find-the-differential-of-each-of-the-given-functions-y2-sqrtx-frac18-x-4

Question

Find the differential of each of the given functions.
$$y=2 \sqrt{x}-\frac{1}{8 x}$$

Amrita Bhasin · Accepted Answer

okay. Part A The derivative using the product rule. We have acts times You did the negative for ox Natural order of E. We know that we&#x27;re taking the derivative of negative four acts. So in this case, this simplifies to negative for X E to the negative or ax plus e to the night of four ax using the chain rule and factoring we end up with Okay, Part B. We know that we have one nice t to the fourth to the 1/2. Therefore we have, if you have tea, is one nice to the fourth than we know that we could have 1/2 you to the negative. 1/2 comes negative for t cubed, which is negative to t cubed over the square root of one minus t to the fourth.

Samuel Hannah · Answer

So this problem we have a differential equation, which I&#x27;ve written in the top left on. I&#x27;ve rewritten this from a question simply by dividing each term by two. This equation is now a form of a binary equation where N equals minus one. Andi Piana, P of X and Q of X functions are is written on the right hand side. As is as visible newly equation, we couldn&#x27;t make V transformation of U equals Y to the power of one minus end on one minus minus. One is to by equation 1.8 point 12. In the textbook, this transforms the differential equation into off utx plus co. Tangent of X over two times you that&#x27;s equal to four times Co sign Cubed of X. Now we&#x27;re just multiply fruit by two to get listen to a nicer form, and this gives us for the u. D. X lost the co tangent of X times. You it&#x27;s equal to eight times but co sign cubed of X. This is a linear differential equation. This is a simple differential equation which we can solve using integrating factor method this integrating fact that which will take we need to find it using the taking the exponential of the integral of co tangent of X with respect to X. Now, the curtain being School of Co Tangent of X can use standard trick in a metric identities. But we just one for it. How would find it using substitution. Here we take substitution. T equals sign of X. Wonderful DT equals the co sign of X DX. This is helpful because we can rewrite the code tangent as cause of X over Sign Lex and we see the under a substitution. This becomes the central of one over t DT, which has simply ha solution the natural locust T which is the natural log the sine X but entrance that&#x27;s one back. Therefore, the integrating factor is the exponential of the natural log of sign of X, The exponential in the natural order of from cancel So integrating factor is simply sign of X. If we put this into other friendship equation, this gives us that the ex derivative off you times of sign of X is equal Teoh eight times the co sign cubed of X times The Summit of X. We want to solve this for me to integrate both sides, so we need to look at the integral on the right hand side. In more depth, we integrate, we get the eight times the integral the the coastline cubed of X time signed of X t X and we want to do this by substitution. This time will be substituting. V equals their sign, Lex and therefore TV equals minus sign of X e X. This transforms the integral into minus eight, the interval of be cubed TV, which when we sold the interval, is equal to minus eight times veto power for over four, which gives us minus two feet too powerful. This is the right hand side for their French equation. When integrated, the left hand side simply becomes you times sign of X, and once we substitute back into the back into the equation, this becomes the co sign. Two are full of X, and then we just need to finish by adding the integration constant C Rewrite this into a form for you by dividing fruit by sign. We get that you is equal to co sign to the powerful of X plus C for some integration constant all over the sign of x Andi If we make the final substitution that you is, why squared? We get our solution in terms of why.

Caelan Thomas · Answer

so to get to you, our first gonna take the derivative with respect to X, which is going to give us two X over to times the square root of X squared plus three y squared T X. And we&#x27;re going to add the derivative with respect to Y, which is going to be six. Why? Over to root X squared plus three y squared. Do you? Why and then this would just simplify by dividing by two here and here to give us X over the square root of X squared plus three y squared de eggs plus three way over the square root of X squared post, rely squared, do you? Why? What?

