Question

answer this anybody​

Answer 1

$\boxed{\textbf{\large{Step-by-step explanation}}}$

$\sqrt{ \frac{y}{x} } + \sqrt{ \frac{x}{y} } = 2$

$\\ \frac{dy}{dx} = ?$

$\sqrt{ \frac{x}{y} } + \sqrt{ \frac{y}{x} } = 2$

$\frac{ \sqrt{ {x}^{2} } + \sqrt{ {y}^{2} } }{ \sqrt{xy} } = 2$

$\sqrt{ {x}^{2} } + \sqrt{ {y}^{2} } = 2 \sqrt{xy }$

◾Squaring on both sides ,

$x^2 + 2xy + y^2 = 4xy$

◾[here we have used $( a + b) ^2 = a^2 + 2ab + b^2$ identity for the squaring of $(\sqrt{ {x}^{2} } + \sqrt{ {y}^{2} })$ ]

⟹ $x^2 + y^2 = 4xy - 2xy$

⟹ $x^2 + y^2 = 2xy$

⟹ $x^2 + y^2 = xy + xy$

⟹ $x^2 - xy = xy - y^2$

⟹ $x (x - y)= y ( x - y)$

⟹ $\frac{(x - y)}{(x - y) }=\frac{y}{x}$

⟹ $\frac{y}{x} = 1$

Therefor ,

$y = x$

◾Now, differentiate wrt x ,we get

$\frac{dy}{dx} = \frac{d}{dx} ( x)$

$\frac{dy}{dx} = 1$

$\boxed{\textbf{\large{ dy /dx = 1}}}$

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Derivatives of some standard composite functions :

1] y =[ f(x) ]^n

➡ dy / dx = n[f(x)]^(n - 1)x f'(x )

2] y = sin f (x)

➡dy / dx = cos f (x) x f' (x)

3] y = √ ( f (x ) )

➡dy / dx =( 1 / 2 √[ f(x)]) x f'(x )

4] y = k

➡ dy / dx = 0 [ Note : it is not a composite function ]

5] y = cos f (x )

➡ dy /dx = -sin f (x) x f'(x )

6] y = e^([f (x) ])

➡dy /dx = e^([f (x) ]) x f'(x)

7] y = a^[f(x)]

➡ dy /dx = a^[ f (x) ] x log a x f'(x)

8] y = log [f (x) ]

➡dy /dx =( 1 / f (x) ) x [f'(x) ]

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