For this particular, we'd have to use logarithmic differentiation, which works as follows:
Let #y = (ln x)^(cos x)#
Taking the natural log (#ln#) of both sides yields
#ln y = ln((ln x)^(cos x))#
#ln y = cos x * ln(ln x)#
Since the next step is to take derivatives, the rules we're going to use is
#d/dx[ln u] = (u')/(u)#
Differentiating both sides gives
#(y')/(y) = -sin x * ln(ln x) + cos x * ((1/x) * (1/ln x))/(((cancel(ln x))/1) * (1/cancel(ln x)))#
#(y')/(y) = -sin x * ln(ln x) + (cos x)/(x ln x)#
But since we're looking for #y'#, we simply move #y# to the right side of our equation to get
#y' = y[-sin x * ln(ln x) + (cos x)/(x ln x)]#
Substituting #y# back into the equation yields
#y' = (ln x)^(cos x)[-sin x * ln(ln x) + (cos x)/(x ln x)]#
