find-the-derivatives-of-the-given-functions-yx2-cos-1-xsqrt1-x2-3

Question

Find the derivatives of the given functions.
$$y=x^{2} \cos ^{-1} x+\sqrt{1-x^{2}}$$

Tyler Moulton · Accepted Answer

All right. We want to find the derivative dy dx for the function Why is equal to X square our cousin X plus the square root of one minus x square. To solve this problem, we&#x27;re going to rely on derivative shortcuts. We&#x27;ve learned throughout single variable calculus in particular the inverse trig derivatives that we&#x27;ve recently learned. So I&#x27;ve noted here the inverse trig derivatives. The most important one we need for this problem is DDX Arcos X equals negative 1/1 minus x squared square rooted. We also are going to have to use the chain rule in the product rule. So why can be written as X squared? Arcos X plus one minus X squared like half. Thus, we can take dy dx using the product in the first term chain rule, A second term as dy dx equals two. X. Archos sex taking the root of X squared minus x squared times one over root, one minus x square. This is derivative. Our coast plus one half, one minus X squared and one half times 92 X. The second two terms have the same denominator of square, one minus X squared plus. We can combine them to obtain final solution at the bottom. Dy dx is equal to X. Archos x minus x squared plus x over the square root of one minus x squared.

Samantha Lincroft · Answer

So for this problem, we&#x27;re going to be solving the this derivative on the right. And so to do that, we&#x27;re gonna use a quotient rule. But we&#x27;re gonna use the version of the quotient rule where we can see that the top F here is just equal to one. And so when that happens, we just have the negative derivative of the bottom on the top and the derivatives squared on the bottom. So that means that D y dxc Why is going to be equal to, so to take the derivative of the top, So it&#x27;s gonna be a negative of this. But to take the derivative of the top, we will start by having basically the top of this quotient rule between these two products here. But we&#x27;re going to have a plus sign, So that will be ah X squared plus X plus one times two x plus X squared, minus one times two x plus one. And all of that will be over this bottom part squared. So X squared minus one squared and x squared plus X plus one squared. And so, when we, uh, sort of multiple ID&#x27;s out will then get so I&#x27;ll put the negative in at the same time about the multiplying this. So we&#x27;re multiplying all of these essentially by negative two x So minus two x cubed minus two X squared minus two x and then world do AA minus two x cubed so we can do Pem Dozier first inside outside last inside will be plus two eggs minus X squared plus one, and that&#x27;s all over keeps squared, and so finally will get that are derivative is going to be when we combine terms minus for X cubed minus three x squared zero x because those cancel plus one all over, Don&#x27;t you write this out this time X squared minus one squared X squared plus X plus one It&#x27;s quick, you know.

Robert Daugherty · Answer

Section three. That six problem number 29 were daily with two first derivatives where you must use the chain role. So in this case, as everything in this math or capital is, it builds. So not only will I need to use the chain rule, I&#x27;m also going to usually use the product role to get some with an integrated. So why prime going to be equal to Let&#x27;s take that 1st 1? The product rules as take the derivative of that first portion two X times a signed Fourth of Eggs plus X squared. Now what&#x27;s the derivative of the sign to the fourth? That&#x27;s gonna be four times the sign cubed of X Times, the derivative of the sign of X, which is just co sign backs. OK, that takes care of the 1st 2 terms. OK, so that&#x27;s two X times signed fourth of X plus X squared times for signed Cube Co sign. Now let&#x27;s go to the next term. The derivatives of Ellicott that the derivative of X is just going to be one times co sign minus two of X plus. Then you&#x27;re gonna have X the river of co sign to the minus two is going to be minus two co signed to the minus three of X and then the derivative of the co sign is just minus sign of X. Make that a little bit smaller. Size fits, so it&#x27;s gonna be minus sign backs. Now, here again, we&#x27;re in the situation where calculus is all done. It&#x27;s just trig identities and algebraic simplification to see if we can get this looking any better. So why prime is going to be what two X signed, Fourth of X Um, plus four X squared sign cubed of X co sign of X and then it looks like I&#x27;m going to have a plus co sign minus two of acts. And then it looks like, um, plus to sign of x co sign minus three of X. Now, if you think about what you can do is I could say, OK, well, look at the 1st 2 terms. So why prime? I can factor out a two x sign cubed of axe, and then I&#x27;m left with sign of X plus two x co sign of X. And then if I look at these last terms, um, co sign to the negative too. That is the seeking squared of X Plus two. I can write this as the sign of X over the co sign of X and then times the secret squared of X. So now you see, you can factor out, um, a seeking squared of X. So when you do that, your final answer is going to be why prime is two X sign cubed of X sign of X plus two x co sign of X plus seeking squared of X and then you&#x27;re left with one plus two tangent of X. So here&#x27;s a case where, you know, you might argue it, but the the algebraic manipulation of these trick functions was more involved in the actual taking of the first derivative. Okay, so this is my final answer. You realize any of these would have been the right answer? Eso any of these is the right answer, but it&#x27;s just getting it down to the the final form that is the simplest

