Question

Hey !!!

iit level question ..

Solve , step by step solution needed ✍

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Best of luck ✌

Answer 1

Hey friend, Harish here.

Here is your answer:

Applying Quotient Rule we get:

$\frac{d}{dy} \bigl( \frac{sinx - xcosx}{xsinx + cosx} \bigr )$

⇒ $\frac{ \bigl [ \bigl (\frac{d}{dx} \bigl (sinx -xcosx \bigr ) \bigr ) \bigl (xsinx + cosx \bigr ) - \bigl( \frac{d}{dx} \bigl (xsinx + cosx \bigr ) \bigr ) \bigl (sinx - xcosx \bigr ) \bigr] }{(xsinx + cosx)^2 }$

⇒  $\frac{ \bigl [xsinx \bigl ( xsinx + cosx \bigr ) - xcosx \bigl (sinx - xcosx \bigr) \bigr ] }{(xsinx + cosx)^2 }$

⇒ $\frac{\bigl [x^2 sin^2x + (xsinx)(cosx) - (sinx)(xcosx) + x^2 cos^2x \bigr ]}{(xsinx + cosx)^2 }$

⇒ $\frac{x^2 (sin^2x + cos^2x) }{(xsinx + cosx)^2 }$

⇒ $\bold{ \boxed{ \frac{x^2}{(xsinx + cosx)^2 } } \ option\ d\ is\ the\ correct \ answer}.$
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Hope my answer is helpful to you.

Answer 2

$\bf HELLO \: \: DEAR,$

$\boxed {\bf \frac{d(\frac{sinx - xcosx}{xsinx + cox})}{dx}}$


$\it {\bf \frac{d(\frac{sinx - xcosx}{xsinx + cox})}{dx}} = \bf \frac{[(xsinx + cosx)* \frac{d}{dx}(sinx - xcosx) - (sinx - xcosx) * \frac{d}{dx}(xsinx + cosx) ]}{(xsinx + cosx)^{2}}$



$\to \: \bf \frac{[(xsinx + cosx) * (cosx + xsinx - cosx) - (sinx - xcosx) * (xcosx + sinx - sinx)]}{(xsinx + cosx)^{2}}$


$\to \bf \frac{[(xsinx + cosx) * (xsinx) - (sinx - xcosx) * (xcosx)]}{(xsinx + cosx)^{2}}$


$\to \bf \frac{[(xsinx)^{2} + (xsinxcosx) - (xsinxcosx) + (xcosx)^{2}]}{(xsinx + cosx)^{2}}$

$\to \bf \frac{x^{2}(sin^{2}x + cos^{2}x)]}{(xsinx + cosx)^{2}}$
$\: \: \: \: \: \: \: \: \: \: \: \therefore \boxed{ \underline { \bf (sin^{2}x + cos^{2}x) = 1}}$


$\to \bf \frac{x^{2}}{(xsinx + cosx)^{2}}$


$\boxed { \bf HENCE, \: \: OPTION \: \: (D) \: \: IS \: \: CORRECT}$


$\underline { \bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}$