Question

DIFFERENTIATE IT w.r.t x​

Answer 1

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Given that,

$\sf{y = ( \sqrt{x} - \frac{1}{ \sqrt{x} }) {}^{2} }$

Of the form,

(a - b)² = a² - 2ab + b²

Now,

$\sf{y = ( \sqrt{x }){}^{2} \: + \: ( \frac{1}{ \sqrt{x} } ) {}^{2} \: - \: 2. \sqrt{x}. \frac{1}{ \sqrt{x} } } \\ \\ \implies \: \boxed{\sf{y = x \: + \: \frac{1}{x} - 2 }}$

Differentiating y w.r.t to x,we get:

$\sf{ \frac{dy}{dx} = \frac{d(x + \frac{1}{x} - 2 )}{dx} }$

Using Chain Rule,

$\implies \: \sf{ \frac{dy}{dx} = \frac{d(x)}{dx} \: + \: \frac{d(x {}^{ - 1}) }{dx} + \frac{d( - 2)}{dx} } \\ \\ \implies \: \: \boxed{\sf{ \frac{dy}{dx} = 1 - x {}^{ - 2} }}$

Answer 2

Given:

$\sf y={(\sqrt{x}-\dfrac{1}{\sqrt{x}})}^2$

To find:

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

Explanation:

$\longrightarrow \: \sf y={(\sqrt{x}-\dfrac{1}{\sqrt{x}})}^2 \:$

let us expand it using the formula

$\boxed{\tt (a-b)^2=a^2+b^2-2ab}$

$\longrightarrow \: \sf \: y = x + \dfrac{1}{x} - 2$

Now,

Differentiating y with respect to x

$\longrightarrow \: \sf \dfrac{dy}{dx} = \dfrac{dx}{dx} + \dfrac{d}{dx} ( \dfrac{1}{x} ) - 0$

Differentiation of constant=0

$\boxed{\sf \dfrac{d}{dx}(\dfrac{1}{x})=\dfrac{-1}{x^2}}$

$\longrightarrow \: \sf \dfrac{dy}{dx} =1 + ( \dfrac{ - 1}{ {x}^{2} }) - 0 \\ \\ \longrightarrow \: \sf \dfrac{dy}{dx} =1 - \dfrac{1}{ {x}^{2} } \\ \\ \longrightarrow \: \boxed{\sf \dfrac{dy}{dx} = \dfrac{ {x}^{2} - 1}{ {x}^{2} } }$