Math

[Guest]

 

 

y = (5x^3 -7x -1) / x^2 

 

I'm not sure if you've learnt this but, when we have y = ax^n, ∴ dx/dy = anx^n -1. You'll need it later on.

 

We can "break" the denominator from the equation and simply multiply it instead. Like so:

 

 1/x^2 (5x^3 -7x -1) / x^2.

 

Assuming you know the laws of indices, the 1/x^2 can be written as x^-2. We use that instead of the unpleasant fraction.

 

x^-2 (5x^3 -7x -1), Now we simply expand by multiplying the outside by the inside. This gives us: 

 

5x -7x^-1 -x^ -2. 

 

 

I guess u r wrong here 

 

1/x^2 (5x^3-7x-1) / x^2

=> x^-2 (5x^3-7x-1) x^-2

=> x^-4 (5x^3-7x-1)

=>5x^-1-7x^-3-x^-4

Derivative=>-5x^-2+21x^-4+4x^-5

 

Thnx btw  :)

Edited July 25, 2018 by Quasar

[Chronicle 164]

I guess u r wrong here 

 

1/x^2 (5x^3-7x-1) / x^2

=> x^-2 (5x^3-7x-1) x^-2

=> x^-4 (5x^3-7x-1)

=>5x^-1-7x^-3-x^-4

Derivative=>-5x^-2+21x^-4+4x^-5

 

Thnx btw  :)

*Facepalm* I'm afraid you are wrong actually.... 

 

Edit: This would make me look twice as bad if it actually turns out that I'm wrong... I'll double check my answers

Edited July 25, 2018 by Chronicle

[Guest]

*Facepalm* I'm afraid you are wrong actually.... 

 

Edit: This would make me look twice as bad if it actually turns out that I'm wrong... I'll double check my answers

welp.. How?

[LOLKILLERTOTHEDEATH 1,921]

*Facepalm* I'm afraid you are wrong actually.... 

 

Edit: This would make me look twice as bad if it actually turns out that I'm wrong... I'll double check my answers

Did he use the product rule? IT doesn't seem like it, which is why I ask.

[Chronicle 164]

No, turns out my answer was correct... simple mistake on his end. 

 

(5x^3-7x-1) / x^2 

 

The x^2 is the denominator, if we break it off of the (5x^3-7x-1) we end up with 1/x^2. We can't simply take it away, it remains in the system, and it would look like this once we break the denominator, 1/x^2 (5x^3-7x-1). 1/x^2 turns out to be x^-2 using the law of indices, it simply makes it easier for us to expand with a whole integer (with a negative power) rather than a fraction. So we transform our fraction into x^-2. Leaving us with x^-2 (5x^3-7x-1), which has the same value as 1/x^2 (5x^3-7x-1) and (5x^3-7x-1) / x^2, just broken down. From there we expand, use the y = ax^n, ∴ dx/dy = anx^n -1 rule and simplify if needed.

[Royalworld 1,269]

 

We can "break" the denominator from the equation and simply multiply it instead. Like so:

 

 1/x^2 (5x^3 -7x -1) / x^2.

The method is correct but here when you write you are multiplying 1/x^2 to the original term you should remove the /x^2 . It looks like you've made it (5x^3 -7x -1) / x^4.

If I was learning, I would've lost you at this step. 

[Chronicle 164]

The method is correct but here when you write you are multiplying 1/x^2 to the original term you should remove the /x^2 . It looks like you've made it (5x^3 -7x -1) / x^4.

If I was learning, I would've lost you at this step. 

I have to leave it as a fraction, I can't remove x^2 from 1/x^2 unless I multiply it by the negative reciprocal which isn't happening I think. I wasn't multiplying 1/x^2 by (5x^3 -7x -1), I was showing you what it would look like if x^2 wasn't the denominator, if I did multiply it we would end up with the same format as before. It's all part of the learning process, if I had a paper and a pen and you were sat next to me, explaining it would be much easier since I can't do the fraction type writing on this platform. Can I?

[Royalworld 1,269]

I have to leave it as a fraction, I can't remove x^2 from 1/x^2 unless I multiply it by the negative reciprocal which isn't happening I think. I wasn't multiplying 1/x^2 by (5x^3 -7x -1), I was showing you what it would look like if x^2 wasn't the denominator, if I did multiply it we would end up with the same format as before. It's all part of the learning process, if I had a paper and a pen and you were sat next to me, explaining it would be much easier since I can't do the fraction type writing on this platform. Can I?

Lets use a simple term: x/2

Either write it as (1/2)*x or just x/2 not as (1/2)*x/2 because that makes it x/4

You multiplied with the negative reciprocal and still didn't remove the fraction. Look at what I quoted earlier, it is technically equal to (5x^3 -7x -1) / x^4. 

And yes, explaining online is harder than explaining on paper. 

 

[Chronicle 164]

Lets use a simple term: x/2

Either write it as (1/2)*x or just x/2 not as (1/2)*x/2 because that makes it x/4

You multiplied with the negative reciprocal and still didn't remove the fraction. Look at what I quoted earlier, it is technically equal to (5x^3 -7x -1) / x^4. 

