solve-the-given-problems-by-finding-the-appropriate-derivatives-if-fx-is-a-differentiable-function-5

Question

Solve the given problems by finding the appropriate derivatives.
If $f(x)$ is a differentiable function, find an expression for the derivative of $y=x^{2} f(x)$.

Gregory Higby · Accepted Answer

So these are some classic calculus problems that actually get you to make sure that you know the product rule, because the last students have figured out how to fake the product rule and things like that. Um So to correctly do this problem again, identify where your product is so that you do the product rule correctly. And a lot of times I&#x27;ll even put in four blanks with a plus in here to help students, especially the first time they&#x27;re learning it. So when you&#x27;re tasked with finding the derivative dy dx, you have to take the derivative of one side. So I like to do the left side first times you leave the right side alone and then plus now you leave the side, you took the derivative of the X squared alone and then times the derivative of the right side or the other side. Um and we don&#x27;t know anything about f just that it&#x27;s differentiable. So we just write down the notation so this is a perfect answer. Remember additions communities. So you can, you can switch the order around if you really want to. I&#x27;m not sure why anybody would do this, but you could also, if you want to factor out an X and the problem. So you&#x27;d have two times ffx plus X times f prime of X. There&#x27;s a lot of different ways that you can write the correct answer and it&#x27;s really up to you what makes most sense to you or your teacher. So these are both correct and it utilizes the product rule.

Runpeng Li · Answer

Okay, so for problem seven were given a fraction. We chased that. The room too. Attraction is X squared. Times X minus one, divided by X plus two and X is none. X is not equal Thio negative too. Okay, So our first party in this problem is to find the critical points. Now, to this fraction, we just we only need to make our numerator to be zero. So that is acts to be zero or X is one. So these two are our Rico points now, Part B. We need to find a by the increasing the rose and increasing levels. So that is still we need to let destruction to be hostage as salt for the inequality. Now, notice that, um in this case, X X squared is always a negative. So to be to make this whole this whole direction, to be positive, our ex cannot be zero. Otherwise, our our entire direction will be zero. And at the same time, we need to make a fraction X minus one and express to to be to be positive. So that is same to say, Ex tremendous one is positive, and X plus two is positive or we can make them both connective. We stood s steel works because connective number divided by a naked number will be positive. So or we can make explain is one to be negative and x plus two to be negative. Okay. Now, by solving these sets off in qualities, what we have is ex bigger than one or sale or here or acts is smaller than connected to. So that means from negative infinity to connect it to union one to infinity dysfunction will increase and from next 2 to 1, dysfunction will decrease. Okay, Now, for part C, our local maximum will be away on. So yeah, it&#x27;s doing this problem. We don&#x27;t have any look Mexico. But we have a local minima because from connected 2 to 1 from the function will decrease now from one to infinity function will increase. So at axes, axes equal to one, we can reach the know Camino and no local maximum

Matt Just · Answer

So you&#x27;re giving the derivative of a function F Crime of X is explain this one times X plus two. And so let&#x27;s find the critical points of that. Well, these are the values of X through its theft crime of X zero Cor, undefined. But this derivatives never going to be undefined Soon X equals one and X equals negative to our critical points who just plug in one a negative to those will be the values vex to make of prime zero. Okay, And then we want to know All right, what values is f increased, their decreasing. Well, that&#x27;s just asking it. What values Andi or what Open intervals is f prime, positive or negative. So you know, the F prime will potentially change sign at the critical points. So put not just to you here and one here and no test f prime so that blogging something smaller the negative to this factor is negative. This factor will be also negative, said F prime will be positive. And then, if I plug in and value between you two and one, this factor will be negative, and this factor will be positive. So the drip it it&#x27;ll connective. And then if X is bigger than one and bigger than to both of these factors were positive. And so what that means for Earth is that ofthis increasing here, Decreasing here, increasing here. So it&#x27;s increasing on this interval. Negative infinity. It&#x27;s a negative, too decreasing between aged two and one and then increasing between one and infinity. All right, And then at what points does f assume local maximum minimum guys So it looks like here have changes from being increasing to decreasing. So hey, we have a local max at X equals negative too. And then we have a local men because of changes from decreasing to increasing at X equals one. Then a pre just sort of visualize what that looks like. It&#x27;s increasing and decreasing that increasing. So we have a local max in the community.

