How do you differentiate $f (x) = x arctan \sqrt{x}$ $f(x)=xarctansqrtx$ ?

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Answer 1 · 2018-04-04T04:52:21

$\Rightarrow arctan (\sqrt{x}) + \frac{\sqrt{x}}{2 (1 + x)}$ $=>arctan(sqrtx) + sqrt(x)/(2(1+x))$

$f (x) = x arctan (\sqrt{x})$ $f(x) = x arctan (sqrtx)$

$f'(x) = (d(x))/(dx) arctan(sqrtx) + x (d(arctan(sqrtx)))/(dx)$

$= arctan(sqrtx) + x(1/(1+(sqrtx)^2))(d(sqrtx))/(dx)$

$= arctan(sqrtx) + x/(1+x)*1/(2sqrtx)$

$= arctan(sqrtx) + sqrt(x)/(2(1+x))$

How do you differentiate f(x)=xarctansqrtxf(x)=xarctan√xf(x)=xarctansqrtx?