Answer
$$\frac{dy}{dx}=x^3\sec\left(\frac{1}{x}\right)\left(5x-\tan\left(\frac{1}{x}\right)\right)$$
Work Step by Step
$$\frac{dy}{dx}=\frac{d}{dx}(x^5\sec(\frac{1}{x}))=5x^4\sec(\frac{1}{x})+x^5\cdot\sec(\frac{1}{x})\tan(\frac{1}{x})\cdot(\frac{1}{x})'=
5x^4\sec(\frac{1}{x})+x^5\sec(\frac{1}{x})\tan(\frac{1}{x})\cdot(-\frac{1}{x^2})=5x^4\sec(\frac{1}{x})-x^3\sec(\frac{1}{x})\tan(\frac{1}{x})=
x^3\sec(\frac{1}{x})(5x-\tan(\frac{1}{x})$$
