once you see it a couple of times, it will make sense what is going on.
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Let us jump right on in.
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We have our first example, xy + 4x - 3x² = 9.
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Here we want you to find dy dx by both implicit differentiation and solving explicitly for y.
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Here we want you to do it both ways, to see if we actually get the same answer.
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Let us see what we have.
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Let us do it explicitly first.
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This is going to be explicit.
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We have our function here.
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Explicit means we want y = some function of x.
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Fortunately, this one, we can do.
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I have the function of xy + 4x - 3x² = 9.
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That gives me, xy = 9 - 4x + 3x², y = 9 - 4x + 3x²/x.
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Now we want the derivative, so y’.
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I’m just going to go ahead and use quotient rule here.
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This × the derivative of that - that × the derivative of this/ the square.
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We are going to have x × the derivative of this which is -4 + 6x - that × the derivative of this, -9 - 4x + 3x² × 1/ x².
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Let me see, this gives me -4x.
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I’m going to distribute the x, + 6x².
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This is going to be -9, this is going to be +4x, this is going to be -3x²/ x².
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-4x and +4x go away.
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6x² and -3x², that is going to be 3x² - 9/ x².
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My explicit, in this particular case, was able to separate the x and y, separate the variables.
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I did my derivative the normal way.
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Let us do it implicitly and see if we get the same answer.
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Let us see, where are we?
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Did I skip anything?
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No, I did not skip anything.
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Let me go ahead and do it implicitly.
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3x² – 9/ x², let me just write that.
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3x² – 9/ x², this was the derivative that we got.
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Now let us do it implicitly.
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I have xy + 4x – 3x² = 9, differentiate everything across the board.
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This is differentiating everything with respect to x because we are looking for dy dx.
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This is product rule, it is going to be this × the derivative of that.
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I’m going to get x × the derivative of y xy‘ + y × the derivative of this + y × 1 + the derivative of 4x is 4.
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The derivative of this is 6x and the derivative of 9 is 0.
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I just differentiated right across the board, left side and right side.
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I'm going to write this as xy’ =, I’m going to move everything over.
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It is going to be 6x - y – 4.
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I get y‘ is equal to 6x - y - 4/ x.
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I found my y‘, my dy dx is 6x - y - 4/ x, implicit.
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It involves both x and y.
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Wait a minute, we are supposed to get the same thing.
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When we did it explicitly, we ended up with 3x² - 9/ x², why are not these the same?
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They are the same.
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Let us see what we have got here.
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We said, remember when we solved explicitly for this function.
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We got y is equal to 9 - 4x + 3x²/ x.
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We solved it explicitly.
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Let us put this expression for y into here, to see what happens.
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Y‘ is equal to 6x - 9 - 4x + 3x²/ x - 4/ x.
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Let us simplify this out.
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I’m going to get a common denominator here.
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This is going to be 6x².
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I’m going to distribute -9, that is a -9 + 4x - 3x².
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The common denominator here is -4x.
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All of this is over the common denominator x and all of this is /x.
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4x - 4x, 6x² - 3x² = 3x² – 9/ x².
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They are the same, the only difference is the implicit differentiation ended up expressing the derivative dy dx, in terms of both x and y.
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In this particular case, because we had an explicit form, we ended up checking it by putting it in here
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and realizing that we actually did get that.
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Again, you are not always going to be able to do that.
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In this particular case, you were able to.
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But most implicitly defined relations, you are not going to be able to solve explicitly for one of the variables.
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You have to leave it in terms of x and y.
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This is a perfectly good derivative.
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Example 2, x³ + x² y + 7y² = 14, find dy dx.
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The biggest problem with calculus is the algebra.
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Go slow, stay calm, cool and collected, and hopefully everything will work out right.
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I say this and yet I, myself make mistakes all the time.
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Hopefully we will keep our fingers crossed and we would not make any algebra mistakes.
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Let us see how we deal with this one.
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This is our function and we want to find dy dx.
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I’m going to work with y’.
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I do not want to write dy dx over and over again.
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The derivative of this is 3x², x² y, this is product rule.
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It is going to be + this × the derivative of that.
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It is going to be x² × the derivative of y which is just y' + that × the derivative of this.
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+ y × the derivative of x² is 2x.
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The derivative of 7y², it is 14y y’.
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Or we do the chain rule, we take care of the power and then we take dy dx.
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We write the y'.
