How do you differentiate #f(x) = 1/sqrt(arctan(2x^3) # using the chain rule?

1 Answer
Nov 5, 2016

Answer:

#f'(g(x))g'(x) = 1/2(arctan(2x^3))*(6x^2)/(1+(2x^3)^2)#

Explanation:

first, take the derivative of the denominator using chain rule:
#f=sqrt(x)#
#f'=1/2x^(-1/2)#
#g=arctan(2x^3)#
#g'=(6x^2)/(1+(2x^3)^2)#

#f'(g(x))g'(x) = 1/2(arctan(2x^3))^(-1/2)*(6x^2)/(1+(2x^3)^2)#

that's it! if you want to simplify it further you can but my calc teacher only requires this.

note: recognize that the derivative of arctan has the chain rule built into it already if that makes sense, as #d/dx[arctan(u)]=(u')/(1+u^2)#