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Math Help - Taking partial derivitaive of a mutivariable function by the definition

  1. #1
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    Taking partial derivitaive of a mutivariable function by the definition

    Well I know that to take a derivative by the definition, the formula is:
    f(x+h) - f(x)/h

    and to take the derivative of a multivariable function with respect to x by the definition is:
    f(x+h,y) - f(x)/h

    I don't know why, but the y in there seems to confuse me.

    Was wondering if anyone could show me how to take the derivative of the following functions with respect to x. Make sure to do it by the definition.

    f(x,y) = x^2 + 2y^2

    f(x,y) = 2x+3y+5



    Thanks for your help! I'm brand new here.
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  2. #2
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    Quote Originally Posted by gocubsgo708 View Post
    Well I know that to take a derivative by the definition, the formula is:
    f(x+h) - f(x)/h

    and to take the derivative of a multivariable function with respect to x by the definition is:
    f(x+h,y) - f(x)/h

    I don't know why, but the y in there seems to confuse me.

    Was wondering if anyone could show me how to take the derivative of the following functions with respect to x. Make sure to do it by the definition.

    f(x,y) = x^2 + 2y^2

    f(x,y) = 2x+3y+5



    Thanks for your help! I'm brand new here.
    Your definition is a bit off. There's a f(x) in place of a f(x,y).

    There was a question like this not long ago, see this thread and scroll down a bit.
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  3. #3
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    when taking partial derivatives in this case with respect to x, hold y constant.

    f(x,y) = x^2 + 2y^2

    \frac{\partial{ f}}{\partial{x}} = 2x

     <br />
\frac{\partial{ f}}{\partial{y}} = 4y<br />
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