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Math Help - Partial derivative

  1. #1
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    Partial derivative

    Find  \frac{dw}{ds}(-2,2) and  \frac{dw}{dt}(-2,2) if  w= xy -yz -5xz, where  x = st, y = e^{st}, z= t^2

    I'm confused since only two coordinates are given but the problem has 3 variables   x,y, z
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chaos787 View Post
    Find  \frac{dw}{ds}(-2,2) and  \frac{dw}{dt}(-2,2) if  w= xy -yz -5xz, where  x = st, y = e^{st}, z= t^2

    I'm confused since only two coordinates are given but the problem has 3 variables  x,y, z
    You need to apply the chain rule: \frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial s}

    Similarly, \frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by chaos787 View Post
    Find  \frac{dw}{ds}(-2,2) and  \frac{dw}{dt}(-2,2) if  w= xy -yz -5xz, where  x = st, y = e^{st}, z= t^2

    I'm confused since only two coordinates are given but the problem has 3 variables   x,y, z
    The "two coordinates" are values of s and t. When s= -2, t= 2, x= (-2)(2)= -4, y= e^{(-2)(2)}= e^{-4}, and z= 2^2= 4. Actually, it is a little ambiguous as to which of "(-2, 2)" is s and which t but it turns out that, in this problem, it doesn't matter!
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