find-the-derivatives-of-the-given-functions-y3-cos-1leftx205right-3

Question

Find the derivatives of the given functions.
$$y=3 \cos ^{-1}\left(x^{2}+0.5\right)$$

Tyler Moulton · Accepted Answer

we want to find the derivative of Y. With respect to X. For the function while equals three. Our coasts of X squared plus 0.5. To solve this, we&#x27;re going to rely on tricking electric or rather inverse trig derivatives that we recently learned. We also need to rely on differentiation shortcuts we&#x27;ve learned throughout our journey and single variable calculus. So I have listed here the three major inverse trig derivatives, Arcos, arc sine and our 10. We need to use the Arcos derivative, which is negative one over square root one minus X squared. Since the argument of our coaches expert plus 10.5, we&#x27;re also going to have a chain rule effect. So why is already written it&#x27;s most easily differential forms we can see to solve. Dy dx is negative three over square root of one minus X squared plus 10.5 squared by the chain rule. We multiply by the derivative X squared plus 0.5 which is two X. Thus multiplying out our constant and simplifying the denominator we obtain. Do I d x equals negative six X over the square root of one minus x squared plus 10.5 squared as presented on the right and boxed in.

Sid Wan · Answer

number eighteen. We have y equals cause I lodge a log to Axe Cube. Absolute value. So still just good general collective. Sign off. That part has a lot to X cube, but fly directive off This part is one over to excuse. Yes, that&#x27;s his sign to execute to you.

Sid Wan · Answer

twenty two we have. Why was three co sign x Our far Mona&#x27;s? Cause I look thiss car looks We just need to use quotient plus Lina&#x27;s cause I square top on Derivative is back You sri sai acts What price for the war. So frak bottom What directive is fine? What a black top one Sorry, because I and that&#x27;s it because somebody five Barton laws like fifteen sign at plus three side at cause I eggs coz some tracks race Okay, he&#x27;s got cancer, fathers, mothers, coastline where, like fifteen

Amrita Bhasin · Answer

this question gives us why is co sign of a huge plus X cubed and asked us to differentiate what we know that the derivative of CO sign you is negative. Sign of you times you prime or the derivative on the inner. Therefore, we have negative sign of a cube plus X cube times the derivative of the inner. This is a constant there for the devil Constant zero What&#x27;s the derivative of X Cubed three X square. Now, lastly, we should simplify this negative three x squared sine of a cubed plus x cubed. And this is our

Robert Daugherty · Answer

section $3 6 Problem number 30 were asked to take derivatives where using the chain rule in several cases, you have to use other rules that you know. So I see two functions multiplied in both cases. I&#x27;m going to use that you need to use the product. Ruelas. Well, so why Prime is going to be equal to, Let&#x27;s just say one over acts now. What&#x27;s the derivative of signed? The negative Fifth could be negative. Five. Um, sign. Um, if you differentiate this so, um to the negative six x And now the derivative of Sine X is just co sign X. Okay, that&#x27;s just the first part of the product rule. And now the next part will be what&#x27;s the derivative of one over acts that&#x27;s minus one over X squared time. Sign negative. Fifth of X. Now let&#x27;s move on to this second term minus X over three at the derivative of three co sign X. That&#x27;s three co sine squared of X and then the derivative of the co sign. It&#x27;s minus sign of X and then minus that river to the X Over three is 1/3 and then you&#x27;re left with co sign Cubed of X. So that&#x27;s it. Now it&#x27;s just seeing what combines eso If I look at this first term, Um, this is what it&#x27;s a Why prime is minus five over X, signed to the minus six of X co sign X And then if you look at the next term, that&#x27;s just gonna be minus one over X squared sign minus fifth of X. And then you look at the next term was here three and three that Hansel&#x27;s. So this is going to be and I&#x27;ve got two minuses. So that&#x27;s going to be plus plus x co sine squared of x sign of X and then the Met last term is his minus 1/3 co sign cubed of X. So it looks quite complicated. Um, and it is. But really I mean, is just product ruin chain rule. It&#x27;s just that the trigger looks quite complex in here, but that is the derivative of the initial function that we started out with

Nicholas Barvinok · Answer

we want to find the derivative of the function a cube plus co sign Cubed of X. So the first property of derivatives that we&#x27;re going to use is the fact that the derivative of a sum is the sum of the derivatives of the individual terms. So we can just differentiate, turn by turn. So we get that the derivative is the derivative of a cube, plus the derivative of co sign Cubed X. Now a cube is just a constant, so it&#x27;s derivative of zero. So this whole thing is really just equal to the derivative. Been cubed X now to take the derivative of coastline Cute X, we&#x27;re going to use the chain rule because we have the composition of functions here. The outermost function is raising something to the third power and the innermost function is co sign of X. So as the chain role dictates, we start with the outside function, raising something to the power of three, and we apply the power rule. So the director of a cubed, as we said, was zero. Now to take the derivative of coastline cubed X, we first get three co sign squared X by the power rule and then by the chain rule. We multiplied by the derivative of the Inside Function Co sign X, which has a derivative of minus sine X, which finally gets us our final answer by just rearranging things a bit. Minus three co sign squared X signed X, and that&#x27;s it.

Problem

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

Problem 10 Easy Difficulty

Find the derivatives of the given functions.
$$y=3 \cos ^{-1}\left(x^{2}+0.5\right)$$

Answer

$\frac{d y}{d x}=-\frac{6 x}{\sqrt{1-\left(x^{2}+0.5\right)^{2}}}$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Kayleah T.

Caleb E.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$\frac{d y}{d x}=-\frac{6 x}{\sqrt{1-\left(x^{2}+0.5\right)^{2}}}$

Basic Technical Mathematics with Calculus

Grace H.

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

Grace H.

Kayleah T.

Caleb E.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus