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Problem:

$$ \text{Find}\: \frac{\partial{u}}{\partial{x}}\:\text{of}\:\:\:\:\: u=x^u+u^y $$
Would $\frac{\partial{u}}{\partial{x}}$ look like: $$ \frac{\partial{u}}{\partial{x}}=ux^{u-1}\frac{\partial{u}}{\partial{x}}+yu^{y-1}\frac{\partial{u}}{\partial{x}}?$$
Or possibly $$ \frac{\partial{u}}{\partial{x}}=x^u\log\left|x\right|\cdot\frac{\partial{u}}{\partial{x}}+y{u}^{y-1}\frac{\partial{u}}{\partial{x}}?$$
Where am I going wrong? All of the terms can't possibly have a $\frac{\partial{u}}{\partial{x}}$ attached to them or I couldn't separate and solve for it...

bjd2385
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  • Do you intend to apply the partial differential operator to both sides of the equation or partially differentiate wrt to x the expression $x^u+u^y$? – UserX Oct 01 '14 at 22:43
  • @UserX It was my assumption that this would be implicit and I would have to chain-along a $\frac{\partial u}{\partial x}$ every time I took the partial derivative of one of the terms with a $u$ in it... – bjd2385 Oct 01 '14 at 22:55
  • Reread my question. – UserX Oct 01 '14 at 22:57
  • What do you mean by wrt? ... I was simply trying to find the partial derivative of the equation above with respect to x...nothing fancy. I'm new to partial derivatives, trying to nail them down so I apologize if my ignorance is annoying in any way... – bjd2385 Oct 01 '14 at 22:59
  • I'm asking whether want to evaluate(doesn't really make sense but that's what the question explicitly states right now) $$\frac{\partial u}{\partial x} u=\frac{\partial u}{\partial x}(x^u+u^y)$$ or $$\frac{\partial}{\partial x} (x^u+u^y)$$ PS:wrt means with respect to, PS2: The mentality of MSE is to be laconic and direct, don't misunderstand it as being rude. – UserX Oct 01 '14 at 23:08
  • My apologies, I really didn't think you were being rude I'm just afraid my ignorance might make someone mad eventually. On the problem: $$ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial{x}}\left(x^u+u^y\right)$$ I think that is what the book is asking for. The book states it exactly as I have, just find $\frac{\partial u}{\partial x}$... – bjd2385 Oct 01 '14 at 23:20

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