Differentiate (cos x)sin x with respect to (sin x)cos x.

Let u = (cos x)sin x and v = (sin x)cos x.


We need to differentiate u with respect to v that is find.


We have u = (cos x)sin x


Taking log on both sides, we get


log u = log(cos x)sin x


log u = (sin x) × log(cos x) [ log am = m × log a]


On differentiating both the sides with respect to x, we get



Recall that (uv)’ = vu’ + uv’ (product rule)



We know and





We know




But, u = (cos x)sin x




Now, we have v = (sin x)cos x


Taking log on both sides, we get


log v = log(sin x)cos x


log v = (cos x) × log(sin x) [ log am = m × log a]


On differentiating both the sides with respect to x, we get



Recall that (uv)’ = vu’ + uv’ (product rule)



We know and





We know




But, v = (sin x)cos x




We have




Thus,


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