If y=cosx1+sinx, then dydx=

# If $y=\frac{\mathrm{cos}x}{1+\mathrm{sin}x}$, then $\frac{dy}{dx}=$

1. A

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

2. B

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

3. C

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

4. D

$\mathrm{tan}x$

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### Solution:

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

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