J Hardin · Answer

let&#x27;s find the differential of this given function. So no here that here you is a function of two variables. So let&#x27;s go ahead and rewrite that usu of X comma y and then the formula for the differential do you? This is the differential of you And on the right side we&#x27;Ll have the partial derivative of you with respect X times the X plus the partial derivative with respect So why d y so we&#x27;LL have to go ahead and find some parts of the road Is here to make that a little easier less Just rewrite this as a rational power So it we can use the power rule with the chain rule So we see here, take the first they&#x27;re evident with your spent X What half comes off to the front? Sory using the power rule right now Subtract one from their power So you get a minus one and then by the chain rule you have to multiply by the derivative of the inside which in our case, is just too X So that gives us x over X square plus three y square onside the radical in the bottom and then similarly weaken do you? Why so use the power ruling The chain rule together here So differentiating this guy right here with their Spencer Why one half comes off in front We&#x27;Ll get that radical in the bottom again And when you multiply by the derivative of the inside with respect So why that will be six y So that gives us three. Why over the radical after we cancel And now we&#x27;Ll just go back and plug all this in. So you ex we just found that down here that will give us X over the radical And then you still have to Don&#x27;t forget your DX, their multiplication and then for you Why we found that over here, Radical back in the bottom. And also hear what we can&#x27;t forget to multiply by the d y So there it is. We have the differential notation to you on the left hand side And then we use the formula to obtain the right hand side. So that&#x27;s our final answer

