find the derivative and simplify f(x) = xln(arctanx) would the answer be 2x^{4} + 2X^{2}+ ln(arctanx)
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Use the product rule, for $\displaystyle y = v\times u \implies y' = vu'+uv'$ In your case $\displaystyle v= x , u = \ln(\arctan x) $ use the chain rule to get $\displaystyle u'$
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