Posted by **Angela** on Monday, October 8, 2012 at 3:34pm.

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

- Calculus -
**Steve**, Monday, October 8, 2012 at 4:14pm
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

- Calculus -
**Steve**, Monday, October 8, 2012 at 4:15pm
oops that's 1-x^2y^2 throughout

- Calculus -
**Angela**, Monday, October 8, 2012 at 4:25pm
when I plug in the ordered pair (.25, 2)I get a zero in the denominator, was I doing that wrong?

- Calculus -
**Steve**, Monday, October 8, 2012 at 4:37pm
√(1-x^2y^2) = 0, all right, but it's not in the denominator.

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

- Calculus -
**Steve**, Monday, October 8, 2012 at 4:42pm
oops: -2/.25 = -8

- Calculus - final correction -
**Steve**, Monday, October 8, 2012 at 5:02pm
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!

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