The derivative of the expression #x.e^(3x)+tan^-1(2x)#
Knowing that :
#(u+v)'=u'+v'# (1)
#(e^u)'=u'e^u# (2)
#(tan^-1(u))'=(u')/(1+u^2)# (3)
#(u.v)'=u'v+v'u#. (4)
Lets find the derivative of #x.e^(3x)#:
#color(blue)(x.e^(3x))'#
#=x'e^(3x)+x.(e^(3x))'# applying above formula (4)
#=e^(3x)+x.3.e^(3x)# applying the above formula (2)
#color(blue)(=e^(3x)+3xe^(3x). name it (5))#
Now let's find the derivative of #tan^-1(2x)#
#color(blue)((tan^-1(2x)))'# applying the above formula (3)
#=((2x)')/(1+(2x)^2)#
#color(blue)(=2/(1+4x^2) name it (6))#
The derivative of the sum #x.e^(3x)+tan^-1(2x)# is :
#color(red)((x.e^(3x)+tan^-1(2x))')#
#=(x.e^(3x))'+(tan^-1(2x))'#. applying the above formula (1)
#color(red)(=e^(3x)+3xe^(3x)+2/(1+4x^2)#substituting (5) and (6)
