Implicit Differentiation Applying Implicit Differentiation Applying Implicit Differentiation Page [1 of 5]

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1 Page [1 of 5] The final frontier. This is it. This is our last chance to work together on doing some of these implicit differentiation questions. So, really this is the opportunity to really try these things, get your hands dirty, get in the mud, splash around, and see if you can have some fun. So, the first question I want to just look at two questions right now and the first question is the following. We are given the following relation. E raised to the power xy equals y. And the question is find the equation of the line tangent to this curve at the point the suspense is building zero, one. So, this is some curve. Now, in fact, what s the graph look like? I don t even know, for sure, but is the point zero one on there? That we can check, because I ll plug in zero for x and one for y and see if this is satisfied. So, if I put in a zero for x, that s zero times one. Zero times one is zero. E to the zero -- we have on this side, and e to the zero is one and notice one does equal one. So, in fact, this point is a point on this curve. But how does the curve go? Well, it s some wiggly thing, and honestly, I don t know what the graph of that looks like. But I m asking you to figure out even though we don t know what the graph looks like figure out the equation for the tangent line, the line that just grazes the curve at that point. Pretty impressive that now, the power of calculus turns out to be more powerful than our own ability to take a look at these functions now. We have now actually exceeded in our abilities what we have already been able to do. Because these functions are so complicated, it s hard to visualize them, but the calculus still works. That s the really cool thing. You don t need to visualize it anymore. The calculus will give you the answer. So, what we need to do is we need to find the equation of the tangent line. Okay, so now, what do we need in order to find the equation of the tangent line? And since this is the last time we re going to be working together on implicit differentiation, I d thought we d really scramble things up and have a lot of interaction here really have a chance for you to participate. So here we go. The first question is, what is it that we are going to need what kind of things are we going to need in order to find the equation of a tangent line? So, I want you to think about that now, and then I m going to want you to make a guess. So, what do we need in order to find the equation of a tangent line? Give it a shot. All right. Well, a lot of possibilities we saw, but as always, all we need is a point on the line and the slope of the line. Well, a point on the line. Well, there s a point that s on the line, because since the line is going to be tangent to the curve at that point, that must be a point in common with both the line and the curve. Well, you already saw that the curve satisfies that point. The point is satisfied at that curve, but also the line must be. So, in fact, the line must contain the point zero, one. Therefore, all that s left to do is find the slope of the line. And what line is it? It s the slope of a tangent line. So, what do we need? We need dy/dx. Surprised? No. Need slope, which equals dy/dx. Okay. So, how are we going to find the slope? Well, I have to differentiate this. What technique will I use to differentiate it? Well, it s not a function, so I m going to use implicit differentiation. So, what I m going to do is I m going to differentiate everything here both sides differentiate with respect to x the left-hand side, and then differentiate with respect to x the right-hand side. Okay. Now, a chance for you to participate in the discussion. What I want you to do is to take a look at the left-hand side. And I want you to make a laundry list the kind of laundry list that I make up in terms of what things we re going to have to need the Product Rule, the Quotient Rule, the Chain Rule, something else to figure out the derivative of this what s holding it together. And then, try to even indent it in terms of what you need to nest it in the right order. So, give that a shot right now, and let s see how you do. Well, let s see. If you look at that, what I see here is e to the blop. So, therefore, this is going to be a Chain Rule where the inside is right here. Now, you know, a lot of students get a little bit confused. Because they say, Gee, Professor Burger, I would have thought that s sort of the outside you know something to the something, and that s the outside sticking out. It s sort of funny to think that s the inside.

