- Thread starter hardiger
- Start date

- Joined
- Dec 11, 2004

- Messages
- 199

first deal with the numerator

y²x

___

xy

divide out one x and on y

and now the fraction looks like

....y

_____

3y - x

and that look like all that should be done

What is it you want done? Simplify?hardiger said:Thank you for receiving this email. I need help with the following problem:

{[(y^2)x]/xy}/(3y-x)

Thank you for getting back to me quickly!

Simplify is all that can be done: it's NOT an equation.

Simplifying:

(y^2)x / (xy) = y

leaves:

y / (3y - x) : that's it; can't be further simplified.

Why were you having a problem with that?

Edit: you're faster than me, CL!

G

ln(xy) = y^3 + 1

Differentiate both sides with respect to x.

d/dx ln(xy) = d/dx (y^3 + 1)

1 / (xy) d/dx xy = d/dx y^3 + d/dx 1

1 / (xy) * y = 3y^2 dy/dx

1 / x = 3y^2 dy/dx

Now solve for dy/dx.

dy/dx = 1 / (3xy^2)

. . . . . . . . . . (1/xy)(xy' + y) .= .3y<sup>2</sup>y'Given: .ln(xy) = y<sup>3</sup> + 1. . Find y' using implicit differentiation.

.

Multiply by xy: . . . . xy' + y .= .3xy<sup>3</sup>y'

And we have: . 3xy<sup>3</sup>y' - xy' .= .y

Factor: . . . . . . .(3xy<sup>3</sup> - x)y' .= .y

. . . . . . . . . . . . . . . . . . . . . . . . . . . y

. . . . . . . Therefore: . . . . y' .= .-----------

. . . . . . . . . . . . . . . . . . . . . . . . .3xy<sup>3</sup> - x

G

Whoops, d/dx xy isn't y because y isn't constant with respect to x. My bad.