How do you find the derivative #f(x)=arctansqrtx#?

Question

995 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-24T19:22:42

The answer is #=1/(2(1+x)sqrtx)#

We need

#tan^2x+1=sec^2x#

#(tanx)'=sec^2x#

Let #y=arctan(sqrtx)#

#:.tan y=sqrtx#

#(tan y)'=(sqrtx)'#

#sec^2ydy/dx=1/(2sqrtx)#

#dy/dx=1/(sec^2y*2sqrtx)#

#sec^2y=1+tan^2y==1+x#

So,

#dy/dx=1/(2(1+x)sqrtx)#