Heref(x)=cos x1+sin x
∴f′(x)=ddx[cos x1+sin x]
=(1+sin x)ddx(cos x)−cos x.ddx(1+sin x)(1+sin x)2
=(1+sin x)(−sin x)−cos x(cos x)(1+sin x)2
=−sin x−sin2x−cos2x(1+sin x)2
=−sin x−(sin2x+cos2x)(1+sin x)2
=−sin x−1(1+sin x)2=−(1+sin x)(1+sin x)2
=−11+sin x