Derivatives of powers secant functions

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Find the derivative $\;\;\;\frac{d}{dx}sec^{8}(2x-2)$

**Answer**

$\;\;\;\frac{d}{dx}sec^{8}(2x-2)$

$=(8)\,sec^{7}(2x-2)\,(sec(2x-2)\,tan(2x-2))\,(2)$

$=16\,sec^{8}(2x-2)\,tan(2x-2)$