Answer
$\frac{dy}{dx} = (sec (\frac{1}{x} + 7x) \times tan (\frac{1}{x} + 7x))\times (\frac{1}{x^{2}} -7)$
Work Step by Step
$\frac{dy}{du} = \frac{d (-sec u)}{du} = -sec u\times tan u$
and,
$\frac{du}{dx} = \frac{d (\frac{1}{x} + 7x)}{dx}$
$\frac{du}{dx} = \frac{-1}{x^{2}} +7$
So,
$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
$\frac{dy}{dx} = (-sec u\times tan u)(\frac{-1}{x^{2}} +7)$
$\frac{dy}{dx} = (-sec (\frac{1}{x} + 7x) \times tan (\frac{1}{x} + 7x))\times (\frac{-1}{x^{2}} +7)$
$\frac{dy}{dx} = (sec (\frac{1}{x} + 7x) \times tan (\frac{1}{x} + 7x))\times (\frac{1}{x^{2}} -7)$