Answer
$$y = x\sin \left( 1 \right) - \sin \left( 1 \right) + 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{\sin x}};\,\,\,\,\,\,{\text{at the point }}x = 1 \cr
& {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& f\left( x \right) = {e^{\sin x\ln x}} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = {e^{\sin x\ln x}}\frac{d}{{dx}}\left[ {\sin x\ln x} \right] \cr
& {\text{Use the product rule}} \cr
& f'\left( x \right) = {x^{\sin x}}\left[ {\sin x\left( {\frac{1}{x}} \right) + \ln x\left( {\cos x} \right)} \right] \cr
& {\text{Multiply}} \cr
& f'\left( x \right) = \frac{{{x^{\sin x}}\sin x}}{x} + {x^{\sin x}}\ln x\cos x \cr
& \cr
& {\text{Evaluate the function at }}x = 1 \cr
& f\left( 1 \right) = {1^{\sin 1}} \cr
& f\left( 1 \right) = 1 \cr
& {\text{Calculate the slope at }}x = 1 \cr
& f'\left( 1 \right) = \frac{{{1^{\sin 1}}\sin 1}}{1} + {1^{\sin 1}}\ln 1\cos 1 \cr
& f'\left( 1 \right) = \sin \left( 1 \right) \cr
& {\text{The equation of the tangent line is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = \sin \left( 1 \right)\left( {x - 1} \right) \cr
& y = x\sin \left( 1 \right) - \sin \left( 1 \right) + 1 \cr} $$
