Question

find dy/dx
$log \sqrt{1 - {x}^{2} }$
​

Answer 1

$\huge \red{\mathfrak{solution}}$

To find :

$\frac{d}{dx} ( log( \sqrt{1 - {x}^{2} } )$

Here we use chain rule :

$\rightarrow \: \frac{d}{dx} ( log( \sqrt{1 - {x}^{2} } )) \times \frac{d}{dx} ( \sqrt{1 - {x}^{2} } ) \times \frac{d}{dx} (1 - {x^{2} }) \\ \\ \rightarrow \: \frac{1}{ \sqrt{1 - {x}^{2} } } \times \frac{1}{2 \sqrt{1 - {x}^{2} } } \times (0 - 2x) \\ \\ \rightarrow \: \frac{ - 2x}{2(1 - {x}^{2}) } \\ \\ \rightarrow \: \cancel \frac{2}{ 2} (\frac{ - x}{1 - {x}^{2} } ) \\ \\ \rightarrow \: \frac{ - x}{1 - {x}^{2} } \\ \\ or \\ \\ \rightarrow \: \frac{x}{ {x}^{2} - 1}$

Formula used :

◆

$\boxed{ \purple{ \bold{\frac{d}{dx} log(x) = \frac{1}{x} }}}$

◆

$\boxed{ \bold{ \green{ \frac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }}}$

◆

$\boxed{ \bold{\frac{d}{dx} {x}^{y} = y {x}^{y - 1} }}$

Answer 2

Question :-

$\sf{find \: \: \frac{dy}{dx} } \\ \implies \sf{ log( \sqrt{1 - {x}^{2} } ) }$

Answer :-

$\rightarrow \boxed{\sf{ \frac{dy}{dx} = \frac{ - x}{1 - {x}^{2} } }} \:$

Formula used :-

$\large \sf{\star \: \frac{d}{dx} log(x) = \frac{1}{x} } \\ \\ \large \star \sf{ \frac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }$

Solution :-

$\sf{ \: \: let \: \: y \: = log( \sqrt{1 - {x}^{2} } ) } \\ \\ \sf{ differentiate \: with \: respect \: to \: x} \\ \\ \rightarrow \sf{ \frac{dy}{dx} = \frac{d}{dx} log( \sqrt{1 - {x}^{2} } ) } \\ \\ \rightarrow \sf{ \frac{dy}{dx} = \frac{1}{ \sqrt{1 - {x}^{2} } } \: \frac{d}{dx} \sqrt{1 - {x}^{2} } } \\ \\ \rightarrow \sf{ \frac{dy}{dx} = \frac{1}{ \sqrt{1 - {x}^{2} } } \times \frac{1}{2 \sqrt{1 - {x}^{2} } } \: \frac{d}{dx} (1 - {x}^{2} )} \\ \\ \rightarrow \sf{ \frac{dy}{dx} = \frac{1}{2(1 - {x}^{2} )} (0 - 2x)} \\ \\ \rightarrow \boxed{\sf{ \frac{dy}{dx} = \frac{ - x}{1 - {x}^{2} } }}$

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