2 Page [2 of 5] Well, you have to remember that we know how to take the derivative of e to the x. It s e to the x. So, we need to really think about this as e to a whole, very complicated exponent. So, really, you might want to write it this way. You might want to write it as e to that power. And now, I think it makes it a little easier to see maybe not outside/inside in the traditional sense. But that is nestled inside the parentheses it s e to the blop. So, we need to take the derivative of e to the blop. The derivative of e to the blop is e to the blop. Then, we have to multiply that the Chain Rule tells us by the derivative of the blop. And the derivative of the blop well, that s a product. So, I hope that your answer was first a Chain Rule, and then within the Chain Rule, we have to do a Product Rule. That s the correct answer. A Chain Rule to peel off that e, and then a Product Rule to untangle that product. That was the answer. Okay, so why don t we give this a shot right now and see how we make out. So, we want to take the derivative of e to the xy. We re taking the derivative with respect to x. So, I have e to the blop. The derivative of e to the blop is, well, e to the blop. I ve now just peeled off the e. Now, I have to multiply this by the derivative of the blop. So now, I ve got to use the Product Rule. The first times the derivative of the second plus the second times the derivative of the first. The first well, that s just x times the derivative of the second. What is the derivative of the second? Answer right now. Dy/dx. Because I m taking the derivative of y with respect to x. And so, the derivative of the second is the derivative of y with respect to x, which is dy/dx. Now, where am I? Well, I forgot. So, I ve got to recap. I m doing a Product Rule here. The first times the derivative of the second plus the second, which is y, times the derivative of the first. And what s the derivative of the first? That s just the derivative of x with respect to x, which is one. So, there s an invisible one there. Okay. So, where am I in this process? Well, am I done? I don t know. I ll have to regroup and take a look. I have e to the blop. The derivative of that is e to the blop times the derivative of the blop. That s the first times the derivative of the second plus the second times the derivative of the first. Yes, this is the completed answer. This is the derivative of all that. Okay. Well, that is just a little teeny portion of the problem. Let me remind you that we saw that that was just the lefthand side. Now, I have to take the derivative of this. And what s the derivative of this? Well, that s the derivative of y with respect to x, which is, as always, dy/dx. So, the derivative of this is equal to dy/dx. We just found the derivative of that here, and so, what we see is we see that e to the xy times x dy/dx plus y is equal to dy/dx. That s what we ve just discovered. Now, we have to solve this for dy/dx. And now, I would like for you to try to solve this for dy/dx. So, just take a look at this expression, and now solve it for dy/dx. Got an answer? Let s see if it agrees with what we ll do together. What I ll do first is I have to isolate the dy/dx. So, I m actually going to distribute this term here, because I have a dy/dx interlocked with a not dy/dx here. So, when I distribute that, what I see is let s see. I see an x e to the xy dy/dx. That s that term. And then, plus a y e to the xy and that equals dy/dx. Now, I removed all the parentheses. I can easily see where the dy/dx s are and where they re not. I ll bring this term over to the left-hand side by subtracting it, this over to the right-hand side by subtracting it. And so, what I would see is x e to the xy dy/dx minus the dy/dx that was formerly on the right, but now I moved over. And that equals and now what I do is that stays. This is now going to migrate. And so, it equals y e to the xy. Now, what s wrong there? Well, if I bring this over to the other side, I have to subtract it. So, there should be a minus sign right there. Well, now I notice a common factor of dy/dx. And so, if I factor that out, what I see is dy/dx is being multiplied by x e to the xy minus one. And that still equals minus y e to the xy. And now, if I just divide both sides by x e to the xy minus one, this side will cancel, and then here will come down as a denominator, and I ll see what dy/dx is. Look how long this is. This is a three-sheet problem, folks. They re getting really long, but as long as we just are careful, we can conquer them.

3 Page [3 of 5] And I see here minus y e to the xy, all divided by -- that thing I divided by -- which is x e to the xy minus one. So, that s the derivative. That s the derivative. So, that s the answer. That s the answer. Actually no, that s not the answer. Remember, we have to concentrate, make sure we answer the question that we re asked. The question was, given this relationship, we want to find the equation of the tangent line at this point. And we needed to find the slope, and we just found the slope. So, that s the slope, but that s just the machine that produces the slope. We want to find the slope at that point. So, I need to plug that point in right here and see what I get. So, let s do that right now. So, now the slope at the point, which I remind you is zero, one, is equal to let s see. So, I m going to put in zero for x and one for y. So, here I see minus that s now a one e raised to the power zero times one, which is zero, all divided by and this is a zero times e to the zero minus one. And so, e to the zero is just one. So, I have a negative one on the top. And this is just zero minus one negative one on the bottom, this is one. So, this is the slope. Well, you already know a point on the line, namely this point, because this is the point of tangency. So, what