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))))