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The derivative of 14 is 0.
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We are going to go ahead and put things that involve the y or the dy dx together.
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We have x² y' + 14y y1' =, I’m going to move this and this over to the other side.
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-3x² -2xy, factor out the y'.
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Y’ × x² + 14y = -3x² - 2xy, and then divide.
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I get y’ is equal to -3x² - 2xy all divided by x² + 14y.
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That is it, that is my derivative, that is dy dx, that is y’.
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Let us move on to next example.
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Here we have x³ y² + y³ x² = 4x, find dy dx.
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Here same thing, this is going to be product rule.
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I have this × the derivative.
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I got x³ × the derivative of y² which is 2y y' + y² × the derivative of x³
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which was 3x² + y³ × the derivative of x² which is 2x + x² × the derivative of y³ which is 3y² dy dx y'.
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The derivative of 4x is 4, put my prime terms together.
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I’m going to write this as 2x³ y y' + 3x² y² y’ = 4 - 3x² y².
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I move this over to that side.
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I’m going to move this over to that side, - 2xy³, factor out the y'.
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Y' × 2x³ y + 3x² y² = 4 - 3x² y² - 2xy³, and then divide.
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Y’ is equal to this whole thing.
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-3x² y² - 2xy³ divided by 2x³ y + 3x² y².
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It is exhausting, is not it?
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It really is, calculus is very exhausting.
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Y e ⁺x² + 14 = √2 + x² y.
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Let us dive right on in.
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Let me go to blue here.
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This is product rule, this × the derivative of that + that × the derivative of this.
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We have y × the derivative of this which is e ⁺x² × 2x + e ⁺x² × the derivative of y which is just y'.
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Just writing y’, dy dx is the derivative, that is what I'm looking for.
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The derivative of 14 is 0, the derivative of this, this is we know is equal to 2 + x² y ^ ½.
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The derivative of that is going to be ½ × 2 + x² y⁻¹/2 × the derivative of what is inside which is 0 + this × the derivative of that.
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X² y’ + this × the derivative of that, x² y’ + y × 2x.
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Putting all this together, we are going to end up with 2xy e ⁺x² + e ⁺x ² y' =,
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I’m going to distribute this thing over that and that.
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This first one is just 0 = ½ x² y' × 2 + x² y⁻¹/2 + this × the other term.
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½ × 2xy × 2 + x² y⁻¹/2.
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Let me see, where am I?
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Now I’m going to go ahead and collect.
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Let me go ahead and do red.
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This is something that involves a y' term.
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This term is what involves a y’ term, I’m going to bring them together.
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I’m going to get e ⁺x² y’ - ½ x² y' × 2 + x² y⁻¹/2 =, I’m going to put everything else on the other side.
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I leave this over here, the 2 cancel.
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I'm left with xy × 2 + x² y⁻¹/2.
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I bring that over to that side, -2xy e ⁺x².
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Factor out the y', I get y' × e ⁺x² - ½ x² × 2 + x² y ^-½ = xy × 2 + x² y ^ -½ -2xy e ⁺x².
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We finally have y’ = xy × 2 + x² y⁻¹/2 - 2xy e ⁺x² all divided by e ⁺x² – ½ 2 + x² y – ½.
00:28:45.600 --> 00:28:48.800
There you go, that is our final answer.
00:28:48.800 --> 00:28:58.100
Very complicated looking, but again, once you have x and y, you just plug them in and you solve it.
00:28:58.100 --> 00:28:59.800
Let us see what we have got.
00:28:59.800 --> 00:29:03.700
That was example 4.
00:29:03.700 --> 00:29:07.700
Example 5, 6 sin x cos y = 1.
00:29:07.700 --> 00:29:16.600
Let me go back to blue here.
00:29:16.600 --> 00:29:21.100
Let me do it down here.
00:29:21.100 --> 00:29:24.800
6 ×, this is product rule, sin x × cos y.
00:29:24.800 --> 00:29:31.600
It is going to be this × the derivative of that + that × the derivative of this.
00:29:31.600 --> 00:29:45.900
Sin x × the derivative of cos y which is –sin y × y' + cos y × the derivative of sin x which is cos x.
00:29:45.900 --> 00:29:51.400
The derivative of 1 = 0.
00:29:51.400 --> 00:30:11.900
We end up with, 6 goes away, we end up with - sin x sin y × y' + cos x cos y = 0.
