#### Answer

$$\,\,\,\frac{{d\left( {{x^{\cos x}}} \right)}}{{dx}} = {x^{\cos x}}\left( {\frac{{\cos x}}{x} - \left( {\sin x} \right)\ln x} \right)$$

#### Work Step by Step

$$\eqalign{
& \frac{d}{{dx}}\left( {{x^{\cos x}}} \right) \cr
& {\text{let }}y = {x^{\cos x}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \,\,\,\ln y = \ln {x^{\cos x}} \cr
& {\text{using logarithmic properties}} \cr
& \,\,\,\ln y = \cos x\ln x \cr
& {\text{differentiate both sides with respect to }}x \cr
& \,\,\,\frac{d}{{dx}}\left( {\ln y} \right) = \frac{d}{{dx}}\left( {\cos x\ln x} \right) \cr
& {\text{product rule}} \cr
& \,\,\,\frac{d}{{dx}}\left( {\ln y} \right) = \cos x\frac{d}{{dx}}\left( {\ln x} \right) + \ln x\frac{d}{{dx}}\left( {\cos x} \right) \cr
& {\text{compute derivatives}} \cr
& \,\,\,\frac{1}{y}\frac{{dy}}{{dx}} = \cos x\left( {\frac{1}{x}} \right) + \ln x\left( { - \sin x} \right) \cr
& \,\,\,\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{\cos x}}{x} - \left( {\sin x} \right)\ln x \cr
& \,\,\,\frac{{dy}}{{dx}} = y\left( {\frac{{\cos x}}{x} - \left( {\sin x} \right)\ln x} \right) \cr
& {\text{replace }}y{\text{ with }}{x^{\cos x}} \cr
& \,\,\,\frac{{d\left( {{x^{\cos x}}} \right)}}{{dx}} = {x^{\cos x}}\left( {\frac{{\cos x}}{x} - \left( {\sin x} \right)\ln x} \right) \cr} $$