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Math Help - Multivariable Calculus - Chain Rule

  1. #1
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    Multivariable Calculus - Chain Rule

    I might have to keep coming back here, this assignment is killing me...

    Given w = xy + yz, y = sin x and z = e^x, find dw/dx
    \frac{dw}{dx} = \frac{\partial w}{\partial y}\frac{dy}{dx} + \frac{\partial w}{\partial z}\frac{dz}{dx}

    = (x+z) \cos x + ye^x
    = x \cos x + e^x \cos x + e^x \sin x

    But if you sub in the values for y and z before taking the derivative, you get:
    x \sin x + \sin x e^x
    \frac{dw}{dx} = (\sin x + x \cos x) + (e^x \sin x + e^x \cos x)
    = x \cos x + e^x \cos x + e^x \sin x + \sin x

    Different from the top answer. Can anyone see where I went wrong?

    Also:
    Given u = x + ay and v = x + by, for what values a and b can you change:
    9(\frac{\partial ^2 f}{\partial x^2}) - 9(\frac{\partial ^2 f}{\partial x \partial y}) + 2(\frac{\partial ^2 f}{\partial y^2}) = 0
    to:
    \frac{\partial ^2 f}{\partial u \partial v} = 0

    At first I tried to factor it into (3n - 2m)(3n - m), but then I remembered that the second derivative doesn't equal the first derivative squared. Also, the middle term isn't necessarily both partial derivatives multiplied together. I found:
    \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}
    and:
    \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}a + \frac{\partial f}{\partial v}b

    But I can't think of a way to sub them in, or take the next derivative without knowing what f is.
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  2. #2
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    <br />
\frac{dw}{dx} = \frac{\partial w}{\partial y}\frac{dy}{dx} + \frac{\partial w}{\partial z}\frac{dz}{dx}<br />

    wants to be

    <br />
\frac{dw}{dx} = \frac{\partial w}{\partial y}\frac{dy}{dx} + \frac{\partial w}{\partial z}\frac{dz}{dx} + \frac{\partial w}{\partial x}<br />
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  3. #3
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    Hello, BlackBlaze!

    You left out a term in your equation . . .


    Given: . w \:=\: xy + yz,\;\; y \:=\: \sin x,\;\;z \:=\: e^x
    Find \frac{dw}{dx}

    The equation is: . \frac{dw}{dx} \;=\; {\color{blue}\frac{\partial w}{\partial x}\!\cdot\!\frac{dx}{dx}} + \frac{\partial w}{\partial y}\!\cdot\!\frac{dy}{dx} + \frac{\partial w}{\partial z}\!\cdot\!\frac{dz}{dx}

    . . . . . . . . . . . . . . . =\;\;\;y\cdot 1 + (x+z)\cos x + y\cdot e^x

    . . . . . . . . . . . . . . . =\;\; \sin x + (x+e^x)\cos x + \sin x\cdot e^x

    . . . . . . . . . . . . . . . =\;\;\sin x + x\cos x + e^x\cos x + e^x\sin x

    . . which matches your other answer . . .

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  4. #4
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    Quote Originally Posted by BlackBlaze View Post
    I found:
    \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}
    and:
    \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}a + \frac{\partial f}{\partial v}b

    But I can't think of a way to sub them in, or take the next derivative without knowing what f is.
    Carrying on from there, but writing it

    f_x = f_u + f_v

    and:

    f_y = a f_u + b f_v

    (if you've no objection), the same reasoning gets

    f_{xx} = f_{uu} + 2f_{uv} + f_{vv}

    and

    f_{yy} = a^2 f_{uu} + 2ab f_{uv} + b^2 f_{vv}

    and

    f_{xy} = a f_{uu} + (a+b) f_{uv} + b f_{vv}

    so that

    9 f_{xx} - 9 f_{xy} + 2 f_{yy} = 0

    means

    (9 - 9a + 2a^2) f_{uu} + (18 - 9(a+b) + 4ab) f_{uv} + (9 - 9b + 2b^2) f_{vv} = 0

    Then you can choose a and b to make the f_{uu} and f_{vv} drop out.
    Last edited by tom@ballooncalculus; March 17th 2010 at 06:28 AM. Reason: removed fudge
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