How do you differentiate $1/cos(x) = x/(x-y^2-y)$ ?

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Answer 1 · 2016-11-03T17:50:02

$dy/dx=-(-y^2-y-(x-y^2-y)^2sinxsec^2x)/(2xy+x)$ or $dy/dx=(y^2+y+(x-y^2-y)^2sinxsec^2x)/(2xy+x)$

$1/cosx = x/(x-y^2-y)$

First put everything on one side

$0=x/(x-y^2-y)-1/cosx$

Then find the derivative with respect to x let's call it $f_x$ while holding y constant

$f_x=((x-y^2-y)-x)/(x-y^2-y)^2+1/cos^2x*-sinx$

$=(x-y^2-y-x)/(x-y^2-y)^2-sinxsec^2x$

$=(-y^2-y-(x-y^2-y)^2sinxsec^2x)/(x-y^2-y)^2$

Next find the derivative with respect to y call it $f_y$ and hold x contant

$f_y=(-x(-2y-1))/(x-y^2-y)^2$

$=(2xy+x)/(x-y^2-y)^2$

To write your final answer use the formula $dy/dx=-f_x/f_y$

$dy/dx=-((-y^2-y-(x-y^2-y)^2sinxsec^2x)/(x-y^2-y)^2)/((2xy+x)/(x-y^2-y)^2)$

$dy/dx=-(-y^2-y-(x-y^2-y)^2sinxsec^2x)/(2xy+x)$

$dy/dx=(y^2+y+(x-y^2-y)^2sinxsec^2x)/(2xy+x)$

How do you differentiate 1/cos(x) = x/(x-y^2-y)1/cos(x) = x/(x-y^2-y)?