Question

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Determine the derivative of cosx/(1+sin x)​

Answer 1

Answer:

Determine the derivative of cosx/(1+sin x)

Step-by-step explanation:

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Answer 2

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

Given function is

$\rm :\longmapsto\:\dfrac{cosx}{1 + sinx}$

Let assume that

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

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

$\rm :\longmapsto\:\dfrac{d}{dx}y \: =\dfrac{d}{dx} \: \: \dfrac{cosx}{1 + sinx}$

We know,

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

So, using this, we get

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

We know,

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

and

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

$\rm \: = \: \dfrac{(1 + sinx)( - sinx) - cosx(0 + cosx)}{ {(1 + sinx)}^{2} }$

$\rm \: = \: \dfrac{ - sinx - {sin}^{2} x - {cos}^{2} x}{ {(1 + sinx)}^{2} }$

$\rm \: = \: \dfrac{ - sinx - ({sin}^{2} x + {cos}^{2} x)}{ {(1 + sinx)}^{2} }$

$\rm \: = \: \dfrac{ - sinx - (1)}{ {(1 + sinx)}^{2} }$

$\rm \: = \: \dfrac{ - sinx - 1}{ {(1 + sinx)}^{2} }$

$\rm \: = \: \dfrac{ - (sinx + 1)}{ {(1 + sinx)}^{2} }$

$\rm \: = \: - \: \dfrac{1}{1 + sinx}$

$\bf\implies \:\dfrac{dy}{dx}= \: - \: \dfrac{1}{1 + sinx}$

Additional Information :-

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