Help with Rational Fractions Please

[hardiger]

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!

 

[ChaoticLlama]

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

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

 

[Denis]

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!

What is it you want done? Simplify?
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!

 

[hardiger]

Thanks Llama,

I thought that was right, but the answer to the whole problem is wrong. Here is the entire problem, please help me understand.

find the derivative of ln(xy)=y^3+1 using implicit differentiation

Thanks

 

[Guest]

Equations don't have derivatives. I think you mean find dy/dx.

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)

 

[soroban]

Hello, hardiger!

Given: .ln(xy) = y³ + 1. . Find y' using implicit differentiation.
.

. . . . . . . . . . (1/xy)(xy' + y) .= .3y²y'

Multiply by xy: . . . . xy' + y .= .3xy³y'

And we have: . 3xy³y' - xy' .= .y

Factor: . . . . . . .(3xy³ - x)y' .= .y

. . . . . . . . . . . . . . . . . . . . . . . . . . . y
. . . . . . . Therefore: . . . . y' .= .-----------
. . . . . . . . . . . . . . . . . . . . . . . . .3xy³ - x

 

[Guest]

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