Answer
$$y = \cosh \left( 1 \right)x - \cosh \left( 1 \right) + 1$$
Work Step by Step
$$\eqalign{
& y = {x^{\cosh x}},{\text{ }}\left( {1,1} \right) \cr
& {\text{Using logarithmic differentiation}} \cr
& \ln y = \ln {x^{\cosh x}} \cr
& \ln y = \left( {\cosh x} \right)\left( {\ln x} \right) \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {\left( {\cosh x} \right)\left( {\ln x} \right)} \right] \cr
& {\text{Use product rule}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \cosh x\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {\cosh x} \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \cosh x\left( {\frac{1}{x}} \right) + \ln x\left( {\sinh x} \right) \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left[ {\cosh x\left( {\frac{1}{x}} \right) + \ln x\left( {\sinh x} \right)} \right] \cr
& {\text{Calculate the slope at the given point }}\left( {1,1} \right) \cr
& m = \left( 1 \right)\left[ {\cosh \left( 1 \right)\left( {\frac{1}{1}} \right) + \ln \left( 1 \right)\left( {\sinh \left( 1 \right)} \right)} \right] \cr
& m = \left[ {\cosh \left( 1 \right) + 0} \right] \cr
& m = \cosh \left( 1 \right) \cr
& {\text{Calculate the equation of the tangent line at the point }}\left( {1,1} \right) \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = \cosh \left( 1 \right)\left( {x - 1} \right) \cr
& y - 1 = \cosh \left( 1 \right)x - \cosh \left( 1 \right) \cr
& y = \cosh \left( 1 \right)x - \cosh \left( 1 \right) + 1 \cr} $$
