How do you differentiate # f(x) =arcsin(2x + 1) #?

1 Answer

It is #f'(x)=2/(sqrt(1-(2x+1)^2))#

Explanation:

Let # u(x) = 2x + 1#, function f may be considered as the composition #f(x) = arcsin(u(x))#. Hence we use the chain rule,
#f '(x) = ((du)/dx) (d(arcsin(u)))/(du)#,
to differentiate function f as follows

#f'(x)=((2x+1)/dx)*(1/(sqrt(1-u^2))) => f'(x)=2/(sqrt(1-(2x+1)^2))#