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How do you differentiate #f(x)=xarctansqrtx#?

1 Answer

Apr 4, 2018

#=>arctan(sqrtx) + sqrt(x)/(2(1+x))#

Explanation:

#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))#

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