If y=cosx1+sinx, then dydx=

y=cosx1+sinxdydx=(1+sinx)ddx(cosx)−cosxddx(1+sinx)(1+sinx)2=(1+sinx)(−sinx)−cosx(cosx)(1+sinx)2=−sinx−(sin2x+cos2x)(1+sinx)2=−sinx−1(1+sinx)2=−(1+sinx)(1+sinx)2=−11+sinx

If y=cosx1+sinx, then dydx=

y=cosx1+sinxdydx=(1+sinx)ddx(cosx)−cosxddx(1+sinx)(1+sinx)2=(1+sinx)(−sinx)−cosx(cosx)(1+sinx)2=−sinx−(sin2x+cos2x)(1+sinx)2=−sinx−1(1+sinx)2=−(1+sinx)(1+sinx)2=−11+sinx

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If $y = \frac{\cos x}{1 + \sin x}$ , then $\frac{d y}{d x} =$

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$\frac{- 1}{1 + \sin x}$

$\tan x$

$\frac{1}{1 + \sin x}$

$\frac{1}{1 + \cos x}$

answer is A.

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Detailed Solution

$\begin{array}{l} y = \frac{\cos x}{1 + \sin x} \\ \frac{d y}{d x} = \frac{(1 + \sin x) \frac{d}{d x} (\cos x) - \cos x \frac{d}{d x} (1 + \sin x)}{{(1 + \sin x)}^{2}} \\ = \frac{(1 + \sin x) (- \sin x) - \cos x (\cos x)}{{(1 + \sin x)}^{2}} \\ = \frac{- \sin x - (\sin^{2} x + \cos^{2} x)}{{(1 + \sin x)}^{2}} \\ = \frac{- \sin x - 1}{{(1 + \sin x)}^{2}} \\ = \frac{- (1 + \sin x)}{{(1 + \sin x)}^{2}} \\ = \frac{- 1}{1 + \sin x} \end{array}$