Question

If y = 1/x+x

, then find dy/dx at x= 1​

Answer 1

Step-by-step explanation:

$y = \frac{1}{x} + x \\ \\ = > \frac{dy}{dx} = - x {}^{ - 2} + 1 \\ \\ = > \frac{dy}{dx} (x = 1) = - \frac{1}{ {1}^{2} } + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 1+ 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0 \\ \\ using \: formula \: \: \: - \: \: \: \: \: \: \: \frac{d}{dx} ( \frac{1}{ {x}^{2} } ) = \frac{d}{dx} ( {x}^{ - 2} ) = - 2.x ^{ - 2 - 1} = - 2x {}^{ - 3} \\ \\ \frac{d}{dx} ( {x}^{n} ) = n. {x}^{n - 1}$