Question

if y=log(
$\sqrt{x} + \sqrt{x - a}$
than dy/dx​

Answer 1

Given:

$\star \: y = log( \sqrt{x} + \sqrt{x - a} )$

To find :

$\star \: \frac{dy}{dx}$

Solution:

$\implies \: \frac{d}{dx} ( log( \sqrt{x} + \sqrt{x - a} ) \\ \\ \implies \: \frac{1}{ \sqrt{x} + \sqrt{x - a} } \times \frac{d}{dx} ( \sqrt{x} + \sqrt{x - a} ) \\ \\ \implies \: \frac{1}{ \sqrt{x} + \sqrt{x - a} } \times ( \frac{1}{2 \sqrt{x} } \frac{d}{dx} (x) + \frac{1}{2 \sqrt{x - a} } \frac{d}{dx} (x - a)) \\ \\ \implies \: \frac{1}{ \sqrt{x} + \sqrt{x - a} } \times ( \frac{1}{2 \sqrt{x} } (1) + \frac{1}{2 \sqrt{x - a} } (1 - 0)) \\ \\ \implies \: \frac{1}{ \sqrt{x} + \sqrt{x - a} } \times ( \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{x - a} } )$

Formula:

$\star \bold{ \frac{d}{dx} log(x) = \frac{1}{x} } \\ \\ \star \bold{ \frac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } } \\ \ \\ \star \bold{ \frac{d}{dx} constant = 0}$

Answer 2

Given ,

The function is

• $\tt y = log( \sqrt{x} + \sqrt{x -a } )$

Differentiating y wrt x , we get

$\tt \implies \frac{dy}{dx} = \frac{1}{ \sqrt{x} + \sqrt{x - a} } \frac{d( \sqrt{x} + \sqrt{x - a} }{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{ \sqrt{x} + \sqrt{x - a} } \{ \frac{d( \sqrt{x}) }{dx} + \frac{d( \sqrt{x - a} )}{dx} \}$

$\tt \implies \frac{dy}{dx} = \frac{1}{ \sqrt{x} + \sqrt{x - a} } \{ \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{x - a} } \frac{d(x - a)}{dx} \}$

$\tt \implies \frac{dy}{dx} = \frac{1}{ \sqrt{x} + \sqrt{x - a} } \{ \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{x - a} } \} \{ \frac{d(x )}{dx} - \frac{d(a)}{dx} \} \}$

$\tt \implies \frac{dy}{dx} = \frac{1}{ \sqrt{x} + \sqrt{x - a} } \{ \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{x - a} } \} \{ (1 - 0) \}$

$\tt \implies \frac{dy}{dx} = \frac{1}{ \sqrt{x} + \sqrt{x - a} } \{ \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{x - a} } \}$

Remmember :

★ $\tt \frac{d \{ log(x) \}}{dx} = \frac{1}{x}$

★ $\tt \frac{d {(x)}^{n - 1} }{dx} = n {(x)}^{n - 1}$

★ $\tt \frac{d( \sqrt{x} )}{dx} = \frac{1}{2 \sqrt{x} }$

★ $\tt \frac{d(constant)}{dx} = 0$

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