Prove that ddx csc x csc x cot x

To prove that the derivative of the expression csc(x) imes csc(x) imes cot(x) can be derived, we will use the product rule and the chain rule from calculus.

The expression can be simplified as follows:

y = csc(x) imes csc(x) imes cot(x) = csc^2(x) imes cot(x)

Now, to find ddx y, we will apply the product rule:

ddx [u imes v] = u'v + uv'

Let u = csc^2(x) and v = cot(x).

First, we need to find the derivatives of u and v:

Step 1: Derivative of u = csc^2(x)

We know that ddx(csc(x)) = -csc(x)cot(x), so using the chain rule:

ddx[csc^2(x)] = 2csc(x)(-csc(x)cot(x)) = -2csc^2(x)cot(x)

Step 2: Derivative of v = cot(x)

We also know that ddx(cot(x)) = -csc^2(x).

Step 3: Apply the product rule

Now we can apply the product rule:

ddx[y] = u'v + uv' = (-2csc^2(x)cot(x)) imes cot(x) + csc^2(x) imes (-csc^2(x))

Replacing u' and v' gives us:

ddx[y] = -2csc^2(x)cot^2(x) - csc^4(x)

So, the final derivative of csc^2(x)cot(x) is:

ddx[csc^2(x)cot(x)] = -2csc^2(x)cot^2(x) - csc^4(x)

This proves the derivative of the given expression.