Differentiate x+cosx/tanx w.r.t.x

1.	Differentiate x+cosx/tanx w.r.t.x
Answer» $\large\underline{\sf{Solution-}}$ $\sf \: \: \: \: Let \: y = \dfrac{x + cosx}{tanx}$ On differentiating both sides W. r. t. x, we get $\sf \: \dfrac{d}{dx}y = \: \: \: \dfrac{d}{dx} \bigg(\dfrac{x + cosx}{tanx} \bigg)$ We KNOW that $\: \: \: \: \: \: \: \: \boxed{ \sf{ \: \dfrac{d}{dx} \dfrac{<klux>U</klux>}{<klux>V</klux>} = \dfrac{v\dfrac{d}{dx} u - u\dfrac{d}{dx} v}{ {v}^{2} } }}$ Here, u = x + cosx and v = tanx On substituting all these VALUES, we get $\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{tanx \: \dfrac{d}{dx} (x + cosx) - (x + cosx)\dfrac{d}{dx} tanx}{ {tan}^{2}x }$ Now, we know that $\boxed{ \sf\dfrac{d}{dx} x = 1} \: and \: \boxed{ \sf\dfrac{d}{dx} cosx = - sinx} \: and \: \boxed{ \sf\dfrac{d}{dx} tanx = {sec}^{2}x }$ So, $\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{tanx(1 - sinx) - (x + cosx) {sec}^{2} x}{ {tan}^{2} x}$ $\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{tanx(1 - sinx)}{ {tan}^{2}x } - \dfrac{(x + cosx) {sec}^{2}x }{ {tan}^{2} x}$ $\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{1 - sinx}{tanx} - \dfrac{x + cosx}{ {sin}^{2} x}$ $\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{1}{ {tan}x } - \dfrac{sinx}{tanx} - \dfrac{x}{ {sin}^{2}x } - \dfrac{cosx}{ {sin}^{2}x }$ $\sf \: \dfrac{dy}{dx} = \: \: \: cotx - cosx - x {cosec}^{2}x - cotx \: cosecx$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional Information :- $\boxed{ \sf\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }$ $\boxed{ \sf\dfrac{d}{dx} logx = \dfrac{1}{x} }$ $\boxed{ \sf\dfrac{d}{dx} {e}^{x} = {e}^{x} }$ $\boxed{ \sf\dfrac{d}{dx} sinx = cosx}$ $\boxed{ \sf\dfrac{d}{dx} cotx = - {cosec}^{2}x }$ $\boxed{ \sf\dfrac{d}{dx} secx = secx \: tanx}$ $\boxed{ \sf\dfrac{d}{dx} k = 0}$ $\boxed{ \sf\dfrac{d}{dx} kf(x) = k\dfrac{d}{dx} f(x)}$ $\boxed{ \sf\dfrac{d}{dx} u.v = v\dfrac{d}{dx} u \: + \: u\dfrac{d}{dx} v}$

Answer»

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

$\sf \: \: \: \: Let \: y = \dfrac{x + cosx}{tanx}$

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

$\sf \: \dfrac{d}{dx}y = \: \: \: \dfrac{d}{dx} \bigg(\dfrac{x + cosx}{tanx} \bigg)$

We KNOW that

$\: \: \: \: \: \: \: \: \boxed{ \sf{ \: \dfrac{d}{dx} \dfrac{<klux>U</klux>}{<klux>V</klux>} = \dfrac{v\dfrac{d}{dx} u - u\dfrac{d}{dx} v}{ {v}^{2} } }}$