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anonymous

  • 5 years ago

If w=x^2+y-z+sin t & x+y=t , then how to find part.dervivative of w wrt y ? plz click to find more.....

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  1. anonymous
    • 5 years ago
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    \[w = x ^{2} + y - z + \sin(t)\] \[x + y = t\] Sorry which derivative are you looking for?

  2. anonymous
    • 5 years ago
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    we have to find \[(dough w/dough y)_{x,z}\].partial derivative

  3. anonymous
    • 5 years ago
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    I'm honestly not sure what you mean by that notation.

  4. anonymous
    • 5 years ago
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    there's no notation here, consider dough = d (dw/dx)\[_{x,z}\]

  5. anonymous
    • 5 years ago
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    (\[(dw/dx)_{x,z}\] what does this x,z mean here?

  6. anonymous
    • 5 years ago
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    sry , it is \[(dw/dy)_{x,z}\]

  7. anonymous
    • 5 years ago
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    use \partial to get \[\partial\] if you like

  8. anonymous
    • 5 years ago
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    thanks eb... can u temme wat foes 'x,z' mean here?

  9. anonymous
    • 5 years ago
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    \[(\partial w/ \partial y)_{x,z}\] --------this is the actual notation

  10. anonymous
    • 5 years ago
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    Okay I think that means find the derivative of w with respect to y, and evaluate at x,z? Not exactly sure

  11. anonymous
    • 5 years ago
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    i wanted to know what this x,z means...i guess those are independent variables? is it?

  12. anonymous
    • 5 years ago
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    w=x2+y−z+sin(t) x+y=t w=x2+y−z+sin(x+y) assuming w=f(x,y,z) (dw/dy) =0+1+0+cos(x+y) =1+Cos(x+y)

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