How do you differentiate #y=(arctan(x))^(1/2)#?

1 Answer
Jim G.
Nov 7, 2017

#dy/dx=1/(2(1+x^2)(arctanx)^(1/2)#

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))#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
2466 views around the world
You can reuse this answer
Creative Commons License