s the equation of the tangent line to that very complicated-looking curve at this point? We have the slope. We have the point, and as always, I use my point/slope form, which is the thing that I think is terrific, and I hope you re growing fonder and fonder of. So, I can just plug right in. And when I plug right in, what do I see? I see y minus the known y point, which is one, equals the slope, which is one, times x minus the known x point, which is zero. And you can actually simplify that if you want. That s just a zero, so it goes away. That s just x. And I could write this as y equals x, and if I bring that over, plus one. And so, there s the equation of the line that is tangent to that very complicated curve at the point zero one. And what does that curve look like? It s hard for us to say without doing a lot of thinking about it, and yet, calculus empowers us to figure out what the tangent line is. In fact, calculus tells us a little bit more, because we know around that curve is that curve sort of slanting up like this somehow, or is it slanting down like this somehow? Well, we know the tangent there is one. So therefore, it s somehow slanting upward. It s either going like this or like this. Right around that point, it has to be going up, because the slope of the tangent is positive. So look, we re actually able to extract information about the function even though we ourselves don t know too much about the function, which is foreshadowing for the fact that later in life, we re going to use these ideas the calculus machinery to actually discover what functions look like. More on that later. Okay. Let s try one last question together, and I m going to allow you to interact with this a lot. And the question is just to find dy/dx given the following relationship. And here s the relationship. I hope you re ready. Sine of xy plus the natural log of x plus y plus y squared plus one cubed equals one. Look at that thing a lot of parts. I want us to now look at this together. The first thing I m going to do is realize that there s no way to untangle the y. So, this is as implicit as you re going to ever get in life a lot of y s and x s all scrambled up there. We ll have to take the derivative of both sides with respect to x. When I take the derivative of the left-hand side, I ll have to take the derivative of this piece plus the derivative of that piece plus the derivative of that piece, and that will equal the derivative of one, and this will be my contribution to the project. I ll take the derivative of the whole right-hand side. That s zero. So, in fact, what I now see is the following. I have to take the derivative with respect to x of sine xy then add that to the derivative with respect to x of the natural log of x plus y plus the derivative with respect to x of the quantity y squared plus one all cubed. And that equals the derivative of this side, which is zero. There are a lot of pieces here. In particular there are three pieces, and I want you to try each one of these. So first, let s begin by having you make a laundry list for what is needed to resolve this problem Product Rule, Quotient Rule, Chain Rule, and in what order. So, right now, make a laundry list up for us about how we re going to attack this problem.

4 Page [4 of 5] Well, let s see. What we ve got here is sine of blop. So, it sounds like we need a Chain Rule. We need a Chain Rule up first. And then, within the Chain Rule, at some point we ll have to take the derivative of the inside, which requires a Product Rule. So, the correct ordering, I think, is Chain Rule, and within Chain Rule, Product Rule. Okay. Well, let s try to execute that. What I d like to actually do, if you ll allow me, is to keep this thing pristine, and let s solve each of these three sub-problems separately, and then we ll just tack them all together. So, this is going to be, folks, a longy, as they say in the business a longy. That means it s a longy. Okay. Let s see how we re making out here. This is now sine of xy. Okay, so first, we re going to do a Chain Rule to peel off the sine. So, the derivative of sine of blop is cosine of blop. And what s the blop? It s xy. When I multiply that whole thing, I just peeled off the onion part that s the sine. I ve got to take the derivative of that. That requires a little bit of a Product Rule. So, what do I do? It s the first, which is x, times the derivative of the second. And what s the derivative of y with respect to x? That s dy/dx plus the second, which is y, times the derivative of the first. And the derivative of x with respect to x is just an invisible one multiple. And so, there we have, in fact, the derivative. You can check. Sine of blop is the derivative of that is cosine blop. Derivative of the blop first times the derivative of the second plus the second times the derivative of the first. So, that s that piece right there. That s that piece right there. So, in fact, shall we insert that in? We really should go back and insert that in. So, that piece is going to be cosine of the blop times x dy/dx plus y. That was what we saw by using the Chain Rule and then the Product Rule. Okay. Now, it s time to deal with this piece. So, what I d like for you to do right now is to make a laundry list up of all the things we re going to need to tackle this problem right now. Give it a shot. Well, I think we have an inside here and an outside, and that s it. So, really, this is just a Chain Rule. We ve got log of blop. And so, what s the derivative of the natural log of blop? Well, we know that the derivative of natural log of blop is one over blop. So, here I see one over blop. And what was the blop? X plus one I m sorry x plus y. And now, I have to multiply that. I just peeled off the natural log. I have to now take the derivative of this piece. And what s the derivative