Runpeng Li · Answer

kept. So four problem. Three. All right, we have derivative off a function which is X minus one squared times X plus two. So similarly, uh, 1st 1st part, we need to find the, uh, critical points. So excuse me. That is to let this derivative to be equal to zero. Now, the solution will be X. What next? A sequel to one and when next is equal to negative too. Okay. No, To find the opening a rose for increasing and decreasing, we need to let the due to to be pause, do first. Now to solve this, um, to solve his inequality. The first thing we need to observe is dead X minus one squared will always be the negative. So that means we don&#x27;t allow acts to be one. And at the same time, you should add in. And so, at the same time, we need to make experts to haunted. So that is, that gives the interval, which is from, uh let&#x27;s see. Yeah, pushes for elective too. 21 Union one to infinity. And on this interval, the function will increase. And other than that, we will have the interval for decreasing, which is from negative. Infinity to elective too. Okay, now, Parsi. Now what are we going to do is to find a local maximum local animals. So it&#x27;s quite simple. We just need to find We just need to check the locally streak. Lovely stream points. But the thing to notice instead, one is not. It&#x27;s not a loco extreme. The reason is because we just exclude one from our increasing interval. And that means when x x equal to one, nothing will happen because he will keep increasing. So if one is not a local extreme, but have connected to will be a no Cole. This should be a local mutual okay and correspondent in there. There is no local man in Mexico.

Runpeng Li · Answer

Okay, So for the first problem, um, we&#x27;re given a function with the derivative, which is X times X minus one X minus one. So the first thing we need to do is to find a critical point. That is to let this derivative to be equal to zero, then our solution in this case album, it&#x27;s very obvious. Justo, he&#x27;s evil. And one or one. Okay, now the second part. What I want what I&#x27;m hoping DeVos is the act increasing or decreasing. So we just need to let this derivative to be positive and solve the inequality. So this is a parabola. We have to, uh, two zeroes. 101 This one and rebel a should be luke line should be like this. So that means when X is in 0 to 1, dysfunction is increasing. Otherwise it is decreasing. So if the the open devil for the decreasing will be a negative, you can t 20 Union 12 positive infinity now for parts. Uh, we want find a local maximum and local minimum. So that is just a function value off. I have zero and f one. These two are the local extremes.

Matt Just · Answer

okay, Silver, given this derivative F crime of X is X minus one squared times X plus, two squared and the first thing we want to do this find the critical points. Well, it&#x27;s easy to see their critical points here. I&#x27;m going to be with derivatives. Zero drift. It&#x27;s never going to be under signed X equals one and X equals negative, too. And then, secondly, we want to give the open intervals where F is increasing or decreasing. So controlling number line you made it to in one we know that is going to change from being increasing that you&#x27;re decreasing when asked prime changes from being positive to negative. So you want to see where f crime is positive and negative. So if I take a number less than negative too well, I mean, I guess it&#x27;s not so hard to see that actually have. Crime is always greater than or equal to zero because it&#x27;s the product of two things that are squared so f prime. It&#x27;s always going to be increasing unless zero so to the left and negative to F crimes going be positive between aged two and one, coming positive and in greater than one. It&#x27;s going to be positive. And so if we look at F, it&#x27;s going to be increasing from negative infinity, the native to it&#x27;s going to be increasing from negative to one and it&#x27;s going to be increasing from one to infinity. So what that means is that F actually has no local extreme because it never changes from being increasing to decreasing or decreasing to increasing no local extra see.

Problem

Solve the given problems by finding the appropria…

Problem 37 Easy Difficulty

Solve the given problems by finding the appropriate derivatives.
If $f(x)$ is a differentiable function, find an expression for the derivative of $y=x^{2} f(x)$.

Answer

$\frac{d y}{d x}=x^{2} f^{\prime}(x)+2 x f(x)$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

Precalculus Review - Intro

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$\frac{d y}{d x}=x^{2} f^{\prime}(x)+2 x f(x)$

Basic Technical Mathematics with Calculus

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Catherine R.

Kayleah T.

Kristen K.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus