how are you

how can I derivative that equation for y : $M=2x+ye^{xy}$

I know the answer is $ye^{xy}x+e^{xy}$

but I need to details, I need to all steps



To get the second partial derivative $\frac{\partial f}{\partial y}$ of your $C^1$ function $f$, all you have to do is to see $x$ as a constant.Then it comes $$\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}(2x+ye^{xy})=\underset{=0}{\underbrace{\frac{\partial}{\partial y}(2x)}}+\frac{\partial}{\partial y}(ye^{xy}),$$ and by using the product's derivation rule you get $$\frac{\partial}{\partial y}(ye^{xy})=\frac{\partial}{\partial y}(y)\times e^{xy}+y\times\frac{\partial}{\partial y}(e^{xy})=e^{xy}+yxe^{xy}$$ which is the answer you are looking for.

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  • $\begingroup$ can you write it step by step? $\endgroup$ – user283144 Oct 25 '15 at 19:59
  • $\begingroup$ Okay I'll make appear other steps ! $\endgroup$ – Balloon Oct 25 '15 at 20:01

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