How do I find the derivative of #y= x arctan (2x) - (ln (1+4x^2))/4#?

1 Answer
Jan 28, 2015

You can use the Product Rule and Chain Rule to solve your problem.
1) Consider first #xarctan(2x)#; you have the product of #x# and #arctan(2x)# and the function of a function, #arctan(2x)# which requires the chain rule (derive the #arctan# and then multiply by the derivative of the argument #2x#).
You get:
#y'=1*arctan(2x)+x*1/(1+(2x)^2)*2#
2) Secondly you have the #-ln(1+4x^2)/4# bit. Here, again, you have the chain rule to deal with the #ln# of a function (the argument #1+4x^2#):
You get:
#y'=-1/4*1/(1+4x^2)*8x=-(2x)/(1+(2x)^2)#

Collecting the two parts you get:
#y'=arctan(2x)+(2x)/(1+(2x)^2)-(2x)/(1+(2x)^2)=arctan(2x)#

Hope it helps