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Thread: Partial Derivative of x^y

  1. #1
    Member mybrohshi5's Avatar
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    Unhappy Partial Derivative of x^y

    For $\displaystyle f(x,y) = x^2cos(y) - \frac{x^3}{y^2} + x^y $

    Find..... $\displaystyle f_x, f_y, f_x_x, f_y_y, f_x_y, f_y_x $

    I know how to do this expect i am running into trouble with the $\displaystyle x^y$ term.

    so the x partial derivative ($\displaystyle f_x$) of $\displaystyle x^y$ is .... $\displaystyle x^yln(x)$ ????

    and the y partial derivative ($\displaystyle f_y$) of $\displaystyle x^y$ is .... $\displaystyle x $ ????

    Thank you
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  2. #2
    MHF Contributor
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    When you are finding $\displaystyle f_x$, you treat $\displaystyle y$ as a constant.

    So $\displaystyle \frac{\partial}{\partial x}(x^y) = x^{y-1}y$.


    When you are finding $\displaystyle f_y$, you treat $\displaystyle x$ as a constant.

    So $\displaystyle \frac{\partial}{\partial y}(x^y) = x^y\ln{x}$.
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  3. #3
    Member mybrohshi5's Avatar
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    Perfect that is exactly what i needed

    Thank you so much!
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