Thread: Find dy/dx Need help again

$\displaystyle x^4+3x^2y^3-y^2=5$

Thanks!

Just follow the same thought process as the last problem you posted.

$\displaystyle x^4+3x^2y^3-y^2=5$
$\displaystyle (4x^3)+[(3x^2)(3y^2dy/dx)+(y^3)(6x)]-2ydy/dx=0$
$\displaystyle (4x^3)+(9x^2y^2)dy/dx+(6xy^3)-(2y)dy/dx=0$
$\displaystyle (4x^3)+(6xy^3)=(2y)dy/dx-(9x^2y^2)dy/dx$
$\displaystyle (4x^3+6xy^3)/2y-9x^2y^2)=dy/dx$

Thanks for the help Pink!

sorry i havnt figured out how to put a set of numbers over another set, so the / is the divide line.

Correct.

To have a fraction like $\displaystyle \frac{a}{b}$, use \frac{a}{b} inside the [tex] tags.

I use {a\over b} instead of frac
This one can be solved by explicit differentiation.
Let $\displaystyle w=x^2$ and then solve the quadractic equation in w.