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Math Help - Implicit Differentiation with 3 Variables

  1. #1
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    Implicit Differentiation with 3 Variables

    Hi all. I've been having problems with the question below.

    z = sin(xyz) + xln(yz)

    find

    \frac{\partial z}{\partial x}

    I've tried two differentiation methods:


    First


    \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}

    which generally gives me:

    \frac{yzcos(xyz) + ln(yz)}{xycos(xyz) + \frac{xy}{yz}}

    I'm not 100% sure about that though. Especially differentiating \frac {\partial T}{\partial z} (the denominator of \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}})


    Secondly
    I dont know the name of this differentiation method or how to describe it, but essentially you differentiate as normal - with respect to x, treating y as a constant.
    This time, however, also add \frac{\partial z}{\partial x} whenever you encounter a z. Or is it whenever you differentiate z? Or is it when z is left from differentiating x - as in z was a coefficient of x? As you can see this is where my confusion comes in.

    If anyone can confirm or help me with this I would appreciate it. Thanks.
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  2. #2
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    Quote Originally Posted by isp_of_doom View Post
    Hi all. I've been having problems with the question below.

    z = sin(xyz) + xln(yz)

    find

    \frac{\partial z}{\partial x}

    I've tried two differentiation methods:


    First

    \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}

    which generally gives me:

    \frac{yzcos(xyz) + ln(yz)}{xycos(xyz) + \frac{xy}{yz}}

    I'm not 100% sure about that though. Especially differentiating \frac {\partial T}{\partial z} (the denominator of \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}})


    Secondly
    I dont know the name of this differentiation method or how to describe it, but essentially you differentiate as normal - with respect to x, treating y as a constant.
    This time, however, also add \frac{\partial z}{\partial x} whenever you encounter a z. Or is it whenever you differentiate z? Or is it when z is left from differentiating x - as in z was a coefficient of x? As you can see this is where my confusion comes in.

    If anyone can confirm or help me with this I would appreciate it. Thanks.
    z = sin(xyz) + xln(yz)

    The implicit function theorem would give the partial derivative as \frac{\partial z}{\partial x}=-\frac{\frac{\partial T}{\partial x}}{\frac{\partial T}{\partial x}}. I used this formula, and obtained the same expression you did. You just need a negative sign.
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  3. #3
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    WAIT!!!

    We're using the theorem wrong, lol. I'm surprised no one has said anything. I just realized this.

    z=sin(xyz)+xln(xy)

    F(x,y,z)=sin(xyz)+xln(yz)-z=0

    NOW you can apply the implicit function theorem. Always move everything over to one side. The expression we obtained earlier would be true only if 0=sin(xyz)+xln(yz).

    ----------------------------------------------------------------------

    Your second method would work too. It is just a little more involved.

    z=sin(xyz)+xln(yz)

    I will write z=z_x since z must be treated as a function of x. So we could go through the following.

    \frac{\partial z_x}{\partial x}=\frac{\partial}{\partial x}sin(xyz_x)+\frac{\partial}{\partial x}\left(xln(yz_x)\right)

    Treat y as a constant, and remember that z is a function. I would write the first term as \frac{\partial}{\partial x}sin(xyz_x)=\frac{d}{du}sin(u)\frac{\partial u}{\partial x} where u=xyz_x. Then calculate \frac{\partial u}{\partial x} remembering that z=z_x is a function, so apply the product rule to get yz_x+xy\frac{\partial z_x}{\partial x}. So the term is \frac{\partial}{\partial x}sin(xyz_x)=cos(xyz_x)\left(yz_x+xy\frac{\partial z_x}{\partial x}\right). You could do something similar with the second term.

    \frac{\partial }{\partial x}xln(yz_x)=ln(yz_x)+\frac{xy}{yz_x}\frac{\partial z_x}{\partial x}

    Adding the terms gives:

    \frac{\partial z}{\partial x}=ln(yz)+\frac{xy}{yz}\frac{\partial z}{\partial x}+cos(xyz)\left(yz+xy\frac{\partial z}{\partial x}\right).

    \frac{\partial z}{\partial x}=\frac{ln(yz)+yzcos(xyz)}{1-xycos(xyz)+\frac{xy}{yz}}
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  4. #4
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    Thanks adkinsjr,

    double checked the question and it actually states:

    A function z = f(x,y) is defined implicitly by sin(xyz) + xln(yz) = 1 Find \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y}.

    In this case your first post would apply, correct?
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  5. #5
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    Quote Originally Posted by isp_of_doom View Post
    Thanks adkinsjr,

    double checked the question and it actually states:

    A function z = f(x,y) is defined implicitly by sin(xyz) + xln(yz) = 1 Find \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y}.

    In this case your first post would apply, correct?
    Yes
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  6. #6
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    Quote Originally Posted by isp_of_doom View Post
    Hi all. I've been having problems with the question below.

    z = sin(xyz) + xln(yz)

    find

    \frac{\partial z}{\partial x}

    I've tried two differentiation methods:


    First


    \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}}

    which generally gives me:

    \frac{yzcos(xyz) + ln(yz)}{xycos(xyz) + \frac{xy}{yz}}

    I'm not 100% sure about that though. Especially differentiating \frac {\partial T}{\partial z} (the denominator of \frac{\frac{\partial T} {\partial x}}{\frac{\partial T}{\partial z}})


    Secondly
    I dont know the name of this differentiation method or how to describe it, but essentially you differentiate as normal - with respect to x, treating y as a constant.
    This time, however, also add \frac{\partial z}{\partial x} whenever you encounter a z. Or is it whenever you differentiate z? Or is it when z is left from differentiating x - as in z was a coefficient of x? As you can see this is where my confusion comes in.

    If anyone can confirm or help me with this I would appreciate it. Thanks.
    If z= sin(xyz)+ x ln(yz) then
    \frac{\partial z}{\partial x}= cos(xyz) \left(yz\frac{\partial x}{\partial x}+ xz\frac{\partial y}{\partial x}+ xy\frac{\partial z}{\partial x}\right)+ ln(yz)\frac{\partial x}{\partial x}+ \frac{x}{yz}\left(z\frac{\partial y}{\partial x}+ y\frac{\partial z}{\partial x}\right)

    Of course, \frac{\partial x}{\partial x}= 1 and \frac{\partial y}{\partial x}= 0 so that becomes
    \frac{\partial z}{\partial x}= cos(xyz)\left(yz+ xy\frac{\partial z}{\partial x}\right)+ ln(yz)+ \frac{x}{yz}\left(y\frac{\partial z}{\partial x}\right).
    Solve that for \frac{\partial z}{\partial x}.
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