00:30:11.900 --> 00:30:31.700
We have -sin x sin y y' = -cos x cos y.
00:30:31.700 --> 00:30:45.200
Therefore, y' is equal to cos x cos y, I will leave the minus sign, that is fine, divided by -sin x.
00:30:45.200 --> 00:30:48.400
Sin y, I can go ahead and cancel.
00:30:48.400 --> 00:30:51.200
I can just leave it as cos x cos y/ sin x sin y.
00:30:51.200 --> 00:30:52.300
The negatives cancel.
00:30:52.300 --> 00:31:01.800
Or I can write it as cot x cot y, either one of these is absolutely fine.
00:31:01.800 --> 00:31:05.700
Hope that makes sense.
00:31:05.700 --> 00:31:17.300
Example 6, same thing, trigonometric, just more complicated.
00:31:17.300 --> 00:31:24.400
Product rule, it is going to be a long differential.
00:31:24.400 --> 00:31:26.400
This × the derivative of that.
00:31:26.400 --> 00:31:33.100
X² × 2 × cos(y) y'.
00:31:33.100 --> 00:31:45.800
The derivative of cos² is 2 cos y × the derivative of what y is, y’.
00:31:45.800 --> 00:31:50.800
Wait, hold on a second, let me do this right.
00:31:50.800 --> 00:31:53.600
This × the derivative of this.
00:31:53.600 --> 00:32:06.000
X² × the derivative of cos² y, that is going to be 2 cos y × the derivative of the cos y
00:32:06.000 --> 00:32:11.100
which is × -sin y × the derivative of y which is y’.
00:32:11.100 --> 00:32:23.700
There we go, that is better. This × the derivative of that + cos² y × the derivative of x² which is 2x.
00:32:23.700 --> 00:32:37.900
Now, this one, + y × the derivative of sin x which is cos x + sin x × the derivative of y which is y’.
00:32:37.900 --> 00:33:16.800
All of this is equal to 2 × this × the derivative of that which is sin x × -sin y y' + cos y × cos x.
00:33:16.800 --> 00:33:21.800
I do not really need to simplify it, I can just go ahead and jump right on into what it is.
00:33:21.800 --> 00:33:30.200
That is fine, I will just go ahead and rewrite it.
00:33:30.200 --> 00:33:38.100
Let me do it in red, I’m making this a little bit simpler.
00:33:38.100 --> 00:33:50.400
-2x² cos y sin y y’
00:33:50.400 --> 00:34:16.000
+ 2x cos² y + y cos(x) + sin x y’.
00:34:16.000 --> 00:34:27.800
Let me make sure I have everything right here.
00:34:27.800 --> 00:34:46.200
This × the derivative of that, cos x.
00:34:46.200 --> 00:34:52.500
Wait a minute, this is 2 sin x.
00:34:52.500 --> 00:34:56.200
Okay, everything looks good.
00:34:56.200 --> 00:35:18.100
+ sin xy, good, all of that is equal to -2 sin x sin y y’ + cos y cos x.
00:35:18.100 --> 00:35:23.600
I’m going to bring everything, I’m going back the blue.
00:35:23.600 --> 00:35:28.100
This term has a y' in it.
00:35:28.100 --> 00:35:30.300
This term has a y' in it.
00:35:30.300 --> 00:35:32.000
This term has a y' in it.
00:35:32.000 --> 00:35:35.700
I’m going to bring all of those over to one side.
00:35:35.700 --> 00:36:02.000
That is going to be -2x² cos y sin y y' + sin x y' + 2 sin x sin y y' and all of that is going to equal,
00:36:02.000 --> 00:36:06.600
I’m going to put them this term and this term, I’m going to bring those over that side.
00:36:06.600 --> 00:36:25.900
= cos y cos x - 2x cos² y – y × cos(x).
00:36:25.900 --> 00:36:26.900
Let me go back to red.
00:36:26.900 --> 00:36:39.600
When I factor out the y’ and I divide, I end up with y’ = this thing on the right.
00:36:39.600 --> 00:37:13.500
Cos y cos(x) – 2x cos² y – y × cos(x) divided by this -2x² cos(y) sin y + sin(x) + 2 sin x sin y.
00:37:13.500 --> 00:37:16.400
Keeping track of it all, that is it.
00:37:16.400 --> 00:37:24.600
The fundamental part is differentiating this very carefully.
