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

[Becky]

hay everyone...need some help on the question

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!

 

[Hurkyl]

We'd be better able to help if you show us what you've done.

 

[Becky]

Ok dz/dx

x = 2y - 4y +y

y = 4x + 3

dont think that is right!

 

[quasar987]

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.

 

[HallsofIvy]

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.

 

[abia ubong]

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...]

 

[abia ubong]

hey
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.