Question

Find the derivative of
√x + 1/√x
by the first principle

Answer 1

Question :-

$\rm find \: the \: derivative \: of \: \sqrt{x} + \frac{1}{ \sqrt{x} } \: \: by \\ \rm \: \: first \: perinciple \:$

Answer :-

$\to \boxed{ \rm \: \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } - \frac{1}{2x \sqrt{x} } } \:$

Solution :-

We have and let ,

$\to \rm \: y = \sqrt{x} + \frac{1}{ \sqrt{x} } \\ \\ \bf \: differentiate \: with \: respect \: to \: x \\ \\ \to \rm \: \frac{dy}{dx} = \frac{d}{dx} \sqrt{x} + \frac{d}{dx} \frac{1}{ \sqrt{x} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \rm \frac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } } \\ \\ \to \rm \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } + \frac{d}{dx} \: \frac{1}{ {x}^{ \frac{1}{2} } } \\ \\ \to \rm \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } + \frac{d}{dx} {x}^{ \frac{ - 1}{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \rm \because \: \frac{d}{dx} {x}^{n} = n {x}^{n - 1} } \\ \\ \to \rm \: \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } - \frac{1}{2} {x}^{ \frac{ - 1}{2} - 1 } \\ \\ \to \rm \: \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } - \frac{1}{2} {x}^{ \frac{ - 3}{2} } \\ \\ \bf we \: can \: write \: it \: \\ \\ \to \boxed{ \rm \: \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } - \frac{1}{2x \sqrt{x} } }$