00:37:24.600 --> 00:37:30.700
There we are, this is calculus, welcome to calculus.
00:37:30.700 --> 00:37:32.600
I have an extra page for that too.
00:37:32.600 --> 00:37:37.600
Great, I’m squeezing everything into one page.
00:37:37.600 --> 00:37:44.600
Example 7, the √xy = 7 + y² e ⁺x.
00:37:44.600 --> 00:37:48.200
We just do the same thing that we always do.
00:37:48.200 --> 00:37:54.600
We know that this is equal to xy¹/2.
00:37:54.600 --> 00:37:55.500
Let me differentiate.
00:37:55.500 --> 00:38:16.800
We end up with, it is going to be ½ xy ^-½ × the derivative of what is inside the xy,
00:38:16.800 --> 00:38:30.300
which is going to be this × the derivative of that xy' + y × the derivative of the x which is 1 = 0 + this × the derivative of that,
00:38:30.300 --> 00:38:39.500
y² e ⁺x + e ⁺x × the derivative of that which is 2y y'.
00:38:39.500 --> 00:38:49.800
Let me distribute this part.
00:38:49.800 --> 00:38:54.500
I’m going to distribute this over that and this over that.
00:38:54.500 --> 00:39:30.900
I'm going to get ½ x y' × xy ^-½ + ½ y × xy⁻¹/2 = y² e ⁺x + 2y e ⁺x y'.
00:39:30.900 --> 00:39:35.100
Here is a y’ term, let me go back to blue.
00:39:35.100 --> 00:39:38.500
Here is y’ term, here is a y’ term.
00:39:38.500 --> 00:39:39.800
Let me put those together.
00:39:39.800 --> 00:39:52.200
I have ½ x y’ xy⁻¹/2 – 2y y’ e ⁺x.
00:39:52.200 --> 00:39:55.400
Let me move this over to that side.
00:39:55.400 --> 00:40:05.700
It equals y² e ⁺x – ½ y xy⁻¹/2.
00:40:05.700 --> 00:40:06.700
Let me go back to red.
00:40:06.700 --> 00:40:35.200
When I factor out the y’, I’m going to get y’ × ½ x × xy⁻¹/2 – 2y e ⁺x = y² e ⁺x – ½ y xy⁻¹/2.
00:40:35.200 --> 00:40:36.500
I will go ahead and divide.
00:40:36.500 --> 00:40:56.700
I’m left with y’ = y² e ⁺x – ½ y × xy⁻¹/2 divided by ½ x × xy⁻¹/2 – 2y e ⁺x.
00:40:56.700 --> 00:41:07.300
There you go.
00:41:07.300 --> 00:41:16.200
Example number 8, same thing, just equation, any other equation that we have to deal with.
00:41:16.200 --> 00:41:17.600
Let us go ahead and differentiate this.
00:41:17.600 --> 00:41:23.400
This is going to be 4 × this thing.
00:41:23.400 --> 00:41:31.900
The derivative of this thing is going to be 2 × this x² + y² × the derivative of what is inside.
00:41:31.900 --> 00:41:53.400
That is going to be 2x + 2y y' = 35 × 2x – 2y y.
00:41:53.400 --> 00:41:59.500
I get 8 ×, I’m going to multiply this.
00:41:59.500 --> 00:42:01.100
This is a binomial × a binomial.
00:42:01.100 --> 00:42:05.600
I’m going to get 2x³.
00:42:05.600 --> 00:42:30.300
This × this is going to be + 2x² y y' + y² × 2x is going to be 2x y² + 2y³ y' = 70x-70y y'.
00:42:30.300 --> 00:42:36.300
Here I have a y' term, here I have a y' term, here I have a y’ term.
00:42:36.300 --> 00:42:39.200
Let me go back to red, put a parentheses around that.
00:42:39.200 --> 00:42:43.300
Let me multiply through, all of these becomes 16, that stays 70.
00:42:43.300 --> 00:42:47.800
I’m going to go ahead and just multiply, and move things around.
00:42:47.800 --> 00:42:51.700
On this slide, I'm going to have 16.
00:42:51.700 --> 00:42:55.400
Let me go to blue.
00:42:55.400 --> 00:43:05.800
I’m going to have 16x² y y' + 16 y³ y'.
00:43:05.800 --> 00:43:20.800
I’m going to bring the 70 y’ over + 70y y' = 70x - 16 x³, that is this one.
