Answer
$$w'\left( x \right) = - \left( {\sin x} \right)\sec \left( {{x^2} - 1} \right) + 2x\left( {\cos x} \right)\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)$$
Work Step by Step
$$\eqalign{
& w\left( x \right) = \cos x\sec \left( {{x^2} - 1} \right) \cr
& {\text{Differentiate}} \cr
& w'\left( x \right) = \frac{d}{{dx}}\left[ {\cos x\sec \left( {{x^2} - 1} \right)} \right] \cr
& {\text{By the product rule}} \cr
& w'\left( x \right) = \sec \left( {{x^2} - 1} \right)\frac{d}{{dx}}\left[ {\cos x} \right] + \cos x\frac{d}{{dx}}\left[ {\sec \left( {{x^2} - 1} \right)} \right] \cr
& {\text{Computing derivatives}} \cr
& w'\left( x \right) = \sec \left( {{x^2} - 1} \right)\left( { - \sin x} \right) + \cos x\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)\left( {2x} \right) \cr
& {\text{Simplifying}} \cr
& w'\left( x \right) = - \left( {\sin x} \right)\sec \left( {{x^2} - 1} \right) + 2x\left( {\cos x} \right)\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right) \cr} $$
