Help me find derivative of 9 ⋅ x^3 ⋅ y^2 − 3 ⋅ x^2 ⋅ y^3 = 0

[hahol]

Help me find derivative of 9 * x^3 * y^2 - 3 * x^2 * y^3 = 0

I must solve the A and B in the next formula:




But I get the answer like this:

But the answer can't be A = -18y and B = 6x.

Can anybody help and explain this?

Thanks!

 

[Dr.Peterson]

I must solve the A and B in the next formula:

View attachment 35087


But I get the answer like this: View attachment 35088

But the answer can't be A = -18y and B = 6x.

Can anybody help and explain this?

Thanks!

Please show the entire problem (and translate it to English). I expect that you were given an equation relating x and y.

 

[MarkFL]

I get the same value for y'.

 

[lookagain]

I must solve the A and B in the next formula:

hahol,

you need to have gotten \(\displaystyle \ \dfrac{dy}{dx} \ \) to look close to this: \(\displaystyle \ \ \dfrac{2xy^3 - 9x^2y^2}{6x^3y - 3x^2y^2} \ \ \ or \ \ \ \dfrac{- (9x^2y^2 - 2xy^3)}{6x^3y - 3x^2y^2}.\)

What would you have left inside the parentheses after you would factor out \(\displaystyle \ xy^2 \ \) (or, depending on which form, \(\displaystyle \ -xy^2)\)
from the numerator, and then \(\displaystyle \ yx^2 \ \)from the denominator?

 

[Dr.Peterson]

I must solve the A and B in the next formula:

View attachment 35087


But I get the answer like this: View attachment 35088

But the answer can't be A = -18y and B = 6x.

Can anybody help and explain this?

Thanks!

Now that I've noticed that you put the equation in the title (which is a bad practice, because it is surprisingly easy to overlook), I see that you did too well. The question is a poor one, as I see it, because it asks for a specific incompletely simplified form, while you did what we normally encourage, and simplified it fully. I very much dislike this sort of problem.

As @lookagain suggested, look back at your work and consider what it looked like before simplifying; then do only the specific factoring they show, without canceling (other than, possibly, a 3). Or, if you prefer, just deliberately "unsimplify" what you got, by multiplying top and bottom by xy.

But also, can you tell us where you got your -18y and 6x? Those suggest you did some illegal combining.

 

[Steven G]

Just multiply the numerator and denominator by xy.

 

[stapel]

Now that I've noticed that you put the equation in the title (which is a bad practice, because it is surprisingly easy to overlook), I see...

Question copied from subject line back into post.

Eliz.

 

[hahol]

I must find y' yes and write what valuables could be A and B in the solution.

 

[Dr.Peterson]

I must find y' yes and write what valuables could be A and B in the solution.

Have you tried following our suggestions? Please show whatever additional work you have done.

Do you understand that A and B will be expressions, not just variables or multiples of variables?

 

[Steven G]

A will be whatever xy^2 is being multiplied by in the numerator.
B will be whatever x^2y is being multiplied by in the denominator.

 

[farhan]

To find the derivative of the equation 9x^3y^2 - 3x^2y^3 = 0 with respect to x, we can use the chain rule and product rule of differentiation.

Taking the derivative of both sides with respect to x, we get:

27x^2y^2 + 18x^3y(dy/dx) - 6x^2y^3(2xy') = 0

Simplifying this expression, we get:

27x^2y^2 + 18x^3y(dy/dx) - 12x^3y^2(dy/dx) = 0

Now, solving for dy/dx, we get:

dy/dx = (9x^2y^2)/(4x^3y - 3x^2y^2)

Therefore, the derivative of the equation 9x^3y^2 - 3x^2y^3 = 0 with respect to x is (9x^2y^2)/(4x^3y - 3x^2y^2).

 

[lookagain]

To find the derivative of the equation 9x^3y^2 - 3x^2y^3 = 0 ...

27x^2y^2 + 18x^3y(dy/dx) - 6x^2y^3(2xy') = 0

The second line of yours in the quote box I showed is not correct, and from thereon, your post is wrong. The derivative of \(\displaystyle \ -3x^2y^3 \ \) is not the third term
in the second line. It will be two terms. One of the terms of the derivative will
have a dy/dx factor, and the other term will not have it.