Answer
$w^{\prime}(x)= 2x \sec x \sec^2 (x^{2}-1)+\sec x \tan x \tan (x^{2}-1)$
Work Step by Step
We have: $w(x)= \sec x \tan (x^{2}-1)$
We differentiate both sides with respect to $x$.
$w^{\prime}(x)=\dfrac{d}{dx} [ \sec x \tan (x^{2}-1)] \\= \sec x \sec^2 (x^{2}-1) \times (2x)+\sec x \tan x \tan (x^{2}-1)$
Simplify to obtain:
$w^{\prime}(x)= 2x \sec x \sec^2 (x^{2}-1)+\sec x \tan x \tan (x^{2}-1)$
