Question

compute all first order partial derv. of z=(xy^2)/(x^2y^3+1)

Answer 1

Well remember, when doing these you want to pick a variable to stay as a variable and pretend all the other variables in the equation are constants. So you would do 2 quotient rules, one with respect to x treating y as a constant and the other with respect to y treating x as a constant. So the first quotient rule, with respect to x, would be:
[y^2(x^2y^3+1)] - [(xy^2)(2xy^3]
(x^2y^3 + 1)^2
And then you could simplify if needed. Think you'd be able to do the same with y, treating x as a constant?

Answer 2

so i can put any number in for z and y

Answer 3

If it gave you an initial condition of some sort, then sure.

Answer 4

Without an actual value given for y or z, you kinda just let those variables tag along as you would with any constant, though.

Answer 5

the problem didnt sat

Answer 6

Then just leave the variables as is. If I have something like x^3y^3 and I want the partial derivative with respect to x, then that y^3 will just be some random constant and will sit there and tag along for the ride, making the parital derivative 3x^2y^3

Answer 7

so is this the final answer[y^2(x^2y^3+1)] - [(xy^2)(2xy^3]
(x^2y^3 + 1)^2

Answer 8

For the partial derivative in respect to x, yes.

Answer 9

and than i have to do it to y

Answer 10

Yes. So you would do the same thing, but pretend x is some random constant that tags along and only do derivatives of y.

Answer 11

so x^2(x^2y^3+1)-xy^2(2xy^40
(x^2y^3+1)^2

Answer 12

Not quite. It is a little unusual pretending those variables are constants, it can trip ya up for sure, even myself. So let's try and be careful this time. So starting with the numerator, I want the derivative of y. So if I ignore x temporarily, the derivative would be 2y. The x, being a constant, is just still sitting there. So the left side of the numerator doing quotient rule will be 2yx(x^2y^3+1). Now I will have - the numerator times the derivative of the denominator with respect to y. So the denominator, ignoring x temporarily, will be 3y^2 withthe constant of + 1 going away. Now Ill just tack back on that x^2, since it was just a constant and tags along anyway. So now the right portion of that numerator is -[(xy^2)(3y^2x^2)]. Putting everything together, including the denominator squared, I have
2yx(x^2y^3+1) - [(xy^2)(3y^2x^2)]
(x^2y^3+1)^2