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 May 25th, 2013, 03:07 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Second derivatives of implicit function Hi, I can't solve this question and would like help. Please explain your answer, as this is hard for me to understand how to derive an implicit function, was never explained to me. Question: let $f(t)$ be a single variable function, that's derivable 2 times. We are given $u(x,y)=f(e^{x}+cosy)$ show that the following equality is true: $(\frac{d^{2}u}{dx^{2}} - \frac{du}{dx})siny= -e^{x}\frac{d^{2}u}{dxdy}$ Please explain the question and the answer...thanks.
 May 25th, 2013, 11:58 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Second derivatives of implicit function I would start with evaluating the terms from LHS and the RHS apart. by chain rule: $\frac{\partial u}{\partial x}= f#39;(e^x + \cos(y)) \cdot e^x$ by product- and chain rule $\frac{\partial}{\partial x} \left( \frac{\partial u}{\partial x} \right)= \frac{\partial^2 u}{\partial x^2} = f'(e^x + \cos(y)) e^x + f'#39;(e^x + \cos(y)) \cdot e^{2x}$ by chain rule again $\frac{\partial}{\partial y} \left( \frac{\partial u}{\partial x} \right)= \frac{\partial^2 u}{\partial x \partial y} = f'#39;(e^x + \cos(y)) \cdot e^{x} \cdot -\sin(x)$ this gives $\sin(y) \cdot ( \frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial x})= f'#39;(e^x + \cos(y)) \cdot e^{2x} \cdot \sin(y) = -e^x \cdot \frac{\partial^2 u}{\partial x \partial y}$
 May 25th, 2013, 02:56 PM #3 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Second derivatives of implicit function Thank you. That was very clear.

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