Question #b8bdb

1 Answer
Feb 28, 2018

#dy/dx = ( 2x)/(absx(1+x^2))#

Explanation:

Let:

#cosy = (1-x^2)/(1+x^2)#

where #y in [0,pi)#

Then, as for #y in [0,pi]# the sine is positive:

#siny = sqrt(1-cos^2y) = sqrt(1- (1-x^2)^2/(1+x^2)^2)#

#siny = sqrt(((1+x^2)^2- (1-x^2)^2)/(1+x^2)^2)#

#siny = sqrt((1+ 2x^2+x^4- 1+2x^2 -x^4)/(1+x^2)^2)#

#siny = sqrt( 4x^2)/(1+x^2)#

#siny = (2absx)/(1+x^2)#

Now differentiate implicitly:

#-siny dy/dx = ( -2x(1+x^2)-2x(1-x^2))/(1+x^2)^2#

#siny dy/dx = ( 4x)/(1+x^2)^2#

#dy/dx = 1/siny ( 4x)/(1+x^2)^2#

#dy/dx = (1+x^2)/(2absx)( 4x)/(1+x^2)^2#

#dy/dx = ( 2x)/(absx(1+x^2))#