Did you mean the derivative of (cos x)^x ? Then, use logarithmic differentiation:
y = (cos x)^x
ln(y) = ln[(cos x)^x]
ln(y) = x * ln[cos x]
(1/y) * y' = 1* ln[cos x] + x * (-sin x)/(cos x)
y' / y = ln[cos x] - (x sin x)/(cos x)
y' = y* [ ln[cos x] - (x sin x)/(cos x) ]
the key step in the process - the substitution.
You had: y = (cos x)^x, so we will substitute for y.
y' = (cos x)^x* [ ln[cos x] - (x sin x)/(cos x) ]
(and this is your final answer)