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Math Help - Partial Differentiation

  1. #1
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    Partial Differentiation

    f(x,y)=2x^3y^2+2y+4x. Find f_x(1,3) (that is, use the definition to find the partial derivative f_x(x,y) at (1,3)).

    The answer has to be f_x(1,3) = 58.

    My Attempt:

    I want to use the following definition to find the partial derivative

     f_x(x_0,y_0)= \frac{d}{dx}[f(x,y_0)]|_{x=x_0} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}

    So,

    \lim_{\Delta x \to 0} \frac{2(x+ \Delta x)3^2 +2 \times 3 +4x-64}{\Delta x}

     \lim_{\Delta x \to 0} \frac{10x-58+6 \Delta x}{\Delta x}

    If I cancel out \Delta x in the denominator with the the term on top I get:

     \lim_{\Delta x \to 0} 10x-52

    which is wrong... can anoyne help?
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  2. #2
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    Quote Originally Posted by demode View Post
    f(x,y)=2x^3y^2+2y+4x. Find f_x(1,3) (that is, use the definition to find the partial derivative f_x(x,y) at (1,3)).

    The answer has to be f_x(1,3) = 58.

    My Attempt:

    I want to use the following definition to find the partial derivative

     f_x(x_0,y_0)= \frac{d}{dx}[f(x,y_0)]|_{x=x_0} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}

    So,

    \lim_{\Delta x \to 0} \frac{2(x+ \Delta x)3^2 +2 \times 3 +4x-64}{\Delta x}

     \lim_{\Delta x \to 0} \frac{10x-58+6 \Delta x}{\Delta x}

    If I cancel out \Delta x in the denominator with the the term on top I get:

     \lim_{\Delta x \to 0} 10x-52

    which is wrong... can anoyne help?
    You just differentiate each term with respect to x, keeping y constant.


    So \frac{\partial f}{\partial x} = 6x^2y^2 + 4.


    Now evaluate this at (x, y) = (1, 3).
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  3. #3
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    Quote Originally Posted by demode View Post
    f(x,y)=2x^3y^2+2y+4x. Find f_x(1,3) (that is, use the definition to find the partial derivative f_x(x,y) at (1,3)).

    The answer has to be f_x(1,3) = 58.

    My Attempt:

    I want to use the following definition to find the partial derivative

     f_x(x_0,y_0)= \frac{d}{dx}[f(x,y_0)]|_{x=x_0} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}
    Where did the '64' come from? f(1, 3)= 2(1)(9)+ 2(3)+ 4(1)= 18+ 6+ 4= 28.

    And does "from the definition" really mean using the difference quotient? By the time you get to partial derivatives, you usually use the definition that Prove It refers to- that the partial derivative of f with respect to x is just the ordinary derivative holding other variables constant.

    So,

    \lim_{\Delta x \to 0} \frac{2(x+ \Delta x)3^2 +2 \times 3 +4x-64}{\Delta x}
    No, this should be
    \frac{2(1+ \Delta x)^3(3^2)+ 2(3)+ 4(1+ \Delta x)- 28}{\Delta x}

    [/quote]  \lim_{\Delta x \to 0} \frac{10x-58+6 \Delta x}{\Delta x}

    If I cancel out \Delta x in the denominator with the the term on top I get:

     \lim_{\Delta x \to 0} 10x-52

    which is wrong... can anoyne help?[/QUOTE]
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  4. #4
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    \lim_{\Delta x \to 0} \frac{2(1+ \Delta x)^3(3^2)+ 2(3)+ 4(1+ \Delta x)- 28}{\Delta x}

    \lim_{\Delta x \to 0} \frac{2(1+3(\Delta x)+3(\Delta x)^2+(\Delta x)^3)9+ 2(3)+ 4(1+ \Delta x)- 28}{\Delta x}

    \lim_{\Delta x \to 0} \frac{18+54(\Delta x)+54(\Delta x)^2 + 18 (\Delta x)^3 +10+ 4 (\Delta x)- 28}{\Delta x}


     <br />
\lim_{\Delta x \to 0} \frac{18(\Delta x)^3+54(\Delta x)^2+58(\Delta x)}{\Delta x}<br />

    So, how exactly do I need to evaluate this limit to get 58?

    Can I cancel out the \Delta x on top with the \Delta x on the denominator to get

    \lim_{\Delta x \to 0} 18 (\Delta x)^2 +54(\Delta x) +58

    And then substitute 0's

    18 (0)^2 +54(0) +58 = 58

    Is this correct?
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