 Hi, I'm Zor. Welcome to Unizor education. We continue talking about properties of derivatives as part of the advanced mathematics course for teenagers presented on Unizor.com. If you watch this lecture from this website, you will have the notes for the lectures, which are very detailed and I do suggest you to do it. And also maybe after the lecture is finished, it would be a good practice to just go through the text of these notes. They basically are like a textbook and read it again, so it's it's better inculcated in your minds. Alright, so today we will talk about the property about how to derive, how to take derivative of the function which is expressed as one over another function. Now let's do exactly the same as the prescription for taking a derivative actually asks for. So this as a function which is equal to this one, then my derivative is actually the limit of the ratio of increment of the function divided by increment of the argument. So what is increment of the function? Well, that's function at value x plus delta x, which is this minus function at value x, right? Alright, so let's use the common denominator. Now, obviously I assume that f at x is not equal to 0 wherever all these calculations are made. So in the denominator, I will have their product and in the numerator, I will have this. Right? So it's f at x divided the product minus f at x plus delta x divided the product, right? Which by the way, if you take a look at the numerator, it's minus delta f at x, right? f at x is this minus this. So this is minus delta f at x. Now, whenever I am making this ratio, increment of the function divided by increment of the argument and take a limit of this as increment of the argument goes to 0, I will have, well, let me consider this separately from this. So that will be a limit. Of minus 1 divided by f at x plus delta x times f at x times limit of delta f at x divided by delta x. So now you see how easy it is because this piece is basically considering f at x is a continuous function. The limit of this is just minus 1 over f at x square, right? Because this goes to f at x and this f at x basically is independent of delta x. So it's like a constant goes out. So that's what we have as the first multiplier. And the second multiplier is obviously derivative of the function f at x. So this is a derivative of 1 over f at x. So 1 over f at x derivative is derivative of the f at x with a minus sign and divided by square of the function. Now we will use this formula to derive a few limits. My first one is second. Okay, now this is basically by definition it's 1 over cosine of x, right? So we have exactly the situation prescribed by the theory which is actually is minus cosine square of x, right? times derivative of the cosine which is minus sign. So this will be plus and then sine of x. So that's the derivative of the second. I'll use this to save some space on the board. Okay, my next example is tangent d tangent of x by dx. Well, tangent is by definition sine over cosine, right? So it's d by dx sine x times 1 over cosine x. Right? So consider this as a product. So it's the first function sine x times derivative of the second one, which we have just did a second ago, right? That was sine x over cosine square x plus the second one times derivative of the first one, which is equal to this is 1 and this is sine square over cosine square, which is tangent square. So it's 1 plus tangent square of x. So here I have used two different rules. One is the product and another is the inverse. Next. Okay, e to the power of minus x. Again, we know what's the derivative of the e to the power of x, which is e to the power of x. But we never actually did e to the power of minus x separately, but let's just recall that e to the power of minus x is 1 over e to the power of x, right? Which basically we can use this rule with f at x is equal to e at x. So it's e to the power of x. I mean, so it's e to the power of x square minus sine and the product and the derivative of the e to the power of x, which is e to the power of x, which obviously is minus 1 over e to the power of x, right? Because we can reduce by e to the power of x which is minus e to the power minus x. Now, what's interesting is and the situation actually is very similar to the previous lecture when I was deriving the derivative of sine of 2x. I can use actually a composition function here. One function is minus x and another is e to the power of x. Now, the composition of these two functions actually also can be used to take a derivative and there are certain rules which I will explain in the next lecture and according to those rules, I will have to have exactly the same result. But we will see. Now, the final formula which I would like to derive is which I actually used already once. I mean, I don't use it. I actually derived it for a specific case of tangent, right? So tangent was sine over cosine, but now I would like to have a general rule of this. How to take the derivative of this ratio of two functions? Well, again, we do exactly the same as I did with a tangent. It's f at x times 1 over g of x equals first times the derivative of the second, right? And here I will have the derivative of g of x, right? According to this rule, plus the second times the derivative of the first one. Well, obviously it would be much easier if we will use the common denominator g of x square and I will have first I will get this one because it's positive, right? I multiply by g of x numerator and denominator to get g of x square. So that's what I will have in the numerator and here minus f at x g plus g derivatives. All right, so in some way it resembles the product actually, except there is a in the product of f times g derivative. I will have derivative times the second one plus the second second one derivative, the first one. Now I have a minus. Now I have a minus because functions are not symmetrical because one is numerator and other is denominator and then I have a remnants from this denominator also in the derivative as well. So it's a little bit more complicated formula than for the product. For the product, let me just remind it's this one. So it's first time derivative of the second and the second time derivative of the first and here we have, you know, in some way similar but with a minus sign and also there is a denominator square. All right, so these are a few examples to basically demonstrate how we can use the rules to take a derivative from one over the function. I do suggest you to go to unison.com and take a look at the notes for this lecture. That's it for today. Thank you very much and good luck.