# Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6 - Page 158: 29

$$\frac{dy}{dx}=x^3\sec\left(\frac{1}{x}\right)\left(5x-\tan\left(\frac{1}{x}\right)\right)$$

#### Work Step by Step

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

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