Solution :
We have, y = log sin x
By using chain rule in differentiation,
Let u = sin x \(\implies\) \(du\over dx\) = cos x
And, y = log u \(\implies\) \(dy\over du\) = \(1\over u\)
Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\)
\(\implies\) \(dy\over dx\) = \(1\over u\) \(\times\) cos x
\(\implies\) \(dy\over dx\) = \(1\over sin x\) \(\times\) cos x = cot x
Hence, the differentiation of log sin x with respect to x is cot x.
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