1. ## Partial Derivatives

f(x,y)=x^2y + xsiny -ln[x-y^2]

Find fy and fxy

I found fy but to get xy you take the x derivative, which I found to be 2xy+siny-1/(x-y^2) and then take the y. The answer is 2x-xsiny-2y/(x-y^2)^2

This is a practice midterm problem and I messed up somewhere because I can't get xsiny from siny. Could someone walk me through this or point me to where I went wrong in calculating fxy?

2. Originally Posted by sfgiants13
f(x,y)=x^2y + xsiny -ln[x-y^2]

Find fy and fxy

I found fy but to get xy you take the x derivative, which I found to be 2xy+siny-1/(x-y^2) and then take the y. The answer is 2x-xsiny-2y/(x-y^2)^2

This is a practice midterm problem and I messed up somewhere because I can't get xsiny from siny. Could someone walk me through this or point me to where I went wrong in calculating fxy?
All I can say is that the "answer", $\displaystyle 2x- xsiny- 2y/(x- y^2)^2$ is wrong.

You can do this either of two ways. Since you were asked to find $\displaystyle f_y$ and had already found that $\displaystyle f_y= x^2+ x cos(y)+\frac{2y}{x-y^2}$ just differentiate that with respect to x: $\displaystyle f_{yx}= 2x+ cos(y)- \frac{2y}{(x- y^2)^2}$.

Or, first find $\displaystyle f_x= 2xy+ siny- \frac{1}{x-y^2}$ and differentiate that with respect to y: $\displaystyle f_{xy}= 2x+ cos y- \frac{2y}{(x- y^2)^2}$.

(As long as the second derivatives are continuous, the "mixed derivatives", $\displaystyle f_{xy}$ and $\displaystyle f_{yx}$, are equal.)

3. Originally Posted by HallsofIvy
All I can say is that the "answer", $\displaystyle 2x- xsiny- 2y/(x- y^2)^2$ is wrong.

You can do this either of two ways. Since you were asked to find $\displaystyle f_y$ and had already found that $\displaystyle f_y= x^2+ x cos(y)+\frac{2y}{x-y^2}$ just differentiate that with respect to x: $\displaystyle f_{yx}= 2x+ cos(y)- \frac{2y}{(x- y^2)^2}$.

Or, first find $\displaystyle f_x= 2xy+ siny- \frac{1}{x-y^2}$ and differentiate that with respect to y: $\displaystyle f_{xy}= 2x+ cos y- \frac{2y}{(x- y^2)^2}$.

(As long as the second derivatives are continuous, the "mixed derivatives", $\displaystyle f_{xy}$ and $\displaystyle f_{yx}$, are equal.)
That's what I got actually but the solutions sheet gave the wrong answer apparently. Thanks!