Shumayal N · Answer

So for this question are inter function is one plus on co tangent squared X and our outer function is, um u to the power half. So in order to get the determinative first, we will take the derivative of the outermost part of the function which is the square root sign. So using the power world, huh? You leave the inner part as it is, people tangent squared X to the power minus half and now you multiply this with the German of of the inter function which is one plus tangent square. Next, the derivative off one is obviously zero and for the co tangent scored X again You have to apply change rule. So first, take the derivative of the power part of the function. So that will give you cold tinge Tico tangent X and then you multiply. This with the inter function, which is co tangent eggs and the derivative of co tangent Eggs is no. I know this because he can&#x27;t squared X so this two will cancel with this too. And you can take the one plus co tension square next to the par minus happen to the denominator. So your final answer becomes Cosi can&#x27;t squared X minus construction square necks tangent, eggs over disk load of one plus co tangent squared X.

Amy Jiang · Answer

Yes. So let&#x27;s take the Durga. So we have X times finished in verse of X, so we&#x27;ll use our product. You&#x27;ll hear that A derivative of accents one times over many time, plus the derivative of singeing gris that&#x27;s equal to, um, remaining term X over describe of one plus squared. And then we have minus the derivative of the square root of the fallen threats 1/2 X squared plus one to the power of negative 1/2 sometimes a derivative over inside. That&#x27;s two X sorry to councils. And we see that, um, this term here, that&#x27;s equal to X over the square root of X squared plus one. So we see that would be captured. This starting with this one on our derivative. It just left with singe in verse of X.

Amrita Bhasin · Answer

Okay. We&#x27;re gonna be deriving implicitly for this. So this means we have two x y plus x squared, Do you? Why? Over d backs equals d y o ver de axe clos de over de axe times x y sign axe plus X y coast Xanax. Okay, now that we&#x27;ve gotten this, we know we can write this as bringing the bottom. So I bring the right side on the left side because eventually we&#x27;re gonna be dividing by this that I&#x27;m writing in parentheses. Order to get the wild rejects by itself.

Problem

Find the derivatives of the given functions. $$v=…

Problem 16 Easy Difficulty

Find the derivatives of the given functions.
$$y=x^{2} \cos ^{-1} x+\sqrt{1-x^{2}}$$

Answer

$\frac{d y}{d x}=\frac{x^{2}}{\sqrt{1-x^{2}}}+2 x \cos ^{-1} x-\frac{x}{\sqrt{1-x^{2}}}$

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Basic Technical Mathematics with Calculus

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$\frac{d y}{d x}=\frac{x^{2}}{\sqrt{1-x^{2}}}+2 x \cos ^{-1} x-\frac{x}{\sqrt{1-x^{2}}}$

Basic Technical Mathematics with Calculus

Anna Marie V.

Kayleah T.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Anna Marie V.

Kayleah T.

Kristen K.

Samuel H.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus