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    How do you differentiate y = (x^0.5 + 1)^5
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    Chain rule

    let u = x^(1/2) + 1

    d/dx = d/du * du/dx
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    (Original post by L'Evil Fish)
    Chain rule

    let u = x^(1/2) + 1

    d/dx = d/du * du/dx
    Not sure if this is the same but I've always been taught to do the following:
    1) Bring the power down in front of the bracket and decrease the power by one e.g. 5(x^0.5 +1)^4
    2) Differentiate inside the bracket e.g. 0.5x^-0.5
    3) Multiply by the differential of bracket e.g. 2.5x^-0.5(x^0.5 +1)^4


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    (Original post by Lauren___e)
    Not sure if this is the same but I've always been taught to do the following:
    1) Bring the power down in front of the bracket and decrease the power by one e.g. 5(x^0.5 +1)^4
    2) Differentiate inside the bracket e.g. 0.5x^-0.5
    3) Multiply by the differential of bracket e.g. 2.5x^-0.5(x^0.5 +1)^4
    Really

    Have you not been taught the chain rule

    How would you differentiate something like \sin ^3(3x^2 - 5)
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    OP, the way of differentiating this function that you have been shown is an application of the chain rule. The reason that they shown you what they did is because it makes for a general result that can be applied where given

    y = [f(x)]^n, dy/dx = n.[f(x)]^n-1 . f'(x).

    Knowing this result is not enough for C3. You must know how to apply the chain rule in full.

    The chain rule states that:

    dy/dx = dy/du x du/dx,

    where y is a function of u, and u is a function of x. Here is the chain rule applied to your function.

    Let u = x^1/2 + 1. Therefore y = u^5. Using the chain rule:

    dy/dx = 5u^4 x 1/2.x^-1/2

    Substituting u = x^1/2 + 1:

    dy / dx = 5/2.x^(-1/2).(x^1/2 + 1)^4

    Practice using the chain rule in full to find derivatives.
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    (Original post by TenOfThem)
    Really

    Have you not been taught the chain rule

    How would you differentiate something like \sin ^3(3x^2 - 5)
    I'd use the product rule. Let u=sin^3 and v=( 3x^2 -5)

    Then using the product rule dy/dx=u(dv/dx) + v(du/dx)


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    (Original post by TenOfThem)
    Really

    Have you not been taught the chain rule

    How would you differentiate something like \sin ^3(3x^2 - 5)
    She just described the use of the chain rule on the instant problem perfectly did she not ? So she obviously has been taught the chain rule. Your sarcastic efforts at superiority are really irritating.
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    (Original post by Lauren___e)
    I'd use the product rule. Let u=sin^3 and v=( 3x^2 -5)

    Then using the product rule dy/dx=u(dv/dx) + v(du/dx)
    Have you just started C3?

    If so, your error is a common one - this is not a product at all it is a function of a function

    Sin(?) is a function and \sin ^3 is not something that you can differentiate

    You need to use the chain rule

    As I said, if you have just begun C3 this will probably be something you will practice a lot more
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    (Original post by Old_Simon)
    She just described the use of the chain rule on the instant problem perfectly did she not ? So she obviously has been taught the chain rule. Your sarcastic efforts at superiority are really irritating.
    Oh dear

    You probably had not read post 7 when you made your rather unpleasant comment
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    (Original post by TenOfThem)
    Oh dear

    You probably had not read post 7 when you made your rather unpleasant comment
    Well after #7 there is obviously an error. But that is not the point. You criticised after #3 which unless I am mistaken is correct and rather well explained in terms of differentiating the outer function then the inner function before multiplying them. The particular style of your question "haven't you been taught the chain rule ?" is very common in domineering people because it is designed to make the recipient feel inadequate or idiotic. Either they have not been taught it which is not their fault or they have been "taught" by someone who equates teaching with learning so they have in fact now forgotten it. Neither of those two things assist the pupil in answering the question. And incidentally pupils being confused about the chain rule which you say is "common" is only common because their "teachers" confused them in the first place.
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    (Original post by Old_Simon)
    ...
    You seem to have some sort of problem with my posts

    I certainly find your posts rude

    I am now going to add you to my ignore list so will no longer be replying to your posts

    You have a number of choices
    • Ignore my posts
    • Complain to the moderation team if you feel my posts are unacceptable/incorrect/rude/etc
    • Continue to post comments directed at me that I will not see
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    (Original post by Emilyrose1210)
    How do you differentiate y = (x^0.5 + 1)^5
    y = (x^0.5 + 1)^5
    dy/dx = 5(x^0.5 +1)^4 * 0.5x^-0.5
    = 2.5x^-0.5(x^0.5 +1)
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    (Original post by TenOfThem)
    Have you just started C3?

    If so, your error is a common one - this is not a product at all it is a function of a function

    Sin(?) is a function and \sin ^3 is not something that you can differentiate

    You need to use the chain rule

    As I said, if you have just begun C3 this will probably be something you will practice a lot more
    Ah this is awkward, I've taken both the C3 and C4 exams already, so have known how to do this for about a year now :P I can see what you are saying and had I actually attempted to differentiate it using the product rule I'm sure I would have realised that it was wrong! Can you tell I was barely awake when I posted that?

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    (Original post by Lauren___e)
    Ah this is awkward, I've taken both the C3 and C4 exams already, so have known how to do this for about a year now :P I can see what you are saying and had I actually attempted to differentiate it using the product rule I'm sure I would have realised that it was wrong! Can you tell I was barely awake when I posted that?

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    That particular question is a function of a function of a function. Can be tricky
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    (Original post by Lauren___e)
    Ah this is awkward, I've taken both the C3 and C4 exams already, so have known how to do this for about a year now :P I can see what you are saying and had I actually attempted to differentiate it using the product rule I'm sure I would have realised that it was wrong! Can you tell I was barely awake when I posted that?

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    ooops
 
 
 
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