Question

7. Find dy/dx when
y=1/x2+1/√x​

Answer 1

Given:

$\rm y = \dfrac{1}{ {x}^{2} } + \dfrac{1}{ \sqrt{x} }$

To Find:

$\rm \dfrac{dy}{dx}$

Answer:

$\rm \implies \dfrac{dy}{dx} \\ \\ \rm \implies \dfrac{d}{dx} \bigg( \dfrac{1}{ {x}^{2} } + \dfrac{1}{ \sqrt{x} } \bigg)\\ \\ \rm \implies \dfrac{d}{dx} ( {x}^{ - 2} + {x}^{ - \frac{1}{2} } ) \\ \\ \rm By \: using \: power \: rule, \\ \rm \dfrac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1} : \\ \\ \rm \implies - 2 {x}^{ - 2 - 1} - \dfrac{1}{2} {x}^{ - \frac{1}{2} - 1 } \\ \\ \rm \implies - 2 {x}^{ - 3} - \dfrac{1}{2} {x}^{ - \frac{3}{2} } \\ \\ \rm \implies - \dfrac{2}{ {x}^{3} } - \dfrac{1}{2 {x}^{ \frac{3}{2} } } \\ \\ \rm \implies - \dfrac{2}{ {x}^{3} } - \dfrac{1}{2 \sqrt{ {x}^{3} } } \\ \\ \rm \implies - \bigg(\dfrac{2}{ {x}^{3} } + \dfrac{1}{2 \sqrt{ {x}^{3} } } \bigg)$

$\therefore$ $\boxed{\mathfrak{\dfrac{dy}{dx} = - \bigg(\dfrac{2}{ {x}^{3} } + \dfrac{1}{2 \sqrt{ {x}^{3} } } \bigg) }}$