# First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

#### InteGrand

##### Well-Known Member
Re: MATH1131 help thread

How do you know what function to pick out for g(x)?
When x is large, f(x) "looks like" x/(2x^3) = 1/(2x^2), because the -1 on the denominator becomes negligible. So we know we want to compare f(x) to 1/(2x^2) (or can just do it to 1/x^2, since all we need is the limit of f(x)/g(x) to be a positive constant), and this is helpful because we know the convergence status of the improper integral of 1/(x^2).

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• kawaiipotato

#### matchalolz

##### Well-Known Member
Re: MATH1131 help thread

Really?

I don't even know how we got to that line in the first place, let alone realise there was a mistake

#### leehuan

##### Well-Known Member
Re: MATH1131 help thread

#### matchalolz

##### Well-Known Member
Re: MATH1131 help thread

oh I see ty c:

#### InteGrand

##### Well-Known Member
Re: MATH1131 help thread

Why are we multiplying cos(x^6) by 3*x^2?

View attachment 33968

Rip finals
We're multiplying it by the derivative of the x^3 in the integral upper limit, because of the chain rule.

#### matchalolz

##### Well-Known Member
Re: MATH1131 help thread

We're multiplying it by the derivative of the x^3 in the integral upper limit, because of the chain rule.
why do we need to do this? can you elaborate on this please?

#### InteGrand

##### Well-Known Member
Re: MATH1131 help thread

why do we need to do this? can you elaborate on this please?

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#### matchalolz

##### Well-Known Member
Re: MATH1131 help thread

is it like y = F(x^3)

and because F(x^3) is a function of a function, to differentiate y with respect to x you go (3*x^2)*F'(x), where F'(x) = f(x) = cos(x^6)

#### InteGrand

##### Well-Known Member
Re: MATH1131 help thread

is it like y = F(x^3)

and because F(x^3) is a function of a function, to differentiate y with respect to x you go (3*x^2)*F'(x), where F'(x) = f(x) = cos(x^6)

• matchalolz

#### matchalolz

##### Well-Known Member
Re: MATH1131 help thread

Sweet! such a lifesaver

#### leehuan

##### Well-Known Member
I know that the answer is . But I don't know how to properly set out a logical answer because they wanted us to explicitly state any tests we use and how we used it. So I was just looking for some guidance on how I can set out a response?

(I don't remember if I already asked this question either)

• kawaiipotato

#### InteGrand

##### Well-Known Member
I know that the answer is . But I don't know how to properly set out a logical answer because they wanted us to explicitly state any tests we use and how we used it. So I was just looking for some guidance on how I can set out a response?

(I don't remember if I already asked this question either)

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• kawaiipotato and leehuan

#### pikachu975

##### Premium Member
I'm back for another semester of struggling through math

View attachment 34159

Can someone explain this?

Sorry if this is so basic but I legit don't get it
Basically use chain rule so d(tan^-1 (2/y))/d(2/y) * d(2/y)/dy = (-2/y^2) / (1+(2/y)^2) and simplify that

#### Sp3ctre

##### Active Member
I'm back for another semester of struggling through math

View attachment 34159

Can someone explain this?

Sorry if this is so basic but I legit don't get it
You could also use the the differentiation: d(tan^-1(x/a))/dx = a/(a^2+x^2).

Although this method is easier when you're differentiating something in the form tan^-1(x/a), you can treat a as 1/2 and you will get:

-1/2/(1/2)^2 + x^2

= -1/2/(1/4 + x^2)

Times numerator and denominator by 4:

= -2/(1+4x^2)

Edit: As suggested by He-Mann, although this method can be significantly easier to differentiate certain functions, it is best avoided if you do not have a conceptual understanding of differentiation of inverse trig functions since you are just plugging in values into a formula

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#### He-Mann

##### Vexed?
You could also use the the differentiation: d(tan^-1(x/a))/dx = a/(a^2+x^2).

Although this method is easier when you're differentiating something in the form tan^-1(x/a), you can treat a as 1/2 and you will get:

-1/2/(1/2)^2 + x^2

= -1/2/(1/4 + x^2)

Times numerator and denominator by 4:

= -2/(1+4x^2)
This is just mindlessly plugging into formulas which should be avoided at all costs for this case. matchalolz does not have a good idea of how to do this fundamental question so giving formulas without explanation/proof is worthless for the purposes of learning which is top priority in this case.

Whenever you make a claim, it is best to provide proof because the reader does not necessarily believe whatever you say unless it's trivial.

• pikachu975

#### Sp3ctre

##### Active Member
This is just mindlessly plugging into formulas which should be avoided at all costs for this case. matchalolz does not have a good idea of how to do this fundamental question so giving formulas without explanation/proof is worthless for the purposes of learning which is top priority in this case.

Whenever you make a claim, it is best to provide proof because the reader does not necessarily believe whatever you say unless it's trivial.
You're right, sorry. I don't know why but I thought for a second that I could somehow help by giving an alternate solution and thought that it would add to his knowledge of how to tackle those types of questions. Apologies for my stupidity 