# Math Help - derivatives question

1. ## derivatives question

$sin(y)=yx^2-x$

Find $y'$

how do i do this?! @_@

2. You'll have to use implicit differentiation.

3. so $sin(y)=yx^2-x$

then for $y'$

$cos(y)(dy/dx)=(x^2(dy/dx)+2xy)-1$

$cos(y)(dy/dx)-x^2(dy/dx)=2xy-1$

$y' = (2xy-1)/cos(y)-x^2$

uhh doesn't look right... but is it right? lol

4. Originally Posted by break
so $sin(y)=yx^2-x$

then for $y'$

$cos(y)(dy/dx)=(x^2(dy/dx)+2xy)-1$

$cos(y)(dy/dx)-x^2(dy/dx)=2xy-1$

$y' = (2xy-1)/cos(y)-x^2$

uhh doesn't look right... but is it right? lol

It's correct