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Math Help - Partial derivative question

  1. #1
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    Partial derivative question

    (a) If
    f(x, y) = log(x + y) + sin(xy) + y^2e^x, determine partial df/dx and partial df/dt

    find df/dt in terms of
    x and y where x = cos t, y = sin t using the chain rule formula. [ 10 marks ]

    (b) If
    f(x, y) = x3 y2 xy, determine partial df/dx and p-artial df/dy

    If f = 0 then express dy/dx in terms of x and y.

    Just a quick answer check for a i got:
    df/dt=(1/x+y+ycos(xy)+y^2e^x)*-y+(1/x+y+ycos(xy)+y^2e^x)*x
    Now im sure this isnt in the simpliest form

    For part b i got dy/dx= (-3x^2+y)/(-2y-x)?

    Many thanks in advance.

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  2. #2
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    In Question (a) are you sure you're not meant to find \displaystyle \frac{\partial f}{\partial x} and \displaystyle \frac{\partial f}{\partial y}, not \displaystyle \frac{\partial f}{\partial t}?
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  3. #3
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    Sorry yes thats correct need to find partial df/dx and partial df/dy then from that df/dt.
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  4. #4
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    Use standard derivative rules, treating all variables as constants except for the variable you are differentiating with respect to...
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    Quote Originally Posted by Prove It View Post
    Use standard derivative rules, treating all variables as constants except for the variable you are differentiating with respect to...
    Yes i done that to obtain partial df/dx= 1/(x+y)+cos(xy)*y+y^2*e^x then for df/dy= 1/(x+y)+cosy(xy)*x+2y*e^x
    Then am i correct to think you can work out dx/dt and dx/dy then to multiply the partial df/dx by dx/dt and partial df/dy by dy/dt to form df/dt?
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    Yes, your partial derivatives are correct.

    Now use the chain rule: \displaystyle \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\,\frac{dx}{dt} + \frac{\partial f}{\partial y}\,\frac{dy}{dt}.

    Alternatively, you could substitute \displaystyle x = \cos{t} and \displaystyle y = \sin{t} into \displaystyle f(x,y) to get a function \displaystyle f(t).
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Yes, your partial derivatives are correct.

    Now use the chain rule: \displaystyle \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\,\frac{dx}{dt} + \frac{\partial f}{\partial y}\,\frac{dy}{dt}.

    Alternatively, you could substitute \displaystyle x = \cos{t} and \displaystyle y = \sin{t} into \displaystyle f(x,y) to get a function \displaystyle f(t).
    I think thats what i may of done with dx/dt=-sint=-y and dy/dt=cost=x therefore getting: (1/x+y+ycos(xy)+y^2e^x)*-y+(1/x+y+ycos(xy)+y^2e^x)*x
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  8. #8
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    Quote Originally Posted by breitling View Post
    I think thats what i may of done with dx/dt=-sint=-y and dy/dt=cost=x therefore getting: (1/x+y+ycos(xy)+y^2e^x)*-y+(1/x+y+ycos(xy)+y^2e^x)*x
    Looks good to me
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