To prove the derivative of the function \( f(x) = \csc x \cdot \csc x \cdot \cot x \), we will use the product rule and the quotient rule.
First, let’s simplify \( f(x) \):
Rewriting it gives \( f(x) = \csc^2 x \cdot \cot x \).
Using the product rule, where if we have \( u = \csc^2 x \) and \( v = \cot x \), then:
\( f'(x) = u’v + uv’ \)
Now, we need to find \( u’ \) and \( v’ \):
- To find \( u’ = \frac{d}{dx}(\csc^2 x) \):
We use the chain rule:
\( \csc^2 x = \frac{1}{\sin^2 x} \implies \frac{d}{dx}(\csc^2 x) = -2 \csc^2 x \cot x \)
- To find \( v’ = \frac{d}{dx}(\cot x) \):
Remember that \( \cot x = \frac{\cos x}{\sin x}, \implies v’ = -\csc^2 x \)
Substituting these derivatives back into our product rule gives us:
\( f'(x) = (-2 \csc^2 x \cot x)(\cot x) + (\csc^2 x)(-\csc^2 x) \)
Now we can simplify that further:
\( f'(x) = -2 \csc^2 x \cot^2 x – \csc^4 x \)
So, the final result is:
\( \frac{d}{dx}(\csc^2 x \cdot \cot x) = -2 \csc^2 x \cot^2 x – \csc^4 x \)