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,