Question

derivative . pls tell ​

Answer 1

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

$\rm :\longmapsto\:y = \dfrac{sinx + cosx}{sinx - cosx}$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - 2}{1 - sin2x}$

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

Given that,

$\rm :\longmapsto\:y = \dfrac{sinx + cosx}{sinx - cosx}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} \: \dfrac{sinx + cosx}{sinx - cosx}$

So,

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx} \: \dfrac{sinx + cosx}{sinx - cosx}$

We know

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

So, on using this, we get

$\rm \:=\dfrac{(sinx - cosx)\dfrac{d}{dx}(sinx + cosx) - (sinx + cosx)\dfrac{d}{dx}(sinx - cosx)}{ {(sinx - cosx)}^{2} }$

We know,

$\red{\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}}$

and

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

So, using this result, we get

$\rm \:=\dfrac{(sinx - cosx)(cosx - sinx) - (sinx + cosx)(cosx + sinx)}{ {(sinx - cosx)}^{2} }$

$\rm \:  =  \:  \: \dfrac{ - {(sinx - cosx)}^{2} - {(sinx + cosx)}^{2} }{ {(sinx - cosx)}^{2} }$

$\rm \:  =  \:  - \: \dfrac{{(sinx - cosx)}^{2} + {(sinx + cosx)}^{2} }{ {(sinx - cosx)}^{2} }$

We know,

$\green{\boxed{ \rm{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

and

$\green{\boxed{ \rm{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}}$

So, using these Identities, we get

$\rm \:  =  \:  - \: \dfrac{2( {sin}^{2} x + {cos}^{2}x) }{ {sin}^{2}x + {cos}^{2}x - 2sinxcosx }$

We know,

$\purple{\boxed{ \rm{ {sin}^{2}x + {cos}^{2}x = 1}}} \: \: and \: \: \purple{\boxed{ \rm{ sin2x = 2sinx \: cosx}}}$

So, on using these two Identities, we get

$\rm \:  =  \: -   \: \dfrac{2}{1 - sin2x}$

Hence,

$\boxed{ \qquad \bf{ \frac{sinx + cosx}{sinx - cosx} =  \: -   \: \dfrac{2}{1 - sin2x} \qquad}}$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx}x = 1}}$

$\boxed{ \rm{ \dfrac{d}{dx}k = 0}}$

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

$\boxed{ \rm{ \dfrac{d}{dx}logx = \frac{1}{x}}}$

$\boxed{ \rm{ \dfrac{d}{dx} {e}^{x} = {e}^{x}}}$

$\boxed{ \rm{ \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) }}$

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

$\boxed{ \rm{ \dfrac{d}{dx}cotx = { - cosec}^{2}x}}$