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    Hi there
    I've been asked by a friend of mine to help with his maths work (he's an engineer apprentice and needs to do maths to get his job)

    I've just completed A Level maths with Edexcel in 12 months so have a good understanding. However....

    The 1 question (with 3 parts) that i cannot do is to do with 'second derivatives'. Now i think on edexcel this is the same as d (squared) y / dx(squared) unless i'm mistaken. Problem is that on the edexcel syllabus there is nothing on the type of question i have here

    Find the second derivatives of the following functions
    a) y = ln (x^2 + 1)

    b) y = arctan x

    c) y = arccoth x

    Now when i did the maths with Edexcel the only application we did with arc etc was for graphs. In short, i just don't know what they are asking for and what 'second derivatives' are. Really stuck on this and need help asap.

    Thanks for reading. Any help would be appreciated by both of us.
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    To find a second derivative just differentiate the first derivative.

    eg
    y=4x^3

    \frac{dy}{dx}=12x^2

    \frac{d^2y}{dx^2}=24x


    For y=arctan(x) start by changing it to tan y = x.
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    To get the 2nd derivative, you just differentiate twice.

    If you don't know how to differentiate arctan x, arccoth x, life will be a little bit harder.
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    A second derivative of a function y of x is, by definition (not just 'on Edexcel'), \dfrac{d^2y}{dx^2}.

    For (a) I'd use the chain rule with u = x^2 + 1 and y = \ln u, then you get \dfrac{du}{dx} = 2x and \dfrac{dy}{du} = \dfrac{1}{u} = \dfrac{1}{x^2 + 1}, so \dfrac{dy}{dx} = \dfrac{dy}{du}\dfrac{du}{dx} = \dfrac{2x}{x^2 + 1}. You can then differentiate that again using the quotient rule to find the second derivative.

    For (b) you could use a substitution x = \tan u (so that y = u) and then find \dfrac{dx}{du} and \dfrac{dy}{du}. Then using the identity 1 + \tan^2 u = \sec^2 u and \dfrac{du}{dx} = \dfrac{1}{\frac{dx}{du}} and then a similar method to above, you can find \dfrac{dy}{dx}. Then it's just a case of differentiating again.

    The same goes for (c), except you can use x = \coth u.

    Alternatively, for finding the first derivatives, you can just use some standard results (if they are given in a formula book or whatever).
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    Hi guys
    Thanks to noudai's help (or nuodai sorry!) i did question a). However the y=arctanx and y = arccothx have stumped me. I did Edexcel Maths but just do not remember doing anything like those two.

    I've tried substituting x = tan u etc but just can't get anywhere, thinking why does arc disappear, what happens to tan x etc. I've looked through all my books looking for a hint on how to tackle them but just can't get far. I've done the chain then quotient rule but it just stops after one or two steps.

    I've scanned them in now to make it a bit clearer. Its 2 b) and c).

    Once again, thanks for all the help so far, and any further help. Its life-saving!
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    (Original post by sean kefw)
    Hi guys
    Thanks to noudai's help (or nuodai sorry!) i did question a). However the y=arctanx and y = arccothx have stumped me. I did Edexcel Maths but just do not remember doing anything like those two.

    I've tried substituting x = tan u etc but just can't get anywhere, thinking why does arc disappear, what happens to tan x etc. I've looked through all my books looking for a hint on how to tackle them but just can't get far. I've done the chain then quotient rule but it just stops after one or two steps.

    I've scanned them in now to make it a bit clearer. Its 2 b) and c).

    Once again, thanks for all the help so far, and any further help. Its life-saving!
    personally, i'd do it implicitly first.
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    y = arctan x
    tan y = x

    Differentiate both sides (doing the left side implicitly). Does this help at all?
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    **reaches for dictionary to find definition of implicitly**

    so tan y = x

    differentiating then this gives sec^2 y = 1

    so 1 + tan^2 y = 1

    is that right so far? wouldn't that then give tan^2 y = 0?

    EDIT: How does y = arctan x become tan y = x???
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    (Original post by sean kefw)
    **reaches for dictionary to find definition of implicitly**

    so tan y = x

    differentiating then this gives sec^2 y = 1

    so 1 + tan^2 y = 1

    is that right so far? wouldn't that then give tan^2 y = 0?

