Find dy/dx in terms of x and y for:xy = tan(xy)

I'm not too good with this so pls dont skip steps pls.

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- Apr 25th 2009, 06:28 AMkangarooFind dy/dx in terms of x and y
Find dy/dx in terms of x and y for:

**xy = tan(xy)**

I'm not too good with this so pls dont skip steps pls. - Apr 25th 2009, 07:15 AMkangaroo
anyways, i've gotten this far and need help

xy=tan(xy)

d(xy)/dx = d(tanxy)/dx

x * dy/dx +y * dx/dx = sec^2(xy) * (d(xy)/dx)

1 = sec^2(xy)

1 = 1/(cos^2(xy))

cos^2(xy) = 1 - Apr 25th 2009, 07:24 AMderfleurer
$\displaystyle xy = tan(xy)$

$\displaystyle xy' + y = sec^2(xy)(xy' + y)$ // chain rule

$\displaystyle xy' + y = sec^2(xy)xy' + sec^2(xy)y$ // expand

$\displaystyle xy' - sec^2(xy)xy' = sec^2(xy)y - y$ // move terms to opposite sides

$\displaystyle y'(1 - sec^2(xy)) = sec^2(xy)y - y$ // factor out y'

$\displaystyle y' = \frac {sec^2(xy)y - y}{1 - sec^2(xy)}$ // divide - Apr 25th 2009, 08:09 AMkangaroo
thanks!

btw, x has been left out of the factorizing line.

which when corrected, and factorised again after dividing, renders y' = -y/x

thanks again!!