find-the-differential-of-each-of-the-given-functions-y6-x-sqrt1-4-x-3

Question

Find the differential of each of the given functions.
$$y=6 x \sqrt{1-4 x}$$

Samuel Hannah · Accepted Answer

seven This problem. We&#x27;ve been given a differential equation, which I&#x27;ve written in the top left. Now, this is a Benue equation where n equals 1/2 p of X is two times x a pair of minus one cure fax is six times the square root of one plus X squared. So as it&#x27;s a bit nearly equation, we can make the substitution if you equals y to the power of one minus and one minus end is just one minus 1/2. So this is why the power of 1/2. So using equation 1.8 point 12 from the textbook, this transforms the differential equation. So it looks like one over 1/2 the utx plus two x to the minus one times white and 1/2 equal to six times one plus six times square root of one plus X squared. Now, if we multiply fruit by ah half, we get that this equation is equal to D u D X plus X to the minus one. Why spare? Half is equal to free times Square root of one plus X squared of this differential equation could be solved using an integrating factor. Now the integrating factor we propose is off the exponential of the integral one over X, which is excellent minus one The X now the Inter court one over axes just for natural log of the absolute value of X. This gives is eat about natural log the absolute value of X, the exponential in the natural World Council. So we get that the integrating factor is the absolute value of X From the question we know that X is always positive. Service is simply X. We use this. This means the differential equation can be reversing is the differential of you X with respect x and that is equal to free X I&#x27;m square root of one plus X squared. Now we want to integrate both sides to get our solution. This means we need to integrate free x times the square root of one plus x squared with respect to X. We do this by making the substitution off. So we make the substitution off T equals one plus X squared and this causes DT to be equal to two x the X make the substitution into the integral. This interval becomes the integral off free over two. I was square root of tea. E T. Andi has solution t to the power of free over two, which is, you know, original X variables, which is equal to one plus X squared times to the power of free over two, plus some integration constant C. So if we go back to the solution of this differential equation, this means. But we have you Times X is equal to want plus X squared about free over two. Let&#x27;s see we divide through by X. This gives us that the right hand side is equal to one over X times one plus x squared to the power of free over two plus C over x onda. We remember that you is equal to y to the power of half. So our solution in terms of why is that? Why, to the power of 1/2 is equal to one over x times one plus X squared forever tip for C over X

Gilbert Deleon · Answer

we&#x27;re gonna take the derivative of this function six over the fourth root of X. To take the derivative of this function, we&#x27;re going to use the power rule which is the derivative of X to any constant exponents, Um is multiplied by the exponents and then decrease the exponents by one. So that time we would take the derivative of the power function. Before we could do that, we need to write this function as a power function. So one thing that we can do is we can rewrite the fourth route as a fractional exponents, ext with 14 and then another thing that we can do this. We can rewrite that positive exponents in the denominator as a negative exponents in the numerator. And now that we have this in a power function form, now we can apply the power function rule to differentiate. So we have a differentiated yet we&#x27;re different shading now before we&#x27;re just rewriting in terms, rewriting our function as a power function. Now I can multiply by the exponents, so it&#x27;s gonna be six times negative. 14 negative, six over four and then decreasing the ex opponent by one&#x27;s negative 1/4 minus one. Uh, subtracting fractions, getting a common denominator, It&#x27;s gonna give us a negative force. So then, after we differentiate now, we can do some simplification. The derivative D y the eggs, bacon simplifying six over four to be three house and then excellent. Negative. Five over four. And then something else we could do is we could rewrite that negative exponents a za positive exponents in the denominator. And I want other thing that we could possibly do is we could rewrite that fractional ex opponent as a route. You were right that factual exponents as a route that fractional extra only applies to the X will be two times before through of X to the fifth.

Amy Jiang · Answer

we were asked to find a derivative with falling so I could be writes my radical as a power. So that&#x27;s six over extreme part of 1/4, and now I&#x27;ll be writes my variable as a negative exponents so we can bring it to our numerator. So that&#x27;s six X to the negative 1/4, and I can take the derivative. So that&#x27;s why time is equal to negative 1/4 times six and a next department of 1/4 minus one. So six is three times to and forced to tax to our choose cancel. We&#x27;re left with a negative 3/2 extra power of one is for over four, so that&#x27;s negative. 5/4.