of that piece? Well, the derivative of that piece is well, the derivative of x with respect to x is one. Plus what s the derivative of y with respect to x? Well, you re getting in the habit, hopefully, of seeing that s just dy/dx, the derivative of y with respect to x. Great. Well, that was the derivative of that piece right there. And now, we have that last piece to contend with. And so now, how are we going to take the derivative of that? Well, that actually is going to require what? Well, I think it s just going to require a Chain Rule. And I d like for you right now to try to take the derivative of this whole piece. So, take the derivative of this whole piece with respect to x. You ll have to use the Chain Rule. See how it goes. All right. Let s see if we can make progress on this. I see blop cubed. So, the derivative of that is three blop squared. And what s the blop? It s y squared plus one. But now, I have to multiply that by the derivative of the inside. I just peeled off that three. But now, I have to take the derivative of the inside with respect to x. So, how am I going to do that? Well, now, I have to take the derivative of that. What do I see? Well, if you take the derivative of y squared, I actually have to do another little, teeny Chain Rule in my mind, because the y is not an x. And so, I see blop squared. And so, the derivative of blop squared is two blop. So, two blop times and I just peeled off that two times the derivative of y with respect to x, again, dy/dx. Do you see how I took the derivative of that y squared? With respect to x, I have to do a little intermediate step blop squared to blop times the derivative of y with respect to x. And then, the derivative of the one is just zero, and so, in fact, that ends this. And all that equals zero. And now, you can see where the dy/dx terms are. They re there. They re there. And they re all over here. And so, in fact, this whole term has the dy/dx term in it. So, what needs to be done now is to distribute this cosine to this term

5 Page [5 of 5] and to that term to free up that piece. Distribute this one over x plus y to here and here to free up that piece. And this piece remains as it is. This little piece right here will be moved over to that side. This little piece right here will be moved over to that side. And you ll notice what remains will have a common factor of dy/dx. We can then factor it out, get a really complicated thing, and then divide through by the complicated thing to solve for dy/dx. I am not going to do that for you right now. I d like for you to work through it right now at your little table there. Wherever you have, I hope it s a chance to do some work there. And then you can compare your answer and see if you got it right. But I m not going to do it now, because you know what? I m sort of pooped and I don t want to do it. But, I do want to close this section on high note, not on a poop note. You don t want to close things on a poop note. But I want to tell you something, because something really exciting has happened, and what has happened is we ve come to a pivotal point in this course and in our discussions together about calculus, because we have now seen all the techniques that we will learn in this course about how to take the derivative. There will be no more techniques. There s not going to be a new thing I m going to throw at you later on. Look, we saw the Product Rule. We saw the Chain Rule. We saw the Quotient Rule. We saw how to implicitly differentiate objects that weren t functions but were merely relationships. We saw how to take derivatives of certain transcendental functions like trigonometric function, and exponential and logarithm function. And that s all we re going to do. So, in some sense, the differential portion of the course if behind us. No more differential techniques. And so, it s important for you to understand where we are intellectually in this course. Where we are is just completing the ability to differentiate and take the derivatives of functions. So, where do we go from here? Well, what I want to do and what I m sure you re doing in your own class and in your own studies is to now step back and say, Okay. We re empowered now to take derivatives of all sorts of things hard things, easy things, middle of the road things. Now, let s return to the fundamental question, which is what does the derivative mean and how can we use it? So, I want us to spend an awful lot of time looking at all sorts of different applications and ramifications to the derivative and really now appreciate the power of the derivative. We now can begin to get a sense of how to differentiate. But it might be, for a lot of people, just this sort of cookbook thing. I do it and so forth. Now, we have to go back and say now that you can do it, let s see what it all means and how we can use it in examples, in applications, in our lives, in our business world everywhere. So, that is what we re going to spend a lot of time doing, looking at the power of the derivative. Well, congratulations for coming this far. You really should be proud of yourself. I know it s frustrating. The road to this kind of mathematical objects in this mathematical journey is, certainly, not without its bumps and jolts and whatnot. You have to work through this stuff, especially the implicit differentiation stuff, especially the Chain Rule. You need to crack your textbook open and do lots of examples and do lots of problems and talk to a lot of people. Talk to people, whether it s in the chat room here or in your classes or in your dorm room. Work with them and don t get frustrated, but really try to attempt to make it your own. It d be a great intellectual triumph. So, try it, have fun, and really be proud of what you ve done. Okay. We turned a chapter, and we re going to pick up the action.