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Math Help - Chain Rules - Partial Derivatives

  1. #1
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    Chain Rules - Partial Derivatives

    Let  z =f(x,y) = (8+25x)y^{tan(x)}) , where:  x = (\frac{12 - e}{2e})(t-1) + \tan^{-1}(t) , and  y = (2 - t)e^{t} - 12(t-1) . Use the chain rule to find  \frac{dz}{dt} at  t = 1 .

     \frac{dz}{dt} = ?

    The Chain rule formula I used is the following:  \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial y}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}

    And, when  t = 1 ,  x = \frac{\pi}{4}
    When  t = 1 ,  y = e

     \frac{\partial z}{\partial x} = (8 + 25x)y^{\sec^{2}(x)} + 25y^{\tan(x)}

     \frac{\partial z}{\partial y} = (8 + 25x)\tan(x)y^{\tan(x) - 1} . This is one correct?

     \frac{\partial y}{\partial t} = (2 - t)e^{t} + e^{t}

     \frac{\partial x}{\partial t} = \frac{6}{e} - \frac{1}{2} + \frac{1}{1+ t^{2}}
    Last edited by Belowzero78; April 13th 2010 at 02:48 PM.
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  2. #2
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    Quote Originally Posted by Belowzero78 View Post
    Let  z =f(x,y) = (8+25x)y^{tan(x)}) , where:  x = (\frac{12 - e}{2e})(t-1) + \tan^{-1}(t) , and  y = (2 - t)e^{t} - 12(t-1) . Use the chain rule to find  \frac{dz}{dt} at  t = 1 .

     \frac{dz}{dt} = ?

    The Chain rule formula I used is the following:  \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial y}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}
    In your first partial, you should an x wrt t not y wrt t but you did the correct derivatives in your work.

    And, when  t = 1 ,  x = \frac{\pi}{4}
    When  t = 1 ,  y = e

     \frac{\partial z}{\partial x} = (8 + 25x)y^{\sec^{2}(x)} + 25y^{\tan(x)}

     \frac{\partial z}{\partial y} = (8 + 25x)\tan(x)y^{\tan(x) - 1} . This is one correct?

     \frac{\partial y}{\partial t} = (2 - t)e^{t} + e^{t}

     \frac{\partial x}{\partial t} = \frac{6}{e} - \frac{1}{2} + \frac{1}{1+ t^{2}}
    Looks fine besides the typo; however, I didn't check your derivatives. I am little confused on what you are asking. You want to know what \frac{dz}{dt} is but you have done everything as long as your math is correct.
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  3. #3
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    Could someone please show me what the partial is for:  \frac{\partial z}{\partial x} ?
    I'm really confused about the y^tan(x) part when taking the derivative of x.

    Would it be:  \frac{\partial z}{\partial x} = (8 + 25x)y^{\tan(x)}\ln(y)\sec^{2}(x) + 25y^{\tan(x)} ?
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  4. #4
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    For example, y^{tan(x)} can be solved like this:

    y=u and tanx=v

    \big[\frac{u'*v}{u}+ln(u)*v'\big]*u^v

    Therefore, 25y^{tan(x)}=25\frac{y^{tan(x)}*ln(y)}{cos^2(x)}

    If this example doesn't help you with, then just ask.
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  5. #5
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    Ok, after this how do i evaluate dz/dt? Like what stuff do i plug in?
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  6. #6
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    Add your solutions  \frac{\partial z}{\partial x}\frac{\partial x}{\partial  t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}<br />
of the partial derivatives you obtain.
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