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Math Help - Find the Derivative

  1. #1
    Member VitaX's Avatar
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    Find the Derivative

    Find the derivative of:

    y = \sqrt{x} Csc(x + 1)^3

    Tbh the Csc is throwing me off here. I'm unsure what to apply here. I know that the chain rule and product rule will be used but I'm stuck.
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  2. #2
    DBA
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    d/dx(csc u) = - csc u * cot u * du/dx


    Your y is that

    y = sqrt(x)* [csc(x+1)]^3 --> csc cubed
    or
    y = sqrt(x)* [csc(x+1)^3] --> x+1 cubed ?
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by DBA View Post
    d/dx(csc u) = - csc u * cot u * du/dx


    Your y is that

    y = sqrt(x)* [csc(x+1)]^3 --> csc cubed
    or
    y = sqrt(x)* [csc(x+1)^3] --> x+1 cubed ?
    Little confusing the way you wrote it out here. When applying the product rule y' = uv' + vu' what would u and v be in the original function. u = \sqrt{x} and v = Csc(x + 1)^3 Correct so far? If so then my problem is finding the derivative of v = Csc(x + 1)^3 I know how to apply the chain rule but the Csc is just confusing me.
    Last edited by VitaX; October 11th 2009 at 03:16 PM.
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  4. #4
    DBA
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    Yes, that would be the product rule.
    But then for v' you need to use the Chain Rule.

    I am still not sure if the CSC function is cubed or the x+1 since it would be a difference.

    v = [csc(x+1)]^3
    v' = 3*[csc(x+1)]^2 * -csc(x+1)*cot(x+1) * 1

    v = csc((x+1)^3)
    v' = -csc(x+1)*cot(x+1) * 3*(x+1)^2 * 1

    ^2 means squared
    ^3 means cubed

    Sorry, I do not know yet, how to write the functions better.
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  5. #5
    Member VitaX's Avatar
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    Quote Originally Posted by DBA View Post
    Yes, that would be the product rule.
    But then for v' you need to use the Chain Rule.

    I am still not sure if the CSC function is cubed or the x+1 since it would be a difference.

    v = [csc(x+1)]^3
    v' = 3*[csc(x+1)]^2 * -csc(x+1)*cot(x+1) * 1

    v = csc((x+1)^3)
    v' = -csc(x+1)*cot(x+1) * 3*(x+1)^2 * 1

    ^2 means squared
    ^3 means cubed

    Sorry, I do not know yet, how to write the functions better.
    I believe the Csc function is not cubed.

    Here's what I did:
    <br />
y = \sqrt{x} Csc(x + 1)^3<br />
    y' = Csc(x+1)^3*\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-CscxCotx)3(x+1)^2
    u=x^{\frac{1}{2}}
    u'=\frac{1}{2}x^{-\frac{1}{2}}
    v=Csc(x+1)^3
    v'=(-CscxCotx)3(x+1)^2
    Note \frac{d}{dx}Cscx=-CscxCotx

    Hows this look?
    I took the derivative of Cscx seperately from (x+1)^3 Was I supposed to do this?

    Edit: y' = Csc(x+1)^3\frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}}(-Csc(x+1)^3)(Cot(x+1)^3)3(x+1)^2
    y' = \frac{1}{2}x^{-\frac{1}{2}} Csc(x+1)^3 + 3x^{\frac{1}{2}}(x+1)^2(-Csc(x+1)^3)(Cot(x+1)^3)
    Last edited by VitaX; October 11th 2009 at 04:40 PM.
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  6. #6
    DBA
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    Note

    Yes, but you have CSC[(x+1)^3] and not CSC(x), so for your v' you need to use the entire term in [ ]
    The x in your case is (x+1)^3

    Example:
    f= csc(x+1)
    f'= -csc(x+1)cot(x+1)

    f= csc[(x+1)^3]
    f'= -csc[(x+1)^3]cot[(x+1)^3]

    "Hows this look?
    I took the derivative of seperately from Was I supposed to do this?"

    So yes but fom csc[(x+1)^3]

    You always have inner and outer function
    Example
    f= (2-x)^3
    outer function ()^3
    inner function (2-x)

    So,
    f'= 3*(2-x)^2 * (0-1)

    Hope that helps
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  7. #7
    Member VitaX's Avatar
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    Quote Originally Posted by DBA View Post


    Note

    Yes, but you have CSC[(x+1)^3] and not CSC(x), so for your v' you need to use the entire term in [ ]
    The x in your case is (x+1)^3

    Example:
    f= csc(x+1)
    f'= -csc(x+1)cot(x+1)

    f= csc[(x+1)^3]
    f'= -csc[(x+1)^3]cot[(x+1)^3]

    "Hows this look?
    I took the derivative of seperately from Was I supposed to do this?"

    So yes but fom csc[(x+1)^3]

    You always have inner and outer function
    Example
    f= (2-x)^3
    outer function ()^3
    inner function (2-x)

    So,
    f'= 3*(2-x)^2 * (0-1)

    Hope that helps
    So would the final answer be as I said above?
    y' = \frac{1}{2}x^{-\frac{1}{2}} Csc(x+1)^3 + 3x^{\frac{1}{2}}(x+1)^2(-Csc(x+1)^3)(Cot(x+1)^3
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  8. #8
    DBA
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    Yes, I think that is right.
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