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Thread: Implicit Differentiation Problem (Chain Rule?)

  1. #1
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    Question Implicit Differentiation Problem (Chain Rule?)

    Find dy/dx:
    y=[sin(x+5)]^(5/4)

    So far I have this:
    y=[sin(x+5)]^(5/4)
    y'={[(5/4)cosx(x+5)]^(1/2)} [cos(x+5)] (1)


    There are only answers to odd numbered questions in the back of my textbook & this is an even numbered problem. I'm not sure where to go from here or if I've used chain rule the right way.

  2. #2
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    Couple of questions:
    1. Why did the first sin() magically turn into cos()?
    2. Why did fourths magically turn into halves?
    3. What about this is "implicit"?


    Please give it another go.

  3. #3
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    Quote Originally Posted by veronicadeno View Post
    Find dy/dx:
    y=[sin(x+5)]^(5/4)

    So far I have this:
    y=[sin(x+5)]^(5/4)
    y'={[(5/4)cosx(x+5)]^(1/2)} [cos(x+5)] (1)


    There are only answers to odd numbered questions in the back of my textbook & this is an even numbered problem. I'm not sure where to go from here or if I've used chain rule the right way.
    Except for the peculiarities that tkhunny asked about, it looks like you did the chain-rule part ok! It is just the power-law that is a "bit" off.
    DrPhil (not the TV guy)
    If we knew what we were doing,we wouldn't have to do it

  4. #4
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    Question

    Quote Originally Posted by tkhunny View Post
    Couple of questions:
    1. Why did the first sin() magically turn into cos()?
    2. Why did fourths magically turn into halves?
    3. What about this is "implicit"?


    Please give it another go.
    1. Because I thought the derivative of sin() was cos(), unless I'm not supposed to mess with the trig in that first part which is what I think you're implying.
    2. I think that is a typing error on my part, should be 1/4 I think, but then again I'm in a math help forum.
    3. No idea, doesn't look implicit to me, that was just the heading in my textbook.
    Thank you!

    Quote Originally Posted by DrPhil View Post
    Except for the peculiarities that tkhunny asked about, it looks like you did the chain-rule part ok! It is just the power-law that is a "bit" off.
    Okay thank you!

    "Another go"
    y=[sin(x+5)]^(5/4)
    y'={[(5/4)cosx(x+5)]^(1/4)} [cos(x+5)] (1)
    y'=
    {[(5/4)cosx(x+5)]^(1/4)} [cos(x+5)]
    Last edited by veronicadeno; 03-18-2013 at 09:29 PM.

  5. #5
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    Quote Originally Posted by veronicadeno View Post
    1. Because I thought the derivative of sin() was cos(), unless I'm not supposed to mess with the trig in that first part which is what I think you're implying.
    2. I think that is a typing error on my part, should be 1/4 I think, but then again I'm in a math help forum.
    3. No idea, doesn't look implicit to me, that was just the heading in my textbook.
    Thank you!



    Okay thank you!

    "Another go"
    y=[sin(x+5)]^(5/4)
    y'={[(5/4) sin(x+5)]^(1/4)} [cos(x+5)] (1)
    y'=
    {[(5/4) sin(x+5)]^(1/4)} [cos(x+5)]
    Good answers.

    1. Yes, the first part is just the power-law - no trig - so still should be sin(x+5)
    2. Yes, the power is 5/4 - 1 = 1/4
    3. Yes, this function is explicit, not implicit

    EDIT: Thank you srmichael for fixing my blunder!

    y'= {[(5/4) sin(x+5)]^(1/4)} [cos(x+5)]
    Last edited by DrPhil; 03-19-2013 at 09:41 AM. Reason: Wronng Answer
    DrPhil (not the TV guy)
    If we knew what we were doing,we wouldn't have to do it

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    Quote Originally Posted by DrPhil View Post
    Good answers.

    1. Yes, the first part is just the power-law - no trig.
    2. Yes, the power is 5/4 - 1 = 1/4
    3. Yes, this function is explicit, not implicit

    Only remaining question is what you mean by cosx(x+5). That extra letter does NOT belong. The answer is

    y'= {[(5/4)cos(x+5)]^(1/4)} [cos(x+5)]
    There's a liittle hiccup in Dr. Phil's answer:

    [tex]y=[\sin(x+5)]^{\frac{5}{4}}[/tex]

    [tex]y'=\frac{5}{4}[\sin(x+5)]^{\frac{1}{4}}\cos(x+5)[/tex]

    "There are 10 types of people in this world - those who understand binary and those who don't."

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