Question

differentiate sin x /1+tanx​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = \dfrac{sinx}{1 + tanx}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}\bigg[ \dfrac{sinx}{1 + tanx}\bigg]$

We know,

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

So, here,

$\red{\rm :\longmapsto\:u = sinx}$

$\red{\rm :\longmapsto\:v = 1 + tanx}$

Now, on substituting the values, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{(1 + tanx)\dfrac{d}{dx}sinx - sinx\dfrac{d}{dx}(1 + tanx)}{ {(1 + tanx)}^{2} }$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}tanx = {sec}^{2}x}}}$

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

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}k = 1}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{(1 + tanx)(cosx) - sinx(0 + {sec}^{2}x) }{ {(1 + tanx)}^{2} }$

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

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

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$