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)#