Question

if y =
$\frac{cosx}{sinx + cosx}$
then find
$\frac{dy}{dx}$
​

Answer 1

$\large \underline \bold{Solution}$:-

$\sf{y = \dfrac{Cosx}{Sinx + Cosx}}$

$\sf{Differential \: with \: respect \: to \: x -}$

$\sf{\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{d}{dx}\bigg(\dfrac{Cosx}{Sinx + Cosx}\bigg)}$

$\large \underline \bold{Usable \: Formula}$:-

$\: \: \: \: \: \:$$\large\boxed{\sf\red{\dfrac{d}{dv}\bigg(\dfrac{u}{v}\bigg) =\bigg(\dfrac{v \dfrac{d}{dv} u - u \dfrac{d}{dv} v }{v^{2}}\bigg)}}$

$\: \: \:$$\large\boxed{\sf\red{\dfrac{d}{dx}(Sinx) = Cosx}}$

$\: \: \:$$\large\boxed{\sf\red{\dfrac{d}{dx}(Cosx) = -Sinx}}$

$\: \:$$\large\boxed{\sf\red{Sin^{2} x + Cos^{2} x = 1}}$

$\sf{So \: ,}$

$\sf{\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{(Sinx + Cosx) \dfrac{d}{dx}(Cosx) - Cosx \dfrac{d}{dx}(Sinx + Cosx)}{(Sinx + Cosx)^{2}}}$

$\sf{\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{(Sinx + Cosx)(-Sinx) - Cosx(Cosx - Sinx)}{(Sinx + Cosx)^{2}}}$

$\sf{\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{-Sin^{2} x - \cancel{Sinx Cosx} - Cos^{2} x + \cancel{Sinx Cosx}}{(Sinx + Cosx)^{2}}}$

$\sf{\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{- Sin^{2} x - Cos^{2} x}{(Sinx + Cosx)^{2}}}$

$\sf{\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{-(Sin^{2} x + Cos^{2} x)}{(Sinx + Cosx)^{2}}}$

$\small \bold{\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{-1}{(Sinx + Cosx)^{2}}}$