00:43:20.800 --> 00:43:27.900
And then, - 16xy².
00:43:27.900 --> 00:43:30.100
Go back to red, there is my y’.
00:43:30.100 --> 00:43:34.300
Y’ and y’, factor out, divided by what is left over.
00:43:34.300 --> 00:43:58.100
I’m going to be left with y' is equal to 70x -16x³ - 16xy² divided by 16x² y + 16y³ + 70 y.
00:43:58.100 --> 00:44:06.700
There is my derivative.
00:44:06.700 --> 00:44:09.400
We have x² + y² = 25.
00:44:09.400 --> 00:44:12.000
This time they want us have to find d² y dx².
00:44:12.000 --> 00:44:16.200
They want us to find y”, the 2nd derivative.
00:44:16.200 --> 00:44:20.200
I will just do it in black.
00:44:20.200 --> 00:44:21.900
X² + y² = 25.
00:44:21.900 --> 00:44:24.800
The derivative of x² is 2x.
00:44:24.800 --> 00:44:33.800
The derivative of y² is 2y dy dx, y’ = 0.
00:44:33.800 --> 00:44:41.800
I have 2y y' = -2x.
00:44:41.800 --> 00:44:50.500
Therefore, y’ = -x/y, that is my y'.
00:44:50.500 --> 00:44:54.000
I want y”.
00:44:54.000 --> 00:44:58.700
Now I take the derivative of this.
00:44:58.700 --> 00:45:01.900
Again, I differentiate implicitly.
00:45:01.900 --> 00:45:04.800
Here, this is going to come down.
00:45:04.800 --> 00:45:15.300
This is going to end up becoming, it is going to be this × the derivative of that - that × the derivative of this/ this².
00:45:15.300 --> 00:45:20.300
Let me go ahead and just keep my negative sign here.
00:45:20.300 --> 00:45:26.100
Let me write y” =, the negative sign that is this negative sign.
00:45:26.100 --> 00:45:39.100
It is going to be this × the derivative of that, y × 1 - this × the derivative of that – x y'/ y².
00:45:39.100 --> 00:45:40.700
That is equal to distribute the negative sign.
00:45:40.700 --> 00:45:51.000
We get x y' - y/ y².
00:45:51.000 --> 00:45:59.700
However, notice the second derivative, not only does it have x and y but it also has the y'.
00:45:59.700 --> 00:46:02.500
I already know what y' is, I already found the first derivative.
00:46:02.500 --> 00:46:05.400
Y' is equal to -x/y.
00:46:05.400 --> 00:46:13.900
I can take this -x/y, stick it into y'.
00:46:13.900 --> 00:46:15.300
Let me do it down here.
00:46:15.300 --> 00:46:31.500
This =, y” = x × y' which is - x/y - y/ y²
00:46:31.500 --> 00:46:53.600
which is equal to -x²/ y - y/ y² = -x² - y²/ y/ y²,
00:46:53.600 --> 00:47:04.400
which becomes -x² - y²/ y³.
00:47:04.400 --> 00:47:10.200
When you take the second derivative, the second derivative is actually going to involve the first derivative.
00:47:10.200 --> 00:47:13.700
But you already found the first derivative, you can put that back in.
00:47:13.700 --> 00:47:19.700
Something like this, reasonably simple and straightforward.
00:47:19.700 --> 00:47:25.500
Not going to be so reasonable and straightforward, when you actually do some of the longer problems like we are going to do in a second.
00:47:25.500 --> 00:47:33.000
You would have to decide the extent to which you want to actually put in.
00:47:33.000 --> 00:47:36.100
The extent to which you want to simplify it, things like that.
00:47:36.100 --> 00:47:44.600
As long as you realize that, y” is going to also contain the first derivative and you would already have the 1st derivative.
00:47:44.600 --> 00:47:50.400
Those are going to be the important parts.
00:47:50.400 --> 00:47:58.000
Example 10, let us see what we can do.
00:47:58.000 --> 00:48:05.700
Example 10, let me go ahead and go back to blue.
00:48:05.700 --> 00:48:09.900
The derivative of sin x is the cos(x).
00:48:09.900 --> 00:48:23.100
The derivative of cos y =, it is negative, + -sin y y1'.
00:48:23.100 --> 00:48:35.700
The derivative of sin 2x is cos 2x × the derivative of 2 which is 2, it is 2 × cos(2x).
