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Math Help - Partial Deriviatives

  1. #1
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    Partial Deriviatives

    I am having trouble with two questions, one is with a quotient rule and the other a product rule I believe. Instructions indicate to compute all first order derivatives of the given function.

    Z= xe^xy

    Here is what I did: fx= (x)(ye^xy) + (e^xy)(1) Is this correct?
    I started using the product rule again for (fy) but then I saw an example in the book where they used the constant multiple rule. Why am I supposed to use this rule and not the product rule? The book shows a similar example and they used the product rule to differentiate fx and used the constant multiple rule to differentiate fy, I would just like to know why this is...

    My second set of problems is:

    1. 2x+3y / y-x

    2. e^2-x / y^2

    I know the quotient rule can be used, but I was not sure how to keep y and x as constants. If someone can illustrate the steps involved, Thanks...
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  2. #2
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    Quote Originally Posted by kdogg121 View Post
    I am having trouble with two questions, one is with a quotient rule and the other a product rule I believe. Instructions indicate to compute all first order derivatives of the given function.

    Z= xe^xy

    Here is what I did: fx= (x)(ye^xy) + (e^xy)(1) Is this correct?
    I started using the product rule again for (fy) but then I saw an example in the book where they used the constant multiple rule. Why am I supposed to use this rule and not the product rule? The book shows a similar example and they used the product rule to differentiate fx and used the constant multiple rule to differentiate fy, I would just like to know why this is...

    My second set of problems is:

    1. 2x+3y / y-x

    2. e^2-x / y^2

    I know the quotient rule can be used, but I was not sure how to keep y and x as constants. If someone can illustrate the steps involved, Thanks...
     f(x,y)=xe^{xy}

    \frac{\partial f}{\partial x}=(1) \cdot e^{xy}+x(ye^{xy})=(1+xy)e^{xy}

      \frac{\partial f}{\partial y}=xe^{xy}\cdot y=xye^{xy}

     f(x,y)=\frac{2x+3y}{y-x}

    let
     g(x,y)=2x+3y and

     h(x,y)=y-x then

    g_x=2 \mbox{ and } g_y=3 and

     h_x=-1 \mbox{ and } h_y=1

     \frac{\partial f}{\partial x}=\frac{g(x,y)}{h(x,y)}= \frac{h\cdot g_x-g\cdot h_x}{h^2}=\frac{(y-x)(2)-(2x+3y)(-1)}{(y-x)^2}

     = \frac{5y}{(y-x)^2}

    This should get you started.

    Good luck
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  3. #3
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    I am having trouble on this one: f(x,y)= e^(2-x) / y^2

    Here is what I did: fx= (y^2)(1e^2-x) - (e^2-x)(0) / (y^2)^2

    I think I have the second term wrong and I am not sure about the 0 term being multiplied, but if we are considering that y is a constant, than it would go to zero, if I'm not mistaken. what am I doing wrong? Thanks...
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  4. #4
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    Quote Originally Posted by kdogg121 View Post
    I am having trouble on this one: f(x,y)= e^(2-x) / y^2

    Here is what I did: fx= (y^2)(1e^2-x) - (e^2-x)(0) / (y^2)^2

    I think I have the second term wrong and I am not sure about the 0 term being multiplied, but if we are considering that y is a constant, than it would go to zero, if I'm not mistaken. what am I doing wrong? Thanks...
    Remember y is a constant so we can pull it out.

    We don't use the product of quotient rule because it is constant. It is not different then say the number "3"


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


    \frac{\partial f}{\partial x}=\frac{1}{y^2}\frac{d}{dx}e^{2-x}=-\frac{e^{2-x}}{y^2}

    I hope this helps.
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  5. #5
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    I'm sorry, I don't follow what you are saying. If y is a constant, then wouldn't it be 0 when differentiated? Could you illustrate how this would look differentiated because I can't conceptualize it, Thanks...
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    Quote Originally Posted by kdogg121 View Post
    I'm sorry, I don't follow what you are saying. If y is a constant, then wouldn't it be 0 when differentiated? Could you illustrate how this would look differentiated because I can't conceptualize it, Thanks...
    How would you take the derivative of

     f(x)=\frac{e^{2-x}}{3^2}

    Well since "3" is a constant we don't need to use the quotient rule so the derivative is

     \frac{df}{dx}=\overbrace{\frac{(-1)e^{2-x}}{3^2}}^{\frac{d}{dx}(2-x)=(-1)}

    Now lets try with this function Remember y is constant just like the "3" above

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

    \frac{\partial f}{\partial x}=\overbrace{\frac{(-1)e^{2-x}}{y^2}}^{\frac{d}{dx}(2-x)=(-1)}

    I hope this clears it up.
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  7. #7
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    Yes, I understand now, because y is a constant, it doesn't need to be differentiated, so we just leave it as is and differentiate the numerator. Last question, I promise, what would happen when X is a constant and we have fy? Do we use the quotient rule in this case? The book has the answer as fy= -2e^2-x / y^3 I don't see how this was obtained.
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  8. #8
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    Quote Originally Posted by kdogg121 View Post
    Yes, I understand now, because y is a constant, it doesn't need to be differentiated, so we just leave it as is and differentiate the numerator. Last question, I promise, what would happen when X is a constant and we have fy? Do we use the quotient rule in this case? The book has the answer as fy= -2e^2-x / y^3 I don't see how this was obtained.
    Same thing x is a constant so we can rewrite the equation as follows

    <br />
f(x,y)=\frac{e^{2-x}}{y^2}=e^{2-x}y^{-2}<br />

     \frac{\partial f}{\partial y}=\overbrace{e^{2-x}}^{constant}(-2y^{-3})=\frac{-2e^{2-x}}{y^3}

    You could use the quotient rule, but I think this is easier.

    Good luck.
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    If I used the quotient rule would it be this: (y^2) (0) - (e^2-x)(2) / (y^2)^2 The numerator appears to be correct, but the denominator is y^4 and it should be y^3, What am I doing wrong?
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    Quote Originally Posted by kdogg121 View Post
    If I used the quotient rule would it be this: (y^2) (0) - (e^2-x)(2) / (y^2)^2 The numerator appears to be correct, but the denominator is y^4 and it should be y^3, What am I doing wrong?
    f(x,y) =\frac{e^{2-x}}{y^2}

    f_y(x,y)=\frac{y^2(0)-2ye^{2-x}}{y^4}=\frac{-2ye^{2-x}}{y^4}=\frac{-2e^{2-x}}{y^3}

    Just keep practicing
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  11. #11
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    Smile

    Thanks again...
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