 Hi, I'm Zor. Welcome to Unizor Education. I would like to talk about specific methodology to take derivative in some not exactly straightforward cases. This lecture is part of the course of Advanced Mathematics for Teenagers. It's presented on Unizor.com. I recommend you to watch this lecture from this particular website because it has very nice notes like your textbook and it has certain functionality. For instance, you can sign on and take exams any number of times you want. Site is free by the way. So what are we talking today is about so-called implicit differentiation. Now implicit differentiation, again, it's nothing like special. It's just an application of known things. We have already started them. The differentiation of compound functions. The chain rule. Let me just remind you the chain rule. If, for instance, you have a compound function f of g of x and you would like to take its derivative, I will use this notation as a derivative. So what you do is let y be g of x. So first you differentiate f of y by y and multiply it by differentiation of g of x. And then you basically substitute y is equal to g of x. So let me just give you an example. If you have something like, for instance, y is equal to 2 to the power of sin x. Something like this. So first let's not use y here. Let's put it f of x. You can actually say that this is y is equal to sin x and f of y is equal to 2 to the power of y, right? That's the same thing. So to differentiate that, you first differentiate this. Now 2 to the power of y would be 2 to the power of y by natural logarithm 2. And then you multiply it by derivative of this inner function. And now you can go back 2 to the power of y, put y back in place. So it's 2 to the power of sin x times natural logarithm 2 times cosine x. So that's your derivative. Now this is basically the known chain rule for differentiation of compound functions. What I would like to say is that it works basically when you have explicit dependency between argument and the function. Like 2 to the power of sin of x. What if you don't? What if your function, so let's say you have function y is equal to some kind of a function of x. But it's not given explicitly like a formula. It's given for instance like this, which is basically an equation of the circle. So there is no y equals some formula of x. You have this. Well, in this particular case, it's actually easy to resolve this for y. It's plus minus square root of r square minus x square. Basically if you have your circle, now this part is plus and this part is minus. Two different functions basically combined together into one graph. So if you want to know what is the tangent of the tangential line at point x, y, for instance. So you can actually have the derivative of this function. So let's do it. d by x of y is equal to, this is a square root. Let's just talk about plus the upper part. So it will be 1 over 2 square root of r square minus x square, right? That's the square root as a outer function. And then inner function is r square minus x square and its derivative is. Now this is the constant, so it's minus 2 times x, right? So we reduce it by 2, so it's minus x divided by square root r square minus x square. So that's the derivative. Now, in this case, as I was saying, you can resolve it. But let's approach it slightly differently and it seems to me easier. Let's just differentiate the whole thing from here without getting into these formulas. Resolution for y. Now, how can we differentiate this? Well, on the right I have a constant, so on the right I have 0, right? Now on the left I have sum of two functions, function x square is 2x, the derivative, right? The derivative of y square, again, is a compounded function. The outer function is square, which is 2y, and inner function is y, so it's y derivative. From which we can derive as minus 2 and 2 will go x divided by y, right? Now, in theory it's exactly the same as this one, right? Because y is square root of r square minus x square. But we kind of derive this much easier, and if you have x and y coordinates on the circle, this gives you kind of an easier and direct answer of what is exactly the derivative. Now, this is basically an example of implicit differentiation. So, without resolving for y is equal to some kind of a formula of x and directly differentiating it, we do this type of differentiation of both sides, thinking about y square as a compounded function. Alright, now, this works both ways, as we were saying. We can resolve for y and we can do it implicitly. Now, there are some cases which are not so simple to resolve. For instance, if you have x square plus y square is not the constant, but something like a sin x, what then? Well, actually, in my case it's even more complicated. It's sin y, right? If it's x, it can be resolved. If it's y, it's not resolvable for y, right? So, what can we do here? Well, let's do exactly the same thing. Let's differentiate, and this and this will be compounded function where inner function is y in here and here, and outer function is square here and sin there. So, whenever we go to a derivative, we will have this. Now, the derivative of a sum is sum of derivatives. So, it's derivative of x square which is 2x plus derivative of y square which is 2y times derivative of inner function y, right? Same thing here. First, we do cosine y. That's the derivative of outer function y by y