Limits Continuity And Differentiability Question 16
Question: The derivative of ln $ (x+sinx) $ with respect to $ (x+cosx) $ is
Options:
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-sinx)(1-cosx)} $
Show Answer
Answer:
Correct Answer: A
Solution:
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)} $
BETA
