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Thread: First Year Mathematics A (Differentiation & Linear Algebra)

  1. #551
    Executive Member matchalolz's Avatar
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    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)
    .

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    Re: MATH1131 help thread

    Quote Originally Posted by matchalolz View Post
    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)


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    Executive Member matchalolz's Avatar
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    Re: MATH1131 help thread

    Quote Originally Posted by InteGrand View Post


    Sweet! such a lifesaver
    .

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    Ancient Orator leehuan's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    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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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by leehuan View Post
    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)
    Last edited by InteGrand; 18 Jun 2017 at 9:48 PM.
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    Executive Member matchalolz's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    I'm back for another semester of struggling through math

    Screen Shot 2017-07-29 at 3.57.58 pm.png

    Can someone explain this?

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

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    I love trials pikachu975's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by matchalolz View Post
    I'm back for another semester of struggling through math

    Screen Shot 2017-07-29 at 3.57.58 pm.png

    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
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    Junior Member Sp3ctre's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by matchalolz View Post
    I'm back for another semester of struggling through math

    Screen Shot 2017-07-29 at 3.57.58 pm.png

    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
    Last edited by Sp3ctre; 29 Jul 2017 at 10:00 PM.
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by Sp3ctre View Post
    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.
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    Junior Member Sp3ctre's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by He-Mann View Post
    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
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    Executive Member matchalolz's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by He-Mann View Post
    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.
    Yeah pretty much I don't even have the basics down

    I think I'll probably ask someone in real life because I'm so clueless rip
    .

  12. #562
    I love trials pikachu975's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by matchalolz View Post
    Yeah pretty much I don't even have the basics down

    I think I'll probably ask someone in real life because I'm so clueless rip
    Read up about the chain rule, it's fundamental to these types of questions.

    E.g. dy/dx = dy/dt * dt/dx
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  13. #563
    Junior Member Sp3ctre's Avatar
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by matchalolz View Post
    Yeah pretty much I don't even have the basics down

    I think I'll probably ask someone in real life because I'm so clueless rip
    This video could be useful, it covers the very basics of the chain rule.

    https://www.khanacademy.org/math/ap-...e-introduction
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Quote Originally Posted by matchalolz View Post
    Yeah pretty much I don't even have the basics down

    I think I'll probably ask someone in real life because I'm so clueless rip
    To be completely honest, most HSC students who come across similar questions apply the chain rule along with the formula of the differentiation of inverse tan, without knowing where central formula actually comes from. You should probably see it at least once...

    I'll offer an elementary derivation of the differentiation of inverse tan formula. I'll denote inverse tan as arctan.

    Okay, so we want to show that d(arctan(x))/dx = 1/(1+x^2).

    First, let y = arctan(x). Since the inverse tan is, well, an inverse, we have that x = tan(y).

    Differentiating with respect to y, we have that dx/dy = dtan(y)/dy = sec^2(y). The last equality (that differentiating tan(y) gives you sec^2(y)) is also something that HSC students normally memorise. If you want to see it, take tan(y) = sin(y)/cos(y) and apply the quotient rule.

    Now you need to note that sin^2(y) + cos^2(y) = 1. This is a famous identity that can be proven in quite a lot of ways (for example, considering aright-angled triangle made by a radius of a unit circle and the axes, and applying Pythagoras' Theorem). If we divide both sides by cos^2(y), you get tan^2(y) + 1 = sec^2(y). Substituting this into dx/dy = sec^2(y), we get that
    dx/dy = 1 + tan^2(y).

    Now remember that $x = tan^2(y)$, so dx/dy = 1 + x^2. And so finally, by taking the reciprocal, we get the formula that darctan(y)/dx =1/(1 + x^2).

    Now in your problem, we need to replace our y with 2/y. This complicates things a bit, because now we need to apply the chain rule, which pikachu975 has stated as dz/dx = dz/dt * dt/dx, where z = arctan(t) and t = 2/y.

    So dz/dx = darctan(t)/dt * d(2/y)/dy.

    From here, apply the inverse tan differentiation we have derived to the first term (you get 1/(1+t^2)), and differentiate the second term (you get -2/y^2). Then substitute $t=2/y$ back in to get your answer.

    Hope this helps.
    Last edited by sida1049; 30 Jul 2017 at 12:37 AM.
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  15. #565
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    Re: First Year Mathematics A (Differentiation & Linear Algebra)

    Omg I get it, thanks so much everyone
    .

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