Question

differentiate x+cosx/tanx w.r.t.x​

Answer 1

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

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

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

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

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,

$\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{tanx(1 - sinx) - (x + cosx) {sec}^{2} x}{ {tan}^{2} x}$

$\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{tanx(1 - sinx)}{ {tan}^{2}x } - \dfrac{(x + cosx) {sec}^{2}x }{ {tan}^{2} x}$

$\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{1 - sinx}{tanx} - \dfrac{x + cosx}{ {sin}^{2} x}$

$\sf \: \dfrac{dy}{dx} = \: \: \: \dfrac{1}{ {tan}x } - \dfrac{sinx}{tanx} - \dfrac{x}{ {sin}^{2}x } - \dfrac{cosx}{ {sin}^{2}x }$

$\sf \: \dfrac{dy}{dx} = \: \: \: cotx - cosx - x {cosec}^{2}x - cotx \: cosecx$

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Additional Information :-

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

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

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

$\boxed{ \sf\dfrac{d}{dx} sinx = cosx}$

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

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

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

$\boxed{ \sf\dfrac{d}{dx} kf(x) = k\dfrac{d}{dx} f(x)}$

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