Question

differentuate sin( 1-x^2/1+x^2)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = sin\bigg(\dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} sin\bigg(\dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)$

We know that

$\boxed{ \bf{ \dfrac{d}{dx}sinx\: = \: cosx }}$

So,

$\rm :\longmapsto\:\dfrac{dy}{dx} =cos\bigg(\dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg) \: \dfrac{d}{dx}\bigg(\dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)$

We know,

Quotient Rule,

$\boxed{ \bf{\dfrac{d}{dx} \frac{u}{v} \: = \: \dfrac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =cos\bigg(\dfrac{1-{x}^{2} }{1 + {x}^{2} } \bigg) \dfrac{(1+{x}^{2})\dfrac{d}{dx}(1 -{x}^{2}) - (1-{x}^{2})\dfrac{d}{dx}(1+{x}^{2})}{ {(1 + {x}^{2})}^{2} }$

We know,

$\boxed{ \bf{ \dfrac{d}{dx}k\: = \: 0}}$

and

$\boxed{ \bf{ \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} }}$

So, using these we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =cos\bigg(\dfrac{1-{x}^{2} }{1 + {x}^{2} } \bigg) \dfrac{(1+{x}^{2})(0 - 2x) - (1-{x}^{2})(0 + 2x)}{ {(1 + {x}^{2})}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =cos\bigg(\dfrac{1-{x}^{2} }{1 + {x}^{2} } \bigg) \dfrac{(1+{x}^{2})(- 2x) - (1-{x}^{2})(2x)}{ {(1 + {x}^{2})}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - 2x \: cos\bigg(\dfrac{1-{x}^{2} }{1 + {x}^{2} } \bigg) \dfrac{(1+{x}^{2} +1-{x}^{2})}{ {(1 + {x}^{2})}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - 2x \: cos\bigg(\dfrac{1-{x}^{2} }{1 + {x}^{2} } \bigg) \dfrac{(2)}{ {(1 + {x}^{2})}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - 4x \: cos\bigg(\dfrac{1-{x}^{2} }{1 + {x}^{2} } \bigg) \dfrac{1}{ {(1 + {x}^{2})}^{2} }$

Additional Information :-

$\boxed{ \bf{ \dfrac{d}{dx}cosx\: = - \:sinx }}$

$\boxed{ \bf{ \dfrac{d}{dx}cosecx\: = - \:cosecx \: cotx}}$

$\boxed{ \bf{ \dfrac{d}{dx}secx\: = \:secx \: tanx}}$

$\boxed{ \bf{\dfrac{d}{dx}tanx \: = \: {sec}^{2}x}}$

$\boxed{ \bf{\dfrac{d}{dx}cotx \: = \: {cosec}^{2}x}}$

$\boxed{ \bf{\dfrac{d}{dx} \sqrt{x} \: = \: \frac{1}{2 \sqrt{x} } }}$

$\boxed{ \bf{\dfrac{d}{dx}u.v \: = \: v\dfrac{d}{dx}u + u\dfrac{d}{dx}v }}$