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)
