derivatives: find dy/dx for x^2 y^(-3) + 3 = y

[Tascja]

Find dy/dx for:

x^2y^-3 + 3 = y

I tried this so many times and i still cant get it can someone show me the steps plz

 

[stapel]

I tried this so many times and i still cant get it....

We'll be glad to help you get un-stuck, but you'll need to show your work.

Please be complete. Thank you.

Eliz.

 

[Tascja]

where im at so far...

x^2y^-3 + 3 = y

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

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

and now i dont know what to do

 

[galactus]

Hello Tascja:

Don't forget the product rule when doing imp. der.


You have:

\(\displaystyle \L\\x^{2}y^{-3}+3-y=0\)

\(\displaystyle \L\\\frac{-3x^{2}}{y^{4}}\frac{dy}{dx}+\frac{2x}{y^{3}}-\frac{dy}{dx}=0\)

Now, solve for dy/dx?.

 

[Tascja]

how do i divided out after:

dy/dx (-3x^2/y^4 - 1) = -2x/y^3