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Thread: Chain Rules - Partial Derivatives

  1. #1
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    Chain Rules - Partial Derivatives

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

    $\displaystyle \frac{dz}{dt} = ?$

    The Chain rule formula I used is the following: $\displaystyle \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 $\displaystyle t = 1 $, $\displaystyle x = \frac{\pi}{4} $
    When $\displaystyle t = 1 $, $\displaystyle y = e $

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

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

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

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

    $\displaystyle \frac{dz}{dt} = ?$

    The Chain rule formula I used is the following: $\displaystyle \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 $\displaystyle t = 1 $, $\displaystyle x = \frac{\pi}{4} $
    When $\displaystyle t = 1 $, $\displaystyle y = e $

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

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

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

    $\displaystyle \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 $\displaystyle \frac{dz}{dt}$ 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: $\displaystyle \frac{\partial z}{\partial x} $?
    I'm really confused about the y^tan(x) part when taking the derivative of x.

    Would it be: $\displaystyle \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, $\displaystyle y^{tan(x)}$ can be solved like this:

    $\displaystyle y=u$ and $\displaystyle tanx=v$

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

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

    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 $\displaystyle \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}
    $ of the partial derivatives you obtain.
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