Solve Z = x^2y - 2xy^2 + y - 8 and y = 2x^2 +3/x!

  • Thread starter Becky
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  • #1
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hay everyone....need some help on the question :cry:

I need to find dz/dx in terms of x and y, if:

Z = x^2y - 2xy^2 + y - 8 and y = 2x^2 +3/x!

tried solving it a few times, but its a difficult one! any help and working out would be great! :confused:
 

Answers and Replies

  • #2
Hurkyl
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We'd be better able to help if you show us what you've done.
 
  • #3
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Ok dz/dx

x = 2y - 4y +y

y = 4x + 3

dont think that is right!
 
  • #4
quasar987
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You are given z as a function of x and y, and y as a function of x. So I suggest that you substitute all the y int he equation of z for their value in terms of x and then calculate dz/dx.
 
  • #5
HallsofIvy
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Use the chain rule. Since y is a function of x, dz/dx= 2xy+ x2y'- 2y2+ 4xy6y' + y'. since y = 2x2 +3/x, you can find y' as a funcion of x and substitute.
 
  • #6
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hey,
i think i can help,dz/dx=x^2y[dy/dx(log[x^2])+2y/x]-2y^2-4xydy/dx+dy/dx
where dy/dx=4x - {3/x!}[1/x+1/x-1+1/x-2+1/x-3.......]
 
  • #7
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hey
hallsofivy ur step was wrong since y is a function of x cant differentiate like that
check ur solution one more time,ur step will be right if the question reads x^2 times y ,but if it is x raised to the power of 2y then mine is correct.also i hope the 6 in the function 4xy6y' was a typing mistake cos if it is not then ur answer is wrong.
 

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