Log-(sin Let y = Scs _ +cot xVsecdy Find the derivative y DO NOT SIMPLIFY YOUR ANSWER AFTER YOU EVALUATE THE DERIVATIVE, ux...

Find $ dy/dx $ by implicit differentiation.
$ e^{x/y} = x - y $

He's probably given any questions that were asked to use interest depreciation to find a video file. Respect, excellence. I tear it off all the terms with respect effects. Let's start from the German left inside. We're gonna support hope. So we have Sign Ex Turns. Why, plus co signed X times you want Begin Since why's function X on the left on the right. Inside we have two X plus two wine again you. And yet, since Y is a function extra less group all the terms with DVDs one side. Unless group over terms, you got a few ideas on your side. Let's write this one as you want the eggs off to while my disco side effects. Is it going to, uh, we have? Well, that would be negative. Selects since they were two courtside inspector sign. So we have negative. All sign next times. Why minus two X. So basically, I took this Theo idea next time and right here into that two extra men riding on the back. So we haven't divine the answer for everything to be negative off. Why time sign Next plus two x divided by two. Why minus co signer. Thanks or that is equal to white time. Sign of X. You know, I don't like to x divided by co sign X minus two. What?

In this problem, we are learning how to use the technique of implicit differentiation. Now, when I learned this, I thought that this was very tricky to understand. So hopefully this problem helps you. And this video helps you understand a little bit more about how we use implicit differentiation. So where we start with this equation, eat the Y times sine X equals X plus X y. Now, just looking at this that looks like a really tricky, um equation to find the derivative of How would I do that if I have excess and wise? So the first thing that we're going to do by implicit differentiation is will take the derivative of each side of our equal sign. So we're basically saying that the derivative of E y sine X is equivalent to the derivative X plus X Y. So we'll take the derivative of each side. And when we do that, we'll get E to the Y times d y d x times Synnex plus e to the Y times a co sign of X equal to one plus one times y plus x times D Y d X. Now this is something that we can work with. We have to differentials d Y d X, and what we need is D Y dx by itself. So now every step we make is going to be is going to be in the goal to get d y dx by itself because that is our derivative, so we can do a little bit of simplification. Remember, we're trying to get do y dx by itself so we can subtract this year the y cosine X term. So once we do that, we'll get it to the Y times sine x times do I d x equal toe one plus Why minus e to the y close in x plus x times Do I d. X Now we can do a little bit of factoring to make this easier. We'll have eat the y minus sine x minus X times d Y d x equal toe one plus Why minus eat the y co sign X And now this is a very good position that we're in because this is an entire term times D Y d X, and we want the Dubai DX by itself. So we're just going to divide this term onto the other side to get you I d. X by itself. So we'll get do I d X equals one plus y minus e to the y co sign ex all over Eat the y sine X minus X and that is our derivative. So I hope that this problem helped you understand a little bit more about implicit differentiation, why we use it and how we can go through the process of using it. I know that it's a little bit tricky, but hopefully this made sense and you learn something from it.

