How do you differentiate y=(4/ln2)(x^(lnx))(lnx+2ln(cosx))?

1 Answer
Oct 23, 2015

y'=4/ln2 [2x^(lnx-1)lnx(lnx+2ln(cosx))+x^(lnx-1)-2x^(lnx)tanx]

Explanation:

Lets find (x^(lnx))' :

g=x^(lnx)

lng = ln x^(lnx)

lng = lnxlnx = ln^2x

1/g*g' = 2lnx*1/x

g' = 2lnx*1/x * g

g' = 2lnx*1/x * x^(lnx) = 2x^(lnx-1)lnx

y=(4/ln2)(x^(lnx))(lnx+2ln(cosx))

y'=4/ln2 [2x^(lnx-1)lnx(lnx+2ln(cosx))+x^(lnx)(1/x-2/cosxsinx)]

y'=4/ln2 [2x^(lnx-1)lnx(lnx+2ln(cosx))+x^(lnx-1)-2x^(lnx)tanx]