How do you differentiate #y=(arctan(x))^(1/2)#?
1 Answer
Nov 7, 2017
Explanation:
#"differentiate using the "color(blue)"chain rule"#
#"given "y=f(g(x))" then"#
#dy/dx=f'(g(x))xxg'(x)larr"chain rule"#
#rArrdy/dx=1/2(arctanx)^(-1/2)xxd/dx(arctanx)#
#color(white)(rArrdy/dx)=1/2(arctanx)^(-1/2)xx1/(1+x^2)#
#color(white)(rArrdy/dx)=1/(2(1+x^2)(arctanx)^(1/2))#