Hi there. In this problem, we were asked to calculate the derivative of this equation X squared times wine plus two x cubed time is why equals X plus y All right, So, to begin, these always we just need to take the derivative of both sides of the equation with respect to X. So we'll do that. And as we go, just remember any time we take the derivative of why will just need to multiply results. Bye. D y d x, or why prime if you prefer. So here we go. This take the derivative. So the derivative of X squared y We're going to need the product rule for this since its one thing times another X squared times. Why? So let's be very careful here. Product rule says the derivative that will be the first thing just x squared times the derivative of the second thing. Okay, close the derivative of the first thing. So the derivative of X squared times the second part, which is why so all this in blue right here. This all came just from the product rule from X squared y. So we still need Thio. Take the second part of that derivative are equal signs are not gonna line up here. Just goes return of the space. But, um, here we go. So the derivative of to execute Why? Once again, it's a product rule. So let's think of the first part and red here to x cubed The second part in blue can be Why so the product rule tells us Take the first thing to X cube times the derivative of the second thing sometimes the derivative of why plus at the derivative of the first thing Excuse me. Time is the second thing. Okay, so in all this came from product rule of this second term on the left. Now we ever equal sign. The right side will be a little shorter the derivative of X plus y. So the derivative of X is one the derivative of why is also one. But then again, as always, they multiply by. Do you r d x? We take the derivative of why. Okay, so now to clean up all these derivatives in this next step, X squared stays the same the derivative of why with respect to X and is one. But then times do you I d x plus. So the derivative of X squared is just too X. And that's why remains the same plus two x cubed. Once again, we need the derivative of why which is one times do I D X class the derivative of two X cubed is six x squared. The wife stays the same and over here on the right, we have one plus just do your i d x Okay, on the end goal at the end Goal this to have our final answer look like d Y d X equals everything else. We want to get the derivative by itself. So that's our dream. In order to do that, we're going toe. Want to take every term that has a d Y de accident, right? So this has a d y d x in it. This is a D i. D x in it, and this is a d y de accident. So the those three terms and green let's get all those on the same side of the equation. Everything else will get on the other side we already have. So that's just move this one on the right. That's the I D. X. Well, really, Let's subtract the why d x from both sides a little. Take care of that. No, but the new green here. So that way we have X squared y d X that stays where it is on the left. This term here that we're underlining that will stand The left was to ex cute. Do you r d x? Like in all the d y t X is we want to just put on the same side Then finally and we subtracted d y d x from both sides so that we have you are the x on the left. Okay, so those three green terms we moved all the way to the laughter kept them all in the left side. Everything else We're going to want to end up on the right side. The opposite side. Right. So this one that was on the right side, you can stay everything else though. So this term and this term, they did not have d i. D. X in them. So we want to send them over to the right side. Really? What we do eat again, Subtract. Of course, he subtract from both sides. So subtract two ex wife from both sides and also subtract six explored wife on both sides. So hopefully we're convinced that the end of the day when we do that all right, they're gone from the left side and they just show up on the right side, but with different signs. Right, So we're minus to X Y minus six x squared y If it looks confusing, don't be intimidated. Just remember all the only thing we did we just rearranges this equation so that the three terms that had d y d x in them and green were on the left side and all the other terms that didn't have d y d X in them ended up on the opposite side. The right side. The reason we did that because now we can factor out. Do I d. X from everything on the left. So if we factor that out, what's left? Well, we have X squared plus two x cubed minus one. All right, that's all the tears and left and then on the right. It's not going to do well. Let's keep that the same and we're one step away. Hopefully, we can see what to do to get this d Y D X by itself, we simply need to divide both sides. Bye. There's parentheses here. So we will do that. We did run left. Let's also do it on the right. And that's your final answer. You can see on the left side and these cancel out so we can erase all that left side ends up just e Y d x. And that is what we want. So you can write it over here. So it's a little closer. U R D X and that is our final answer. All right, hopefully that helped.

52 were supposed to differentiate this implicitly and find expression for developer DX. So if you differentiate using the power rule 3/2 comes down, the power decreases by once. That's one or two. Two or three years again, power rule comes down, power decreases by one, and then we go inside and we differentiate wire with prospective works, which is give our debts on different station of the constant is zero. Let's take this term over to the right. So what have we lived for? Those 2/3 via days to the par minus one of the three D via or D X Z equal to negative three or two X rays to the power one or two if you multiply both sides by if you take. If you divide both sides by the stone, then we get diva or the excess negative three over to X rays to the ball. When we're doing something, that route takes over two or three wires to the power Negative. Wonderful treat. This can be for the simplified us. Uh, two and two gets multiplied. We get four, three and three gets multiplied. We get nine and we have root techs over virus to the par minus one or three. And since the power of Y is negative, it can be taken in numerator. So the final answer comes Orders minus nine. Route text Times wire is to the Paul one or three over four. So this is the final answer.