A) \[\frac{1+\cos x}{(x+\sin x)(1-\sin x)}\]
B) \[\frac{1-\cos x}{(x+\sin x)(1+\sin x)}\]
C) \[\frac{1-\cos x}{(x-\sin x)(1+\cos x)}\]
D) \[\frac{1+\cos \,\,x}{(x-sin\,\,x)(1-cos\,\,x)}\]
Correct Answer: A
Solution :
[a] In \[{{(x+\sin \,x)}^{1}}=y\] (say) \[\frac{dy}{dx}=\frac{1}{(x+\sin x)}(1+\cos x)\] \[=\frac{(1+\cos x)}{(x+\sin x)}\] \[x+\cos x=z(say)\] \[\frac{dz}{dx}=(1-sinx)\] derivative of ln \[(x+\sin x)\] w.r.t \[(x+\cos x)\] is \[\frac{dy}{dz}=\frac{(1+\cos x)}{(x+\sin x)(1-\sin x)}\]
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