And yes, explaining online is harder than explaining on paper. 

 

​I didn't actually multiply with the negative reciprocal, I used the laws of indices to turn 1/x^2 into x^-2. I moved the 1/x^2 from the right side so that it's in front of the bracket, I can leave it on the right hand side if that makes it easier.

the writing you quoted isn't in my explanation, either I'm not seeing it or it isn't there to begin with.

 

Edit: My mistake, I've just spotted it. I wrote it wrong that's all.

 

Should have looked like this:

 

(5x^3 -7x -1) / x^2 ---To--->  1/x^2 (5x^3 -7x -1). 

Edited July 25, 2018 by Chronicle

[Royalworld 1,269]

Should have looked like this:

(5x^3 -7x -1) / x^2 ---To--->  1/x^2 (5x^3 -7x -1). 

I was just pointing this out. :p 

[Chronicle 164]

I was just pointing this out. :P

And I'm appreciative, so thank you!  

[LOLKILLERTOTHEDEATH 1,921]

I was wondering if there is or there isn't any relation, but does the derivative of a function in any way relate to the the roots or solutions of the original function? I know the roots/solutions to the derivative represent turning points for any given function, but is it possible to use the derivative of a function to calculate the roots?

 

I feel like there isn't, but I don't really know much about calculus compared to many other people. 

[Total_SkiIl 1,983]

this is eye-cancer

[ThirdOnion 4,898]

I was wondering if there is or there isn't any relation, but does the derivative of a function in any way relate to the the roots or solutions of the original function? I know the roots/solutions to the derivative represent turning points for any given function, but is it possible to use the derivative of a function to calculate the roots?

You can, to an extent (see Newton's method), but the question is why? There are different methods more specifically tailored specifically for that purpose.

Edited August 3, 2018 by ThirdOnion

[LOLKILLERTOTHEDEATH 1,921]

You can, to an extent (see Newton's method), but the question is why? There are different methods more specifically tailored specifically for that purpose.

I see. I don't understand that right now, but maybe I will in some time. What is the easiest method to calculate the roots of a function which is neither linear nor quadratic?

[Total_SkiIl 1,983]

I see. I don't understand that right now, but maybe I will in some time. What is the easiest method to calculate the roots of a function which is neither linear nor quadratic?

solve it

---

btw curious if anyone knows how to calculate all the zeroes of a irrational polynomial with some of the answers in roots of a given fraction

 

this thing hurts my eyes the most

Edited August 7, 2018 by Total_SkiIl

[LOLKILLERTOTHEDEATH 1,921]

solve it

---

btw curious if anyone knows how to calculate all the zeroes of a irrational polynomial with some of the answers in roots of a given fraction

 

this thing hurts my eyes the most

Isn't it very difficult to do without a Graphing calculator/desmos? You could try the factor theorem, but that isn't the fastest way.

[Total_SkiIl 1,983]

Isn't it very difficult to do without a Graphing calculator/desmos? You could try the factor theorem, but that isn't the fastest way.

Somehow, we have to do that in exams. .-.

 

Wouldn't the reciprocals also be zeroes of that polynomial?

[LOLKILLERTOTHEDEATH 1,921]

Somehow, we have to do that in exams. .-.

 

Wouldn't the reciprocals also be zeroes of that polynomial?

Reciprocals? Correct me if I'm wrong, but do you mean with equations like "1 + (1/x) = 0"? I'd say working these kind of solutions is somewhat easy. The one that I'm worried about is " x⁴ + 27x³ + 17x² + 115x + 910" where I don't understand what to do. Naturally, the factor theorem is not going to be of much help.

[Total_SkiIl 1,983]

Reciprocals? Correct me if I'm wrong, but do you mean with equations like "1 + (1/x) = 0"? I'd say working these kind of solutions is somewhat easy. The one that I'm worried about is " x⁴ + 27x³ + 17x² + 115x + 910" where I don't understand what to do. Naturally, the factor theorem is not going to be of much help.

no, the numerator and denominator turned upside down

[LOLKILLERTOTHEDEATH 1,921]

no, the numerator and denominator turned upside down

Could you give me an example of what you mean?

 

Edit: I think I may have misspoken. I do understand what a reciprocal is, but just not what kind of equation you refer to.

[Guest]

Does your maths syllabus require you to find the roots of degree 4 polynomials or are you just interested in knowing a method? 
Also did you make up " x4 + 27x3 + 17x2 + 115x + 910" yourself or is it a question you are given in a textbook or something?

[LOLKILLERTOTHEDEATH 1,921]

Does your maths syllabus require you to find the roots of degree 4 polynomials or are you just interested in knowing a method? 

Also did you make up " x4 + 27x3 + 17x2 + 115x + 910" yourself or is it a question you are given in a textbook or something?

Not required, I just want to know if there is another way. I made up the question too, just to express that I mean to try to understand complicated polynomials.

[giant05 207]

I hate that topic name :/

[LOLKILLERTOTHEDEATH 1,921]

I hate that topic name :/

Do you prefer Mathematics to Math? Or perhaps it is Maths you prefer.