find-the-derivative-assume-a-b-c-k-are-constants-y-3-x4-4-x3-6-x

Question

Find the derivative. Assume $a, b, c, k$ are constants.
$$y=-3 x^{4}-4 x^{3}-6 x$$

Yaw Asomani · Accepted Answer

Okay, So you have this one and still gonna be the same thing. Ah, the very wealthy interests X they&#x27;re going to do I d x, and we&#x27;re gonna do it on turn by turn bases. So I started to Why the X take the first time? Uh, this is for right powers, for it&#x27;s gonna drop down here to multiply whatever is here and whatever is endless. Negative three. So four times negative three is negative. 12. Right? And then you bring your ex, and then you take away one from the power here. Right? So what is the power to Paris for? Four minus one is gonna be three. Same thing is happening here. This is a power, right? Didn&#x27;t drop down to multiply. Whatever is here about every serious four. So it&#x27;s just gonna be 12 because four times three is 12 right? And they bring your ex What is a partner here? Three. Take away one from it. It&#x27;s just gonna be two left. What is that power here? It&#x27;s one rate. It&#x27;s gonna drop down to multiply. Whatever is here and it&#x27;s 66 times were still six para years. When you take away one from it. It becomes zero in any number over a recruiter. Power Zero is gonna be one. So I&#x27;m not gonna write the X again. Right? Consisted of prominent zero, and it&#x27;s gonna be one exited power. Zero is one and one time. Six stills, six. That&#x27;s why I don&#x27;t have to write the X again. So this becomes the answer to the question.

Yaw Asomani · Answer

Yes. You got this one at the seven of follow on from a previous tutorial. So if not, if you have not watched the previous tutorial on creature to go watch it and you&#x27;re going to see exactly how we&#x27;re flowing together, right? Eso do I? Yes, this is This is a part right now is the end. So she&#x27;s gonna drop down here to multiple. So you gonna have for over three down here and the next. Then you take away one from that, right? So the decision is gonna be 4/3 extra, the 1/3.

Yaw Asomani · Answer

So we&#x27;re trying to find a derivative of those function. And so you have why it goes three eggs, right? You have a usually why is the same as every vase and that is three eggs, right? So if you have every face to be a polynomial, for example, we have the X to the end, right? Let&#x27;s make it more general a exit, and then to find a derivative, you&#x27;re just gonna multiply those in here by this day said a derivative, Which is why prime usually do why the X? This isn&#x27;t a multiply. This is gonna be end times, eh? Then you write the eggs and then you take away one from this ends, there&#x27;s gonna be 10 minus one, and that is the derivative formula for the derivative. So over here they have. Why here? And that is the same as our sacrifice. Right? So trying to find F bride and our ex prime of eggs, which is the same as to why the eggs now the power&#x27;s year this end over here is one right, So multiply one by three is gonna be three times one, which is still three eggs. And then when we take away one from there&#x27;s gonna be one minus one. Right? And that is the same as three extra power zero and anything to the partners. Zero is one. Right? So finally, you have three times one, which is three. So the derivative off, Why equals three eggs? You just do I d eggs. It&#x27;s just three, right?

Will Erickson · Answer

neither in this problem were asked to find this derivative here. So I&#x27;m gonna show you two ways only because sometimes it&#x27;s good to be able to do it a fast way. And, ah, not necessarily the way you first think of. So this isn&#x27;t the quotient rule chapter. So the first way I&#x27;m gonna show you is actually using the quotient rule, which I think is probably what the book has in mind. So notice we certainly are in the position to use the quotient rule we have, huh? Quotient with the top function X squared plus three. The bottom function is X. So if we use the quotient rule and just follow, uh, the rule that the quotient rule tells us, let&#x27;s notice that we have We&#x27;ll call this F, and we&#x27;ll call the bottom function G just as your textbook does. And then we will, uh, go from there. So according to the caution rule, we first right the function g. So in this case, it&#x27;s G is the bottom, and we multiply by the derivative of the top, minus the top times the derivative of the bottom, then all over the bottom squared. Okay, let&#x27;s just clean up the top here. The bottom will stay X squared. So we have X times. The derivative of X squared plus three we know is two X and over here in the second term that stays the same. The derivative of X is just one. So we end up on the top here with two x squared minus X squared, minus three K. Make sure we&#x27;re a little careful with that. The plus one didn&#x27;t do anything, so we didn&#x27;t have to write it. And then we should distribute that minus sign across these parentheses and to a squared minus. X court is just X squared. So the X squared minus three over X squared. That&#x27;s one way of writing the derivative or you prefer to break it up. For whatever reason, we could break it up into two different factions explored over X squared minus three over X squared, which, of course, is one minus three over X squared. So either way to write this is fine. They&#x27;re both the same answer. That&#x27;s what we get using the quotient rule. If you want to turn off the video right now, you certainly can. But I would like to show you that sometimes the question rule is not the smartest thing to Dio. Harley&#x27;s the fastest so really quickly here. Let&#x27;s rewrite what we started with. Unless notice in this case, we can pretty easily split this up like we did at the end of problem. If we just split it up into two separate fractions X squared over exit just X, and we can rewrite that like expose three times X to the minus one. Now, the derivative is actually pretty easy. Begin this. Use the power rule in both of these Derivative of X is one driven of three extra minus one is minus three X to the minus two and notice That&#x27;s exactly what we got doing it the quotient rule.