and times derivative of inner function by x. From which we can derive dx of y is equal to 2x on the left. On the right, I will have cosine y minus 2y. Alright, so that's an expression. I cannot use this to go into the dependency of x because I cannot resolve this for y. So, it remains as it is, but again, how can we use it? Very simply, if this represents some kind of a graph and you would like to know the derivative at some point on that particular graph, so what you do is you establish your x coordinate and then use some kind of a numerical methodology, computers, whatever else to resolve this for y, not as a formula but as some kind of an approximate value. So, you have x and you have y and you substitute it here and you get the tangent. So, that's just an example where you cannot really have this explicit formula even if you try. So, it's still a useful method, this implicit differentiation. Let me give you an example when this method is really used to simplify something which we actually did but much more difficult way. Remember, when we were talking about the definition of derivatives, we basically had it defined as a limit of incremental function divided by increment of argument where increment of argument goes to 0. Remember this, right? Now, using this straight methodology, we found out that if you want a derivative of this function, it's 1 over x. We really did it like logarithm of x plus delta x minus logarithm x divided by delta x and had a limit using certain properties of the limits, etc., etc. and we derived this formula. Let me explain you how it can be done easier if you already know about exponential functions. So, let's just think about what is the function like this? What is the definition of the logarithm? Well, if you remember, logarithm is, in this case, a natural logarithm. The base is e. So, it's such a number y which is used as an exponent for e it should equal to x. So, this is basically, this is definition of that. So, instead of differentiating this using limits, etc., etc., let's use this implicit definition of function y and use this implicit differentiation methodology which means we'll just differentiate this and use this as a compound function and use the chain rule. So, the derivative of this should be equal to, that's y. It should be equal to derivative of this, right? Now, what is derivative of this? So, first it's exponential function e to some kind of a power and we know that e to the power of something being differentiated gives you exactly e to the power of that something. Now, we have to multiply it by inner differentiation and that is the derivative of the left part. Derivative of the right part is 1, right? Derivative of function x. Now, what is e to the power of y? That's the definition basically of logarithm, right? So, e to the power of y is equal to x. So, x times gx of y is equal to 1. So, gx of y is equal to 1 over x where y is logarithm x. So, we got exactly the same thing but without resorting to all these complicated things about limits. Very easy, right? So, that's kind of a power of this approach if used properly and at the appropriate time. And let me just give you another example where it seems to be the only way, actually. Limits will not help. Like in this case, for instance, when I was talking about x squared plus y squared is equal to r squared we can directly resolve it for y and then take a derivative. In case of logarithm x we can actually use some limits and properties of the limits and then derive with the same formula. So, let me just give you another example where I have no idea how to do it without using this implicit differentiation. Here is the formula for y. Well, it's explicitly defined function but I have no idea how to differentiate it directly. So, what I will do is first let me just logarithm this. So, let's just assume that everything is positive. So, at least for certain positive x where everything makes actually sense I'm not really worrying about the domain of the function, etc. Let's just skip it for a while. So, what I will do is I will do logarithm y is equal to sine x times logarithm x, right? I took logarithm at both sides and since I have this exponent then the logarithm goes like this. Remember this property, right? That's what I was using, where a is sine. Okay, so we know that. Now, that is easier. Now, let me differentiate this because I know how to differentiate it. First of all, the differentiation of logarithm y gives me what? First of all, outer function is logarithm inner function is y. So, derivative of logarithm is 1 over y, right? And I have to multiply by derivative of the y. So, let me use prime number as a derivative. It's easier. What do I have on the right? I have a product of two functions. So, the derivative of the product is first function times derivative of the second plus derivative of the first times second. From which I can derive y prime, the derivative of y, is equal to y times this. And I know what y is. So, it's x sine x. And here I have sine x divided by x plus cosine x times logarithmic. So, that's the formula. And I hardly could derive this formula without this simple technique. Alright? Okay, that's it for today. Try to read the notes for this lecture. It's very useful. And I do encourage you to sign on to the website to Unizor.com and take exams as much as you possible can. Alright. Thank you very much and good luck.