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 July 2nd, 2008, 06:38 PM #1 Newbie   Joined: May 2008 Posts: 5 Thanks: 0 using partial derivatives simple but i think its complicated: f(x,y)=sin(xy^2)/x i used quotient rule for first partial of Fx and Fy, but its seems very complicated to do it a second time for Fxy
July 2nd, 2008, 08:27 PM   #2
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Re: using partial derivatives

Quote:
 Originally Posted by tbdevilrays simple but i think its complicated: f(x,y)=sin(xy^2)/x i used quotient rule for first partial of Fx and Fy, but its seems very complicated to do it a second time for Fxy
So just to put in latext $f(x,y)f=f=\frac{\sin(xy^2)}{x}$

So then

$\frac{\partial{f}}{\partial{x}}=\frac{x\cdot\frac{ \partial[\sin(xy^2)]}{\partial{x}}-\frac{\partial[x]}{\partial{x}}\sin(xy^2)}{x^2}=\frac{xy^2\cos(xy^2 )-\sin(xy^2)}{x^2}$

and then

$\frac{\partial{f}}{\partial{y}}=\frac{\cos(xy^2)}{ x}\cdot(2yx)=2y\cos(xy^2)$

 July 2nd, 2008, 09:17 PM #3 Newbie   Joined: May 2008 Posts: 5 Thanks: 0 Re: using partial derivatives [quote="Mathstud28"][quote="tbdevilrays"]simple but i think its complicated: f(x,y)=sin(xy^2)/x $\frac{\partial{f}}{\partial{x}}=\frac{x\cdot\frac{ \partial[\sin(xy^2)]}{\partial{x}}-\frac{\partial[x]}{\partial{x}}\sin(xy^2)}{x^2}=\frac{xy^2\cos(xy^2 )-\sin(xy^2)}{x^2}$ ok, i got there, then how would you take the second partial derivative of y?
July 9th, 2008, 04:34 PM   #4
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Re: using partial derivatives

Quote:
Originally Posted by tbdevilrays
Quote:
Originally Posted by Mathstud28
Quote:
 Originally Posted by tbdevilrays simple but i think its complicated: f(x,y)=sin(xy^2)/x $\frac{\partial{f}}{\partial{x}}=\frac{x\cdot\frac{ \partial[\sin(xy^2)]}{\partial{x}}-\frac{\partial[x]}{\partial{x}}\sin(xy^2)}{x^2}=\frac{xy^2\cos(xy^2 )-\sin(xy^2)}{x^2}$ ok, i got there, then how would you take the second partial derivative of y?
Just take the derivative in respect to y holding x as a constant

 July 13th, 2008, 08:34 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,753 Thanks: 2136 $\frac{\partial{f}}{\partial{y}}=2y\cos(xy^2),$ so $\frac{\partial^2f}{\partial{x}\partial{y}}=-2y^3\sin(xy^2).$ Differentiating with respect to y the expression found for $\frac{\partial{f}}{\partial{x}}$ gives the same result.

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