Will Erickson · Answer

Hi there. This problem were asked to find the derivative of a X squared plus B, although the third Power A and B are just Constance were told before all these problems. So just treat and be like numbers. Important thing to realize here is that we have a chain rule. We have some function of X. Uh, call it G. We have some function of X, although the third power, so the chain rule says, is that the derivative will be well. We&#x27;ll do the outer derivatives look at three times that inter function squared. But then don&#x27;t forget we&#x27;ll have to multiply by the derivative, the inter function. So this is just a game plan here on the right, because it&#x27;s a chain room. So let&#x27;s go ahead and start taking the derivative, the derivative of this whole function. Well, they had green to the third power. Let&#x27;s apply the power old of the whole thing like we said. So we&#x27;ll get three times and all that&#x27;s green stuff squared. Go ahead and copy that in there. But then, by the chain rule, we do need to multiply by the derivative of the inner stuff. Sometimes the derivative of a X squared. Speak. Okay, so you&#x27;re three. Nothing changes in this beginning. Now let&#x27;s take the derivative. The derivative of X squared, plus B, Remember A and B or just Constance. So the derivative of a X squared that&#x27;ll just be two times a times X, the derivative of bees. Just zero. So that is it right there. We can simplify maybe a little bit by combining this three with the number at the end. So look, att. Two times three is six times a times X times. Everything in that parentheses squared. And that is our final answer. Hopefully this.

Yaw Asomani · Answer

So who I remember from a previous just previous, immediately previous tutorial they want We just ended before this one. He said that a differential operator is a linear operator. And so we have a function. We has more than one term. We can actually differentiate each term on a term by term basis. Right. So you said the same thing here. You have two terms. You have this one that you have that one, right? So it gives in a find a derivative, uh, of each one in isolation. Right. So what is the derivative off? Just this one, the first term? Well, from previous tutorial, you know, that is four is gonna drop here and multiply. The three says he&#x27;s gonna be 12 right? And t. And they take away one from, uh, the power. The power is four. So four minus one is just three. Then you&#x27;re minus. Sign is gonna be here. Now, you deal with this one too, uh, in isolation. Right. Because of the linearity off the differential operator, which we explain in a period. Vittoria. So, uh uh So this one, it&#x27;s so gonna follow on from what we&#x27;ve been doing, the power here is gonna multiply the constant here. So two times two is gonna be four, right? And they bring your tea, and then when you take away one from this power to hear two minus one is just one. So just teaching. Our one said this becomes the derivative off those function, all right?

Problem

Find the derivative. Assume $a, b, c, k$ are cons…

Problem 15 Easy Difficulty

Find the derivative. Assume $a, b, c, k$ are constants.
$$y=-3 x^{4}-4 x^{3}-6 x$$

Answer

$$-12 x^{3}-12 x^{2}-6$$

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Deborah Hughes-Hallett, Patti Frazer Lock, Andrew M. Gleason

Applied Calculus

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Anna Marie V.

Caleb E.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$$-12 x^{3}-12 x^{2}-6$$

Applied Calculus

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

Deborah Hughes-Hallett, Patti Frazer Lock, Andrew M. GleasonApplied Calculus

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

Deborah Hughes-Hallett, Patti Frazer Lock, Andrew M. Gleason

Applied Calculus