00:48:35.700 --> 00:48:48.100
Let us solve, -sin y y’ is equal to 2 cos 2x – cos x.
00:48:48.100 --> 00:49:02.300
Therefore, we get y' is equal to 2 cos 2x - cos x/ -sin y.
00:49:02.300 --> 00:49:05.200
I’m going to go ahead and take this negative sign, bring it up top.
00:49:05.200 --> 00:49:17.800
Flip these two and I’m going to write this as cos(x) -2 cos(2x)/ sin y.
00:49:17.800 --> 00:49:18.600
I hope that made sense.
00:49:18.600 --> 00:49:20.500
A negative sign actually can go top or bottom.
00:49:20.500 --> 00:49:21.500
It does not really matter.
00:49:21.500 --> 00:49:23.900
I went ahead and brought it up here.
00:49:23.900 --> 00:49:29.600
When I distributed over this, the negative of this and this, this one becomes positive, I put it first.
00:49:29.600 --> 00:49:30.900
This one is negative, I put it second.
00:49:30.900 --> 00:49:33.300
That is all I have done.
00:49:33.300 --> 00:49:37.200
There we go, we have y’.
00:49:37.200 --> 00:49:41.700
Now, y”.
00:49:41.700 --> 00:49:48.100
Y”, let me do this one in red.
00:49:48.100 --> 00:49:55.200
Y”, I'm actually going to be differentiating this one implicitly.
00:49:55.200 --> 00:50:01.100
It is going to be this × the derivative of that - that × the derivative of this/ this².
00:50:01.100 --> 00:50:09.900
That = sin y, this × the derivative of this.
00:50:09.900 --> 00:50:17.300
The derivative of cos x is -sin x.
00:50:17.300 --> 00:50:24.900
The derivative of this cos 2x is going to be -2 sin 2x.
00:50:24.900 --> 00:50:32.000
-2 × 2 is 4, it is going to be +, it is going to be +4, × sin(2x).
00:50:32.000 --> 00:50:59.400
That is this × the derivative of that - this cos x - 2 cos 2x × the derivative of this which is cos y y’/ sin² y.
00:50:59.400 --> 00:51:04.900
Let us take a look at this, before we do anything.
00:51:04.900 --> 00:51:10.200
It involves sin x, it involves sin y, or cos x and cos y.
00:51:10.200 --> 00:51:15.500
It looks like the only y' term that we have is here.
00:51:15.500 --> 00:51:17.600
That is this one.
00:51:17.600 --> 00:51:20.400
If you were going to simplify it, you do not have to.
00:51:20.400 --> 00:51:27.000
If you actually have to, for y', this is y'.
00:51:27.000 --> 00:51:31.200
You take this expression, you put it into here.
00:51:31.200 --> 00:51:33.800
And then, you simplify the expression as much as possible.
00:51:33.800 --> 00:51:35.900
I’m not going to go ahead and do this.
00:51:35.900 --> 00:51:42.700
For this particular problem, something that is going to end up looking like this, I would just leave it alone.
00:51:42.700 --> 00:51:46.200
If your teacher is going to have you do this, where you actually have to substitute in,
00:51:46.200 --> 00:51:49.700
only express things in terms of x and y, then you have to do something simple.
00:51:49.700 --> 00:51:52.900
Otherwise, you will be here for 4 days doing this.
00:51:52.900 --> 00:51:58.700
But for something like this, you are going to deal with it at some point in your career.
00:51:58.700 --> 00:52:03.200
You have y', you have y”, if you need to, you go ahead and put it in.
00:52:03.200 --> 00:52:05.700
Again, most of the time, when you are working,
00:52:05.700 --> 00:52:10.100
you are going to be working with mathematical software so it is going to be doing this for you.
00:52:10.100 --> 00:52:14.100
That is it, straight differentiation.
00:52:14.100 --> 00:52:15.600
You differentiate the first one implicitly.
00:52:15.600 --> 00:52:18.800
You get this, you differentiate the second one implicitly.
00:52:18.800 --> 00:52:26.700
It is going to get progressively more complicated, simply by virtue of the fact that this happens to be a quotient rule.
00:52:26.700 --> 00:52:28.300
I hope that that made sense.
00:52:28.300 --> 00:52:30.600
Thank you so much for joining us here at www.educator.com.
00:52:30.600 --> 00:52:31.000
We will see you next time, bye.