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

#### InteGrand

##### Well-Known Member

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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#### matchalolz

##### Well-Known Member

I'm not following how we went from the first step to the second step

#### leehuan

##### Well-Known Member

I'm not following how we went from the first step to the second step

View attachment 33967
$\bg_white \text{Their line one is wrong}\\ \text{It should read }4(z\overline{z}+3i(z-\overline{z})+9)=z\overline{z}\boxed{-3i(z-\overline{z})}+9$

#### matchalolz

##### Well-Known Member

$\bg_white \text{Their line one is wrong}\\ \text{It should read }4(z\overline{z}+3i(z-\overline{z})+9)=z\overline{z}\boxed{-3i(z-\overline{z})}+9$
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

$\bg_white \text{Using the identity they supplied}$
\bg_white \begin{align*}|z+3i|^2 &= (z+3i)\overline{(z+3i)}\\ &= (z+3i)(\overline{z}-3i)\\ &= z\overline{z} + 3i\overline{z} - 3iz - 9i^2\\ &= z\overline{z} - 3i(z-\overline{z}) + 9\end{align*}

#### matchalolz

##### Well-Known Member

$\bg_white \text{Using the identity they supplied}$
\bg_white \begin{align*}|z+3i|^2 &= (z+3i)\overline{(z+3i)}\\ &= (z+3i)(\overline{z}-3i)\\ &= z\overline{z} + 3i\overline{z} - 3iz - 9i^2\\ &= z\overline{z} - 3i(z-\overline{z}) + 9\end{align*}
oh I see ty c:

#### matchalolz

##### Well-Known Member

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

Rip finals

#### InteGrand

##### Well-Known Member

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

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

why do we need to do this? can you elaborate on this please?
$\bg_white \noindent In other words, let F(u) = \int_{0}^{u} \cos \left(t^{2}\right)\, \mathrm{d}t. Then we are asked to find \frac{\mathrm{d}y}{\mathrm{d}x}, where y = \int_{0}^{x^{3}} \cos \left(t^{2}\right) \, \mathrm{d}t = F\left(x^{3}\right). You should be able to get the answer via chain rule and Fundamental Theorem of Calculus from here. (Remember in the FTC, the upper limit of the integral is the same as the variable with respect to which we are differentiating.)$

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#### matchalolz

##### Well-Known Member

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

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)
$\bg_white \noindent Right idea, but the chain rule calculation is \frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}F'\left(x^{3}\right) (rather than F'(x)).$

$\bg_white \noindent Since F'(u) = \cos \left(u^{2}\right) (by the FTC), we have F'\left(x^{3}\right) = \cos \left(\left(x^{3}\right)^{2}\right) = \cos \left(x^{6}\right) (replacing u with x^{3} in the equation F'(u) = \cos \left(u^{2}\right)).$

#### matchalolz

##### Well-Known Member

$\bg_white \noindent Right idea, but the chain rule calculation is \frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}F'\left(x^{3}\right) (rather than F'(x)).$

$\bg_white \noindent Since F'(u) = \cos \left(u^{2}\right) (by the FTC), we have F'\left(x^{3}\right) = \cos \left(\left(x^{3}\right)^{2}\right) = \cos \left(x^{6}\right) (replacing u with x^{3} in the equation F'(u) = \cos u^{2}).$
Sweet! such a lifesaver

#### leehuan

##### Well-Known Member
I know that the answer is $\bg_white \alpha=6$. 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?

$\bg_white \text{What is the largest value of }\alpha > 0\text{ for which the following improper integral diverges?}\\ I=\int_0^\infty\frac{1+x^2}{\sqrt{x^\alpha+2}}dx$

#### InteGrand

##### Well-Known Member
I know that the answer is $\bg_white \alpha=6$. 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?

$\bg_white \text{What is the largest value of }\alpha > 0\text{ for which the following improper integral diverges?}\\ I=\int_0^\infty\frac{1+x^2}{\sqrt{x^\alpha+2}}dx$

$\bg_white \noindent We can show this using a limit comparison test and the p-test. \Big{(}Note that the integrand is continuous on [0,\infty) and is asymptotically equivalent to \frac{1}{x^{\frac{\alpha}{2}-2}} as x\to \infty, for \alpha > 0.\Big{)}$

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#### matchalolz

##### Well-Known Member
I'm back for another semester of struggling through math

Can someone explain this?

Sorry if this is so basic but I legit don't get it

#### pikachu975

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.

#### 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