anonymous
  • anonymous
(xlny+xy) if i differentiate that with respect to y... i think i get 0
Mathematics
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anonymous
  • anonymous
(xlny+xy) if i differentiate that with respect to y... i think i get 0
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
When you differentiate with respect to y, all you do is treat the \(x\) variable as a constant. Since \(\ln{y}\) doesn't differentiate to zero, and neither does \(y\), you shouldn't get zero.
anonymous
  • anonymous
ok wat do i get
anonymous
  • anonymous
cn you show me

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anonymous
  • anonymous
how to do it den
anonymous
  • anonymous
So the derivative of \(\ln{y}\) is \(\frac{1}{y}\), and the derivative of \(y\) is \(1\). Does that help get you moving?
myininaya
  • myininaya
He didn't say partial derivative
anonymous
  • anonymous
i got the answer x/y+x
myininaya
  • myininaya
This could be implicit differiention
anonymous
  • anonymous
is that rite
anonymous
  • anonymous
Yep, that's right. Unless myininaya is right about it being implicit differentiation. Are you learning about implicit differentiation now, or partial derivatives?
myininaya
  • myininaya
If we aren't doing partial differentiation, the answer is x'lny+x/y+x'y+x
anonymous
  • anonymous
implicit
myininaya
  • myininaya
Since x is a function of y
myininaya
  • myininaya
Have you heard of partial derivatives?

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