Answer
$-\frac{ \cos ^{3} (1/x) }{x^2}.$
Work Step by Step
By making use of Theorem 1 (FTC II), we have
$$
\frac{d}{d x} \int_{1}^{1 / x} \cos ^{3} t d t=\left(\frac{d}{d u} \int_{0}^{u} \cos ^{3} t d t\right)\frac{du}{dx}= \cos ^{3} (1/x) \ (-\frac{1}{x^2})\\
=-\frac{ \cos ^{3} (1/x) }{x^2}.
$$
