What is the derivative of $arctan sqrt x$ ?

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Answer 1 · 2015-08-29T07:28:24

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.

$y = arctansqrtx$

$tany = sqrtx$

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

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

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

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

What is the derivative of arctan sqrt xarctan sqrt x?