Deriving Y=(cot x)/(1 cot x)

Step 1: Simplify the given expression:

We know that cot x = cos x/sin x

Therefore, Y = (cos x/sin x)/(1-cos x/sin x)

Simplifying the denominator, we get:

Y = cos x/(sin x - cos x)

Step 2: Use Quotient Rule:

To differentiate Y, we will use the quotient rule:

(d/dx)[f(x)/g(x)] = [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2

Let f(x) = cos x

And g(x) = sin x - cos x

Therefore, f'(x) = -sin x

And g'(x) = cos x + sin x

Now, applying the quotient rule, we get:

Y' = [(sin x - cos x)(-sin x) - (cos x)(cos x + sin x)] / (sin x - cos x)^2

Simplifying the numerator, we get:

Y' = [-sin^2 x + cos^2 x - cos^2 x - cos x sin x] / (sin x - cos x)^2

Y' = [-sin^2 x - 2cos^2 x - cos x sin x] / (sin x - cos x)^2

Step 3: Simplify the derivative:

Now, we can simplify Y' by using trigonometric identities:

- 2cos^2 x = 1 - sin^2 x - cos^2 x
- cos x sin x = sin 2x/2

Substituting these values, we get:

Y' = [(sin^2 x - cos^2 x) - (sin^2 x + cos^2 x - 1) - sin 2x/2] / (sin x - cos x)^2

Simplifying further, we get:

Y' = [1 - sin 2x/2] / (sin x - cos x)^2

Thus, the derivative of Y is [1 - sin 2x/2] / (sin x - cos x)^2.