Question

differentiate y = 1/✓2x​

Answer 1

Answer:

dude ur question is not clear whether the x will be in root or not, but I did both refer to the attachment

Explanation:

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Answer 2

Given :

$\sf y = \dfrac{1}{ \sqrt{2x} }$

To find :

$\sf \: \dfrac{dy}{dx}$

Solution :

$\sf \implies y = \dfrac{1}{ \sqrt{2x} }$

or

$\sf \implies y = \bf\dfrac{ {x}^{ \frac{ - 1}{2} } }{ \sqrt{2} }$

Now differentiating both sides :-

$\sf \implies \dfrac{dy}{dx} = \dfrac{d(\dfrac{ {x}^{ \frac{ - 1}{2} } }{ \sqrt{2} } )}{dx}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{1}{\sqrt{2}} \times \dfrac{d({ {x}^{ \frac{ - 1}{2} } })}{dx}$

We know that :-

$\underline{ \boxed{ \bf \dfrac{d({ {x}^{ n} })}{dx} = n {x}^{n - 1} } }$

Therefore :-

$\sf \implies \dfrac{dy}{dx} = \dfrac{1}{\sqrt{2}} \times \dfrac{ - 1}{2} \times ({ {x}^{ \frac{ - 1}{2} - 1 } })$

$\sf \implies \dfrac{dy}{dx} = \dfrac{ - 1}{2\sqrt{2}} \times ({ {x}^{ \frac{ - 3}{2} } })$

$\sf \implies \dfrac{dy}{dx} = \dfrac{ - { {x}^{ \frac{ - 3}{2} } } }{2\sqrt{2}}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{ - 1 }{2\sqrt{2}{ \: \: {x}^{ \frac{ 3}{2} } }}$

Answer :

$\sf \dfrac{ - 1 }{2\sqrt{2}{ \: \: {x}^{ \frac{ 3}{2} } }}$