find-fprimex-fxcsc-x-cot-x

Question

Find $f^{\prime}(x)$
$$
f(x)=\csc x \cot x
$$

Prakash Hampole · Accepted Answer

F Prime of X is the derivative f of X, so it&#x27;s equal to the derivative of coast. He can&#x27;t x Times co tenant does. And, um, because this multiplying two functions so weaken use the product rule, which tells us this is equal to the derivative of co seeking Tex 10 times coat and index plus Cosi context times the derivative of CO tange Index. So on the next line, in place of the derivative of co seek index weekend right negative co seek and X co tangent ticks and then we have another go tanning tax from the previous line and we&#x27;ll add and I&#x27;m gonna move everything over. So yet more space. Now we have cosi contexts times the derivative of CO Tansi Index, which is negative, of course, Eakins Ah, squared picks. And lastly, we can use exploding laws to combine this with this and this with this. So our final answer is negative. Co second x co tensions squared. Okay, co tangent squared X, um minus. And I think I&#x27;m just taking this minus sign to the outside here who have minus ako seek and Cubed X. That&#x27;s our finalists here for crying of X

Allan Hungria · Answer

that looks is equal two x to the negatives, too. Just look for a prime for that. I think the negative, too. We put it in front. It&#x27;s attract one. You get negative two extra negative three.

Alex Roush · Answer

okay. Sora task in this problem is to find the 1st 2nd and third derivatives. So it&#x27;s exactly as it sounds. We all know to take the first derivative. I&#x27;m assuming so I&#x27;m gonna rewrite this. Of course I can flip this as long as I change the exponents negate it. So one over X becomes X to the negative first power and then the first derivative is just an application of the power rule. Bringing this exponents down is a coefficient in decreasing the exploded by one. The second derivative is gonna be the exact same thing. It&#x27;s just now we&#x27;re differentiating this function f prime of axe. So again, I mean, use the power rule to take negative one times negative two and change the sign. This is very important in this is a positive two times X now raised to the negative third power. Which means that third derivative third derivative would be the derivative of the second derivative. Now, differentiating this function and with the power rule, we again bring this down. Change the sign toe a negative. This will be a negative six x to the negative fourth power

Amy Jiang · Answer

so we&#x27;re us or were given that our function is equal to one over X so we can get that as X a grenade one and I&#x27;ll take the first derivative. Don&#x27;t you go to negative one extra crowd of native to and in our second derivative, we&#x27;ll bring down or, in the year two times negative one. That&#x27;s two extra apartment to minus one to its way three and in order dripping negative three times to That&#x27;s making six exit cognitive for

Prakash Hampole · Answer

We&#x27;re looking for F Prime Ness, which is the derivative of ffx. So is the derivative of this function. So notice we have one function divided by another function so we can use kosher, which tells us this is equal to the derivative of the top and that&#x27;s multiplied by the bottom. Then we subtract and take the top once played by the river of the bottom and that&#x27;s all over the bottom squared. So first you&#x27;ll input those derivatives derivative of co tangent is negative co seek and square. I mean the derivative of one Klesko second ex. The one can be ignored because we&#x27;re taking the derivative and constant. So we have up negative Cosi Contexts, co Tangent X and the denominator can see the same now. Ah, it will be helpful, Teoh. The common factor out of this expression and that common factor would be co seek and X, so Oh, and firstly, of course, these two negatives cancel out So across one of the mouth, the other when we come to us and we will factor out co seek in. So we have post seek index. Then we have from the first term negative coast. He can&#x27;t x times one plus co seek and ah x And then in the next part, we have plus co tangent Times co attention which is just contention squared and the denominator can say the same. So this is, um, this is equal to the derogative. However, we can, of course, simplify the stop more. So, um, firstly, it&#x27;ll be helpful to expand these brackets. So we have, um Yeah, you have as follows. First negative co seeking X and then minus co. Second squared X on and I can keep that co tangent Square decks No. Now remember, it could be derived from the Pythagorean identity that, uh, co Seacon cose Eakins squared. You do one plus contention squared. And, um, we&#x27;re going use that so it&#x27;s gonna be equal to Well, first, it will be helpful to factor to take this negatives These negative science city outside. I like this. We&#x27;re gonna beef acting on a negative one. Essentially, we have negative Kaseke, an accent, and first we have co seeking Tax co seek and Squared X, then for factoring out and they give and this is gonna have to each is gonna have to undergo a sign change. So this will come minus co tangent Squared X. And again that denominator could see the same. Now, um, the reason that a fractured out this negative so that we can actually use this identity. First, we need to rearrange it and show that so we&#x27;re gonna subtract proceeding squared from both sides, which tells us that wine is equal to We&#x27;re sorry. Um, we&#x27;ll subtract co Tangin squared from both sides. Which shows us that, uh, my co ck in squared minus coat and in squared is equal to one. And ah, notice we have co Seacon squared minus containing squared. So this just one. So what we&#x27;re left with is after a bit busy during it is equal to this&#x27;ll. Some brackets, we have one plus, um, go second. It&#x27;s this coast he connects, and that denominators stays the same. And look at that. We have co seek one close co CEO Connex in the denominator and the numerator. So making cancel out like this usually those with our final answer that f prime of X is equal to negative co seeking and X over one plus go seek and x

David Mccaslin · Answer

So in this problem we are asked to find f prime at a when half of x is X to the -2. Well, by definition, if private A Is the limit as H approaches zero of F. Of a plus H minus F of a over H. All right. So that means that this is now the limit as h approaches zero F of a plus age and in this function F of X. So that is one over a plus H squared minus one over a squared all that is over eight. All right. So that means I have a limit as H goes to zero then of what? When I get a common denominator in the top. All right. And so I&#x27;ll have a squared coming on top and then combine the fractions. That means I&#x27;ll have a squared minus a squared plus two. A H plus H squared all over a squared times A plus H squared times H from the from the denominator previously. All right. So like this a squared minus at a square is but those two are gonna be gone, aren&#x27;t they? So that means I have the limit as h goes to zero of minus to a H plus H squared over a squared times A plus h squared H. But I have an H in each term in the numerator so I can cancel one of those out. Can&#x27;t I like that? So, that means I now have the limit As a joke goes to zero of minus to a plus H over a squared times A plus H squared. Okay, so taking now the limit as H goes to zero in the numerator, I&#x27;m just left with minus two way, aren&#x27;t I? And the denominator. I have a squared times A squared, right? Because H goes to zero. So A plus H just goes to a square that let&#x27;s just a square. So that means I&#x27;m left with -2 A over eight of the 4th. Well I can cancel 1, 1 of those AIDS out. and so that means I&#x27;m left with -2 over a cute is our derivative If prime it ain&#x27;t.

Problem

Find $f^{\prime}(x)$ $$ f(x)=\frac{\cot x}{1+\csc…

Problem 12 Easy Difficulty

Find $f^{\prime}(x)$
$$
f(x)=\csc x \cot x
$$

Answer

$$f^{\prime}(x)=-\csc x \cot ^{2} x-\csc ^{3} x$$

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$$f^{\prime}(x)=-\csc x \cot ^{2} x-\csc ^{3} x$$

Calculus Early Transcendentals

Kayleah T.

Caleb E.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Howard Anton, Irl Bivens, Stephen DavisCalculus Early Transcendentals

Kayleah T.

Caleb E.

Kristen K.

Joseph L.

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals