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Math Help - Partial derivative confused.

  1. #1
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    Partial derivative confused.

     2x^2y^(\frac{1}{2})

    Is  f_y = 2x^2\frac{1}{2}y^{-\frac{1}{2}}?

    If im wrong can you please explain why and how to find it properly.

    Thanks.
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  2. #2
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    Quote Originally Posted by el123 View Post
     2x^2y^(\frac{1}{2})

    Is  f_y = 2x^2\frac{1}{2}y^{-\frac{1}{2}}?

    If im wrong can you please explain why and how to find it properly.

    Thanks.
    Are you asking to find \frac{\partial f}{\partial y} if f(x, y) = 2x^2y^{\frac{1}{2}}?


    If so, leave x constant and differentiate with respect to y.


    So \frac{\partial f}{\partial y} = \frac{1}{2}\cdot 2x^2 y^{-\frac{1}{2}}

     = \frac{x^2}{\sqrt{y}}

     = \frac{x^2\sqrt{y}}{y}.
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  3. #3
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    yeah thats what i wanted to know but i'm confused why you don't bring the 2 down as well.
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    Quote Originally Posted by el123 View Post
    yeah thats what i wanted to know but i'm confused why you don't bring the 2 down as well.
    If x is treated as a constant, so is x^2.


    Maybe I should write it out like this

    f(x, y) = 2x^2 y^{\frac{1}{2}}

    \frac{\partial f}{\partial y} = 2x^2 \frac{\partial}{\partial y} \left(y^{\frac{1}{2}}\right), since 2x^2 is a constant.

     = 2x^2 \cdot \frac{1}{2}y^{-\frac{1}{2}}

     = x^2 y^{-\frac{1}{2}}.
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