arcsinxy = 2/3 arctan4x find dy/dx at the point (1/4, 2)

Answer 1

using implicit differentiation,

(xy'+y)/√(1+x^2y^2) = 8/3 * 1/(1+16x^2)

a little rearranging of terms yields

y' = 8√(1+x^2y^2) / (3x(1+16x^2)) - y/x

Answer 2

oops that's 1-x^2y^2 throughout

Answer 3

when I plug in the ordered pair (.25, 2)I get a zero in the denominator, was I doing that wrong?

Answer 4

√(1-x^2y^2) = 0, all right, but it's not in the denominator.

So, all you end up with is -y/x = -16

Answer 5

oops: -2/.25 = -8

Answer 6

Rats! I keep being inconsistent in the squaring! √(1-x^2y^2) is not zero.

at (1/4,2)

y' = 8√(1-1/4)/(3*1/4*2) - 2/.25
= (8√6)/3 - 8

judging from my other posts, you'd better double-check this one too!