    EDIT: How does y = arctan x become tan y = x???
    err. almost

    differentiating gives sec^2y.dy/dx=1 rearange for dy/dx in terms of x then differentiate again

    y=arctanx
    take tan of both sides
    so tany=tan(arctanx)=x
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    "differentiating gives sec^2y.dy/dx=1 rearange for dy/dx in terms of x then differentiate again"

    is that sec^2y (multiplied by) dy/dx = 1 ???

    that has thrown me a bit because....what is the value of dy/dx??

    am i supposed to get something like x = dy/dx * 1??
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    (Original post by sean kefw)
    "differentiating gives sec^2y.dy/dx=1 rearange for dy/dx in terms of x then differentiate again"

    is that sec^2y (multiplied by) dy/dx = 1 ???

    that has thrown me a bit because....what is the value of dy/dx??

    am i supposed to get something like x = dy/dx * 1??
    it does mean multiplied.

    you're supposed to get something like dy/dx=1/sec^2 y which you can then rearrange for x then differentiate again (normally). You say you've just completed A level maths yet you're struggling with these standard questions...
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    sec^2y.dy/dx = 1

    So solve for dy/dx (ie divide by sec^2y):

    dy/dx = 1/sec^2y

    As you said sec^2y = 1 + tan^2y so this becomes

    dy/dx = 1/(1 + tan^2y)

    What have we defined tan y to be earlier?
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    (Original post by around)
    sec^2y.dy/dx = 1

    So solve for dy/dx (ie divide by sec^2y):

    dy/dx = 1/sec^2y

    As you said sec^2y = 1 + tan^2y so this becomes

    dy/dx = 1/(1 + tan^2y)

    What have we defined tan y to be earlier?
    little bit too much information maybe...
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    ah thought so but im wondering if i have

    dy/dx = 1 / sec^2y

    how do i fit x= into that?

    I'm guessing this gives like dy/dx = 1/ 1 + tan^2 y (or tan y can't remember now)

    then dy/dx = 1 / 1 + x
    clear fractions (?)

    so dy/dx * 1 + x = 1 + x

    dy/dx = 1 is that my answer??
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    (Original post by sean kefw)
    ah thought so but im wondering if i have

    dy/dx = 1 / sec^2y

    how do i fit x= into that?

    I'm guessing this gives like dy/dx = 1/ 1 + tan^2 y (or tan y can't remember now)

    then dy/dx = 1 / 1 + x
    clear fractions (?)

    so dy/dx * 1 + x = 1 + x

    dy/dx = 1 is that my answer??
    where do you keep getting these random results from.

    think about how you go from one line to the next as apposed to wildly guessing everything is equal to one.

    see the post 3 above for a better idea on what you should be aiming for.
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    "dy/dx = 1/(1 + tan^2y)" is copied from the suggested post.

    thats what i got though. If i knew exactly what to do i wouldn't ask. But dy/dx = 1 / 1 + tan^2 y

    surely gives

    dy/dx * 1 + x (or x^2) = 1 + x (or x^2)

    i'm not too sure if its 1 + tan y = sec ^2 y or 1 + tan ^2 y = sec ^ 2 y
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    (Original post by sean kefw)
    "dy/dx = 1/(1 + tan^2y)" is copied from the suggested post.

    thats what i got though. If i knew exactly what to do i wouldn't ask. But dy/dx = 1 / 1 + tan^2 y

    surely gives

    dy/dx * 1 + x (or x^2) = 1 + x (or x^2)

    i'm not too sure if its 1 + tan y = sec ^2 y or 1 + tan ^2 y = sec ^ 2 y
    right. maybe writing things out properly will help.

    we have \frac{dy}{dx}=\frac{1}{1+tan^2y} correct? Now, earlier on we said that tany=x so what happens if we square both sides?

    On a side note, we have the identity 1=cos^2x+sin^2x try dividing both sides of the identity by cos^2x and see what you get.
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    so we get tan^2 y = x^2 is that what you are asking for?

    then next stage is dy/dx = 1 / 1 + x^2?
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    (Original post by sean kefw)
    so we get tan^2 y = x^2 is that what you are asking for?

    then next stage is dy/dx = 1 / 1 + x^2?
    Yes, as  \tan y = x , so simply squaring,  \tan^2 y = x^2 .

    But since  \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sec^2 y} = \dfrac{1}{1 + \tan^2 y} , then  \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1+x^2}

    Now you just have to differentiate this again, to get the second derivative. You can do this using quotient rule:

     \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{f(x)}{g(x)} \implies \dfrac{\mathrm{d}^2y}{\mathrm{d}  x^2} = \dfrac{f'(x) g(x) - g'(x)f(x)}{(g(x))^2}

    where f(x) and g(x) are functions of x. Or, you can do this with chain rule:

     \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{h(x)} = (h(x))^{-1} \implies \dfrac{\mathrm{d}^2y}{\mathrm{d}  x^2} = -1 \times h'(x) \times (h(x))^{-2}

    where h(x) is a function of x.
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    Just another thing, if you're not very familiar with calculus on hyperbolic functions, it might be worth writing \mbox{arcoth } x as \dfrac{1}{2}\ln \dfrac{x+1}{x-1}, or even \dfrac{1}{2} \Big( \ln (x+1) - \ln (x-1) \Big) to make it easier.
 
 
 
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