find slope of the line at 3x^2+2xy+6y^2-3x-8y=0 at (1,1) I'm so lost, this review is killing me *leafs through notebook*
also, the integral of tan^2x+sinx , my brain is melting before finals. Help would be much appreciated.
$\displaystyle 3x^2+2xy+6y^2-3x-8y=0$
differntiating implicity we get
$\displaystyle 6x+2[xy'+y]+12y\cdot{y'}-3-8y'=0$
Substituting $\displaystyle x=1$ and $\displaystyle y=1$
we get $\displaystyle 6(1)+2[y'+1]+12\cdot{y'}-3-8y'=0\Rightarrow{5+6y'=0}\Rightarrow{y'=\frac{-5}{6}}$