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

Question

Charles Arocha · Accepted Answer

#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)#

Christopher Agresta · Answer

undefined To differentiate the given equation $ \frac{1}{\cos(x)} = \frac{x}{x - y^2 - y} $ with respect to $ x $, we use the quotient rule:

$$ \frac{d}{dx} \left( \frac{1}{\cos(x)} \right) = \frac{(x - y^2 - y) \frac{d}{dx}(1) - 1 \cdot \frac{d}{dx}(\cos(x))}{(x - y^2 - y)^2} $$

Differentiating $ \frac{1}{\cos(x)} $ yields $ \frac{\sin(x)}{\cos^2(x)} $ and $ \frac{d}{dx}(\cos(x)) = -\sin(x) $.

Plugging these derivatives back into the quotient rule formula:

$$ \frac{(x - y^2 - y) \cdot 0 - 1 \cdot (-\sin(x))}{(x - y^2 - y)^2} = \frac{\sin(x)}{\cos^2(x)} \cdot \frac{1}{x - y^2 - y} $$