Answer
$\dfrac{(2x-1) \sec \sqrt {x^2-x} \tan (\sqrt {x^2-x}}{ 2 \sqrt {x^2-x}}$
Work Step by Step
Our aim is to solve the function $f(x)=\sec \sqrt {x^2-x}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx}[\sec \sqrt {x^2-x}]$
or, $=\sec \sqrt {x^2-x} \tan (\sqrt {x^2-x}) \dfrac{d}{dx} [ \sqrt {x^2-x}]$
or, $=\sec \sqrt {x^2-x} \tan (\sqrt {x^2-x}) \times \dfrac{1}{2} (x^2-x)^{-1/2} \times (2x-1)$
or, $=\dfrac{(2x-1) \sec \sqrt {x^2-x} \tan (\sqrt {x^2-x}}{ 2 \sqrt {x^2-x}}$
