Calculus
posted by Angela .
arcsinxy = 2/3 arctan4x find dy/dx at the point (1/4, 2)

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

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

√(1x^2y^2) = 0, all right, but it's not in the denominator.
So, all you end up with is y/x = 16 
oops: 2/.25 = 8

Rats! I keep being inconsistent in the squaring! √(1x^2y^2) is not zero.
at (1/4,2)
y' = 8√(11/4)/(3*1/4*2)  2/.25
= (8√6)/3  8
judging from my other posts, you'd better doublecheck this one too!
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