Patrick Vaughan · Answer

Hello. So today we&#x27;re giving a problem and were asked to solve it. So looking at it, we see that we have Ah, y prime. So what do you I over Deac. And that should make us think that perhaps we&#x27;re going to use separation variables because we have an X component and a like a boat. Okay, so let&#x27;s start working away at this problem. So first, what I just like to do right off the bat is take that next to the negative one and just put it, um, in the denominator, that just makes it a little nicer to work with in this, uh, scenario. So let&#x27;s just remind ourselves that why equals ive x, Times X and that the equals Why, Rex? So let&#x27;s put this all in the expression of white crime, which involves us trashing that expression over to the right hand side. So they get why prime or de y over DX equals six y squared X to the fourth minus two times why over X, And then we can plug in our, uh, these are what in our we can use our expressions later to plug in. So do you already acts pulling in for? Why there we get vivax times acts equals six times the x squared or B squared, X squared times X to the forest minus to B. Now, we&#x27;ve done this a few times, so I&#x27;m just gonna give the left side, um, derivative just straight away. So we get race that little quick until mark so we get X times DV over the X plus the and that equals six The squared extra, the sex because X squared times excellent forth this four plus two, which is sex and then minus to be Now we can subtract the the from both sides and making a brake line. This gives us TV over the axe equals six squared x 2 to 6 minus three d. And then we can take that ax from the left hand side. And, uh, well, let&#x27;s divide the entire discretion by one of her ex to you. Transfer the extreme left hand side of the right hand side expression. And then that gives us TV over the X equals six the squared extra benefits minus three d overact. So you might be looking at this and you&#x27;re thinking, Well, how are we supposed to separate variables. I don&#x27;t really see how that could be done. And, well, it&#x27;s gonna be a really tricky unless you recognize that this is actually a for newly equation. And if you want to, like, learn more about Bernoulli&#x27;s equation, they use about a lot in fluid dynamics and then also thermodynamics. But for our just strict mathematical purposes, um, I&#x27;m just gonna show you how you can solve this type of equation. So first, let&#x27;s put it into our generalized format and our generalised form, um is gonna look something like y prime plus some function p times why equals some function que, um, times wide some power. Well, let&#x27;s see if our expression looks like that. So we have DV ever Deac switches are the prime. And then if we add this, um, three D over acts Teoh the left, this gives us plus three over acts the equals six x to the fair, the square. So if you look at it, we have our we have RV component are derivative r p prime. We have our be in our the two some power. And then if we isolate our function piak on our function que axe. In our end, we can pull out three over acts, which is our He attacks our Q backs and then our and well, what was the purpose of doing that? Well, the purpose of doing that well, first that&#x27;s right, These out. So P vivax equals three overact que backs equals six extra and a new equals two. And then this might get a little confusing. But we then have a little substitution rule the equals wide, the one minus end. And, um so in our case, our the original be an hour problem set is the why in the solution problems that and that&#x27;s being such a newbie. So I&#x27;ll just put the, uh, slash just for future reference that it&#x27;s different from our original de. Anyways, this solving this, um, we one RV are the or the quote unquote. Why to be substituted in for and to do that are the is then going to be equal to 1/1 minus, which allows us to create this, um, substitution formula, which is really the part that we&#x27;re interested in 1/1 minus on the prime plus p vivax, the equals our Q of axe and then Well, you substitute in. We get 1/1 minus two, which is just negative one. So we get negative. The prime plus three v over acts equals six X to the fifth. And the interesting part of this is that, um once we&#x27;re down to the form, we can take this expression right here. The three D acts in particular, the three x, the Kiev ex parte. And we have, um, something that we&#x27;re gonna get We&#x27;re gonna define a new function on your backs equals e to the integral of p of X. And I should know that to get this formula, um, correct. We need two months by the whole thing by negative ones that we don&#x27;t have that negative one attached to the the prime term. So we have the prime now minus three x three over axity equals negative six x to the fifth power. So new of X equals e to the integration of r p vivax the X. So we plug in our PM backs are three over acts. In this case, we get e to the integral of negative three, divided by X. The X Well, we know that the integral of Mega three over ax. Well, that is, they get three natural log of axe. So when he saw for this, um, the negative three becomes the exponents, and then the e in the natural log cancel. So we&#x27;re left with X, and then they get the native third power, and then I should also have on a side note. So we&#x27;re interested in pee vacs because that&#x27;s related to you. And then this expression, our q of acts will be defined as our inch are function of interest. Are JIA backs? Um, and just remember that it&#x27;s this one down here and that one down there. Okay, so now that we have all these different parts, we can use the generalized, formulaic expression for solving it, which is the integration of our Novaks, which we just saw for times are vivax. This is our be carrot or B slash of acts. Crime. The axe equals the integration of new backs times our G of axe or that Q of X from before. Yes. So plugging in we have, um, are it&#x27;s gonna give third times are the flashbacks. So just be slash we&#x27;ll take the derivative of that times DX equals the integration of X to the negative three times the negative six x to the feds. And now we just have to solve for it. So when you have an integral in a derivative in the same expression, those just cancel each other out. So the left side is, um, X the negative three times our B slash equals Well, what&#x27;s excellent? Five times the native three times except fit. Well, that&#x27;s just going to be are integral of negative six x squared. Well, that&#x27;s pretty straightforward. And a girl. So the integration of X squared is X cubed, uh, divided by three times are negative sex, and that&#x27;s B plus C. So this is gonna reduced under two. We were left with X to the negative Third, he slash equals negative to you. X cubed what I see. And then we can So this is basically is the same as saying one over X cube so we can multiply X cubed by the entire expression, leaving us with our B slash equals, uh, negative to you. Next, The sex plus, um, see X cubed. And now what we have to do to plug in for R B slash Well, well, we set that equal. Teoh, Our, um the two. The one minus end, which is even the negative one. So this son becomes one of her be and therefore inverting the entire expression that gives us our original V equals one over negative two x six plus C x cubed, which is our final answer. All right, um, we could always plugged back in for R B, which equals why, over at so that&#x27;s going to be some form of our final expression. And that&#x27;s how we can use Bernoulli&#x27;s equation to solve something like this. And I&#x27;m gonna try to do him out so you can see everything that we did to get to this point. Um, there we go. Wasn&#x27;t Sheriff. I was gonna be able to seem out far enough. Okay, you men a little too much. Just right. Okay. And that&#x27;s what

Amrita Bhasin · 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.

Problem

Find the differential of each of the given functi…

Problem 14 Easy Difficulty

Find the differential of each of the given functions.
$$y=6 x \sqrt{1-4 x}$$

Answer

$d y=\frac{6(1-6 x)}{\sqrt{1-4 x}} d x$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Anna Marie V.

Kayleah T.

Caleb E.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$d y=\frac{6(1-6 x)}{\sqrt{1-4 x}} d x$

Basic Technical Mathematics with Calculus

Anna Marie V.

Kayleah T.

Caleb E.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Anna Marie V.

Kayleah T.

Caleb E.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus