Answer
$$g'\left( x \right) = \frac{1}{{{x^2}}}\cos \left( {\frac{1}{x}} \right) - \frac{1}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \int_\pi ^{1/x} {\left( {1 - \cos t} \right)} dt \cr
& {\text{Differentiate }} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\int_\pi ^{1/x} {\left( {1 - \cos t} \right)} dt} \right] \cr
& {\text{Using the formula }}\left( {18} \right){\text{ }}\frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( t \right)} dt} \right] = f\left( {g\left( x \right)} \right)g'\left( x \right) \cr
& g'\left( x \right) = \left( {1 - \cos \left( {\frac{1}{x}} \right)} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr
& g'\left( x \right) = \frac{1}{{{x^2}}}\cos \left( {\frac{1}{x}} \right) - \frac{1}{{{x^2}}} \cr
& \cr
& {\text{Evaluating the integral}} \cr
& g\left( x \right) = \int_\pi ^{1/x} {\left( {1 - \cos t} \right)} dt = \left[ {t - \sin t} \right]_\pi ^{1/x} \cr
& g\left( x \right) = \left( {\frac{1}{x} - \sin \left( {\frac{1}{x}} \right)} \right) - \left( {\pi - \sin \left( \pi \right)} \right) \cr
& g\left( x \right) = \frac{1}{x} - \sin \left( {\frac{1}{x}} \right) - \pi \cr
& {\text{Differentiating}} \cr
& g'\left( x \right) = - \frac{1}{{{x^2}}} - \cos \left( {\frac{1}{x}} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr
& g'\left( x \right) = \frac{1}{{{x^2}}}\cos \left( {\frac{1}{x}} \right) - \frac{1}{{{x^2}}} \cr} $$
