What is the derivative of #arctan sqrt x#?

1 Answer
Aug 29, 2015

You can do it two ways. If you remember the actual derivative:

#d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))#

#d/(dx)[arctansqrtx] = 1/(1+x)*1/(2sqrtx) = color(blue)(1/(2sqrtx(1+x)))#

Or, you can implicitly derive it.

#y = arctansqrtx#

#tany = sqrtx#

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

#(1 + tan^2(arctansqrtx))((dy)/(dx)) = 1/(2sqrtx)#

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

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