2

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To start I find the derivative with respect to fx = 2y + x

derivative with respect to fy = 2 x - e^x/((e^x + 2)^2 (1/(e^x + 2)^2 + 1)) + y

Then to find the answer multiply fx by fy.

However the correct answer is 1 and apparently it can be found without writing anything down. How is that possible?

sawreals2
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  • You don't have to multiply, it's a derivative of order 2. Note you can also write $\frac{\partial ^2 f}{\partial y \partial x} f(x, y)$ – userr777 Apr 20 '18 at 20:10

1 Answers1

1

Take advantage that the arctan function depends only on x

when you differentiate according to y it equals zero

So we have that:

$$g(x,y)=\arctan(\frac 1 {e^x+2})+x^2+xy+y^2$$ differentiate wrt x

$$\partial_x g=\partial_x (\arctan(\frac 1 {e^x+2})) +2x+y$$ differentiate wrt y $$\partial_{yx} g=0+0+1=1$$

user577215664
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  • Where has arctan disappeared ? – sawreals2 Apr 20 '18 at 20:06
  • Well it's $f(x)$ we take advantage by the fact that it depends only on x...when we differentiate wrt y its equal zero @sawreals2 – user577215664 Apr 20 '18 at 20:07
  • Where does it say that arctan is f(x)?? – sawreals2 Apr 20 '18 at 20:08
  • look at the function it depends only on x – user577215664 Apr 20 '18 at 20:10
  • Yes there is an x in the arctan portion in the form of e^x – sawreals2 Apr 20 '18 at 20:11
  • when you differentiate wrt y it's zero then – user577215664 Apr 20 '18 at 20:11
  • Sure I understand that, but when you are differentiating wrt x you have simply written it as h(x) instead of actually finding the derivative. – sawreals2 Apr 20 '18 at 20:13
  • @sawreals2 i ahve edited hope it's more clear now...dont calculate the arctan derivative take advantage it will be zero – user577215664 Apr 20 '18 at 20:18
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    To summarize, because you know that eventually when you go to calculate the partial derivative of y after partial derivative wrt x that the arctan portion will be 0 you do not bother to find its derivative? Ok, I think that makes sense. Technically it would still differentiate in the step wrt x but since it will eventually end up as 0 wrt y it is just an extra step that is not necessary. – sawreals2 Apr 20 '18 at 20:21
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    @sawreals2 you got it now....no need to evaluate the derivative its a function of x ...and we end diffrentiating wrt y...so it's zero...no need to calculate – user577215664 Apr 20 '18 at 20:23