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

  • Thread starter Becky
  • Start date
  • #1
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
We'd be better able to help if you show us what you've done.
  • #3
Ok dz/dx

x = 2y - 4y +y

y = 4x + 3

dont think that is right!
  • #4
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
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
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
hallsofivy ur step was wrong since y is a function of x can't 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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