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

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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,

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

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