#f(x)# is a composite of two functions #color(blue)(g(x)=arcsinx)# and #color(brown)(h(x)=e^(3x))#
We have: #f(x)=2g@h(x)#
Then chain rule is applied: #color(red)(f'(x)=2*g'(h(x))*h'(x)#
Let us compute #color(red)(g'(h(x)))#
#color(blue)(g(x)=arcsinx)#
#color(blue)(g'(x))=1/(sqrt(1-x^2)#
Then,
#color(red)(g'(h(x)))=1/(sqrt(1-(h(x))^2)#
#color(red)(g'(h(x)))=1/(sqrt(1-color(brown)((e^(3x))^2)#
#color(red)(g'(h(x)))=1/(sqrt(1-color(brown)(e^(6x)#
Let us compute #color(red)(h'(x))#
#color(brown)(h(x)=e^(3x))#
Knowing that the derivative of exponential function is as follows:
#(e^(u(x)))'=u'(x)e^(u(x))#
#color(red)(h'(x))=(e^(3x))'#
#color(red)(h'(x))=3e^(3x)#
#color(red)(f'(x)=2*g'(h(x))*h'(x)#
#color(red)(f'(x))=2*1/(sqrt(1-color(brown)(e^(6x))))*3e^(3x)#
#color(red)(f'(x))=(6e^(3x))/(sqrt(1-color(brown)(e^(6x))))#
