# Find dy/dx in terms of x and y

• Apr 25th 2009, 06:28 AM
kangaroo
Find 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 AM
kangaroo
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 AM
derfleurer
$\displaystyle xy = tan(xy)$

$\displaystyle xy' + y = sec^2(xy)(xy' + y)$ // 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 AM
kangaroo
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!!