# Thread: Integral and a slope problem

1. ## Integral and a slope problem

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.

2. Originally Posted by 333arizona
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.
$\int\tan^2(x)+\sin(x)dx$

$\tan^2(x)+1=\sec^2(x)\Rightarrow\tan^2(x)=\sec^2(x )-1$

So $\int\tan^2(x)+\sin(x)dx=\int\sec^2(x)-1+\sin(x)dx=\tan(x)-x-\cos(x)+C$

3. Originally Posted by 333arizona
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.
$3x^2+2xy+6y^2-3x-8y=0$

differntiating implicity we get

$6x+2[xy'+y]+12y\cdot{y'}-3-8y'=0$

Substituting $x=1$ and $y=1$

we get $6(1)+2[y'+1]+12\cdot{y'}-3-8y'=0\Rightarrow{5+6y'=0}\Rightarrow{y'=\frac{-5}{6}}$