Patrick Vaughan · Answer

So today we&#x27;re given a question we&#x27;re asked to solve for it using our standard techniques. So this isn&#x27;t a very straight forward problem. And I just want to show you how you should be thinking about these types of problems. So I&#x27;m gonna start out using an incorrect method of how to solve this problem. Or at least it wasn&#x27;t a way that I found particularly useful using our typical y equals V X, um, solutions that our substitution that, um so to the left, you&#x27;ll see I have a skip to solution in a little box later on, I&#x27;m gonna edit in where you can the time to skip to when I start through the solution. But I just wanted to make it so that you can sort of see how we can start to think about these more complex problems. That&#x27;s where we&#x27;re looking at it and being like, Oh, well, I&#x27;m not seeing that this Methodist particularly working. This is how I&#x27;m gonna work. Maybe I should try something else. There&#x27;s something new, So I just want to sort of explain to you how I think about these problems when I get stuck or challenged and hopes that hopefully it will help you to try different things in a person here. Um, when you encounter things like this All right, So let&#x27;s recall that we have been typically taking our substitution as why equals r vivax time or written up another way. Are vivax equals? Why demanded by acts? So starting off with this problem that we want to isolate the wide prime so we can divide why, by both sides Or just most play the whole thing by one of her wife and I give bless our why prime or R d y Over the X equals our radical X squared plus y squared, divided by why minus X over y And then we take our live value where it&#x27;s being x, times X, and we substitute that infer wherever we have a wide so that he over the act time vivax time blacks equals radical X squared plus B squared X squared over all over. Why, um And then, uh, excellent. Why is the inverse of be so that&#x27;s one Overbey. Okay, so taking the derivative of the left side that makes it so that we have are the prime times acts, plus, uh, the which equals. And then on the right side will do. Ah, step will retake our X square out. We pull it out and hopes that this will help us. So that becomes X squared times one plus he squared over. Why? Minus what ever be, um, and then the left side. So we&#x27;ll have a brake line, and then we&#x27;ll have X times are devi over the act plus B equals X over y. So this right here becomes X over y because radical ax squared just acts. So again, I&#x27;m gonna have whatever be radical every time radical one plus squared minus one of her be and then also attracted me from both sides. And I&#x27;ll get TV over. The X equals well, this is radical. One plus b squared might be minus one minus. B squared over one. About it might be because to get that, uh, denominator and B, we multiply by the Overbey. Okay, so now we&#x27;re left with this complex expression on the right, and then we&#x27;re left with, um, access TV yet. So if you, uh, do our separation of variables, we get, um, TV times The over are very large denominator, which would be one plus B squared, minus B squared, minus one times like TV equals, um de X over X. And then you can try integrating mess. But I had difficulties. I couldn&#x27;t really figure out how to make this simpler. So I want you in culture, this type of situation, there&#x27;s to think that you could try the first is to check your assumptions. I didn&#x27;t start a problem that I was gonna be able to use that, um the substitution, nothing. And, uh, looks like that it didn&#x27;t really hope me because I could see integration part. It&#x27;s pretty this cold. I couldn&#x27;t really figure out how to go from there, so well, the right side pretty straightforward with left side. Um, that was a little more complex, and I didn&#x27;t like working with that. So one, uh, check your assumptions to you should, if your assumptions are correct, figure out a different way. Maybe try. Um, so see, it&#x27;s father methods in this. We&#x27;ll put a question mark better because we don&#x27;t know that&#x27;s something we have to investigate. Sometimes it will have other pre established methods for solving these types of problems. Sometimes it&#x27;s just you made the wrong assumption beginning problem. And in this case, I made the wrong assumption. Great being in my problem and I&#x27;ll show you what I mean. So I view this part and great silent create Green is good. Okay, So after a little bit of heaven on I decided to try you substitution where I said you equal to x squared plus y squared And then my d&#x27;you is to act Plus you Why times d y over the x So you might look at me wondering Well, how can I make that assumption? Well, we&#x27;re assuming that is our variable of interest or independent variable and that why is the output very well associated with facts? So that&#x27;s why I could do this. Okay, so now that we&#x27;ve established that way, can then rewrite this as the prime or the derivative of you equal to times X plus why and what&#x27;s another way to write? Do you everyday acts? Well, that&#x27;s why prime and something interesting happens if you use this, you substitution. So I&#x27;m gonna use red and under it highlights. So I have X squared And why squared? I squared y squared and then in my you if we divide by two. So we get you prime over to equals X plus. Why? Why prime? Well, if we look and if we look, we see that we have ah x y y prime we have Why comes like crap. And then if we took this, Max brought it over. So let me redirect this. Um, a little I&#x27;m equals radical. I could quitter plus wife minus X was the original expression and taking the original expression we can make it in. Why I just adding actual side explain. Plus why squared? And then all of a sudden, we see that we have I fly squared And then our x squared plus y squared this original expression that gives us you prime over to you equals radical. Well, what is you prime? Well, that&#x27;s just do you over the X. So do you Over the x times that to you equals radical you. Well, let&#x27;s separate power variables going into a little more space that gives us, uh, you to the negative 1/2 which is just one over radical. You do. You is equal to two e x. So now we can integrate. Well, that&#x27;s a pretty straightforward. Integral. On the right side, we just have to x plus C and on the left side. Well, what&#x27;s the in a role of you to the negative 1/2? That&#x27;s just you with 1/2 time here because between derivative of you to the 1/2 that two and 1/2 cancels out, we&#x27;re left with you to the next 1/2. Okay, so now let&#x27;s continue salt into the problem. What&#x27;s divide by two? Just get this in expression of you. So that gives us you the 1/2 equals plus C over to and then what was you? Let&#x27;s remind ourselves, growing up you was X squared plus y squared. So putting this back in for are you substitution? We get X squared plus y squared one equals X. What? See bottom by two and that&#x27;s your final answer. And now just seem out. There we go. And then, as a reminder, this not dead method on the left did not work and that the U substitution method was the one that ended up being successful

Problem

Find the differential of each of the given functi…

Problem 8 Easy Difficulty

Find the differential of each of the given functions.
$$y=2 \sqrt{x}-\frac{1}{8 x}$$

Answer

$d y=\left(\frac{1}{\sqrt{x}}+\frac{1}{8 x^{2}}\right) d x$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Samuel H.

Michael J.

Joseph L.

Precalculus Review - Intro

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$d y=\left(\frac{1}{\sqrt{x}}+\frac{1}{8 x^{2}}\right) d x$

Basic Technical Mathematics with Calculus

Grace H.

Samuel H.

Michael J.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Samuel H.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus