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)]#