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Mathematics Question on limits and derivatives

Find the derivative of cos x1+sin x\frac{cos\,x}{1+sin\,x}.

A

−(1+sinx)2-(1 + sinx)^2

B

1(1+sin x)\frac{1}{\left(1+sin\,x\right)}

C

(−11+sin x)\left(\frac{-1}{1+sin\,x}\right)

D

None of these

Answer

(−11+sin x)\left(\frac{-1}{1+sin\,x}\right)

Explanation

Solution

Let y=cos x1+sin xy=\frac{cos\,x}{1+sin\,x} Differentiating both sides with respect to xx, we get dydx=ddx(cos x1+sin x)\frac{dy}{dx}=\frac{d}{dx}\left(\frac{cos\,x}{1+sin\,x}\right) =(1+sin x)ddx(cos x)−cos xddx(1+sin x)(1+sin x)2=\frac{\left(1+sin\,x\right) \frac{d}{dx}\left(cos\,x\right)-cos\,x \frac{d}{dx}\left(1+sin\,x\right)}{\left(1+sin\,x\right)^{2}} =(1+sin x)(−sin x)−cos x(cos x)(1+sin x)2=\frac{\left(1+sin\,x\right)\left(-sin\,x\right)-cos\,x\left(cos\,x\right)}{\left(1+sin\,x\right)^{2}} =−sin x−sin2 x−cos2 x(1+sin x)2=\frac{-sin\,x-sin^{2}\,x-cos^{2}\,x}{\left(1+sin\,x\right)^{2}} =−(1+sin x)(1+sin x)2=−11+sin x=\frac{-\left(1+sin\,x\right)}{\left(1+sin\,x\right)^{2}}=\frac{-1}{1+sin\,x}