2cosx - 6siny y' - 6cos^2(y) + 6sin^2(x) y' + 1 = 0
at (3pi,7pi/2) we have
2(-1) - 6(-1)y' - 6(0) + 6(0)y' + 1 = 0
-2 + 6y' + 1 = 0
6y' = 1
y' = 1/6
Find the slope of the tangent line to the curve
2sin(x) + 6cos(y) - 6sin(x)cos(y) + x = 3pi
at the point (3pi, 7pi/2)
Thank you very much for your help.
2 answers
Oops. Make that
2cosx - 6siny y' - 6cosx cosy + 6sinx siny y' + 1 = 0
2(-1) - 6(-1)y' - 6(-1)(0) + 6(0)y' + 1 = 0
-2 + 6y' + 1 = 0
y' = 1/6
got the right answer before, but only because all the wrong stuff evaluated to 0!
2cosx - 6siny y' - 6cosx cosy + 6sinx siny y' + 1 = 0
2(-1) - 6(-1)y' - 6(-1)(0) + 6(0)y' + 1 = 0
-2 + 6y' + 1 = 0
y' = 1/6
got the right answer before, but only because all the wrong stuff evaluated to 0!