Partial Derivatives…?

You are told that there is a function f whose partial derivatives are fx(x,y)=x+4y and fy(x,y) =3x − y. Do you believe it? If so, what is the function f? If not, explain.

I have no idea what this is asking...our teacher barely covered partial derivatives, let alone how to go back and figure them out.

alwbsok, thank you very much! I had f(x,y) but not with the g(y)! I did not think to differentiate that with respect to y afterwards to see if it would be equal to what it was initially stated. Thank you again!

3 Answers

  • Ted s is right, but, IMHO, a little unclear. The question is asking, "Is it possible to find a function f with those partial derivatives?" The answer is no.

    In order to prove this, you have to integrate a partial derivative, either fx by x or fy by y. Let's say fx:

    fx(x, y) = x + 4y

    f(x, y) = (1/2)x^2 + 4xy + g(y)

    Notice the extra g(y) at the end. This is necessary, because, when differentiating, there could be whole functions of the other variable differentiated to 0. Basically, our "constant" of integration is now our "function" of integration.

    Now that we have f(x, y), we may differentiate it with respect to y.

    fy(x, y) = d/dy f(x, y)

    = d/dy ((1/2)x^2 + 4xy + g(y))

    = 4x + g'(y)

    If f can exist, we need it to satisfy the expression for the y derivative, that is:

    3x - y = 4x + g'(y)

    g'(y) = -y - x

    But, this is a contradiction, since g'(y) is a function of y, not x. So f cannot exist, i.e. no function can ever have those particular partial derivatives.

  • deriviative is 4x-(5xy'+5y+xy) use product rule for 5xy, use x's easily spinoff and write spinoff of y as y' and go away it on my own after that. in case you didnt understand product rule is derivitive of one term circumstances the different plus the derivitive of the different term circumstances the 1st term thats frustrating to understand so der. of xy is x'y+xy' that is frequently written as y+xy' in terms of x on the grounds that spinoff of x is a million you additionally can use the chain rule after pulling an x out so as that it somewhat is x(2x-5y) wich then you could use the product rule besides and get (2x-5y)+x(2+5y') that is 4x -5y+2x+5y' which with a splash extra paintings you are able to instruct equals the spinoff of the 1st occasion i confirmed you.

  • from your given δf / δx find f(x,y)...don't forget the constant of integration C(y)

    now compute δf/δy using your work answer and see if it can be the same as the given δf/δy....{no}

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