$$ 2.1 \quad y=\cot ^{2} 5 x $$

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We start with the function
$$
y = \cot^2(5x).
$$
This can be seen as a composition of functions. Let
$$
u = \cot(5x)
$$
so that
$$
y = u^2.
$$

**Step 1. Differentiate $$y$$ with respect to $$u$$.**

Since
$$
y = u^2,
$$
its derivative with respect to $$u$$ is
$$
\frac{dy}{du} = 2u.
$$

**Step 2. Differentiate $$u$$ with respect to $$x$$.**

We have
$$
u = \cot(5x).
$$
Recall that the derivative of $$\cot u$$ with respect to $$u$$ is
$$
\frac{d}{du}[\cot u] = -\csc^2 u.
$$
Since $$u = 5x$$, by the chain rule, we also differentiate the inner function:
$$
\frac{d}{dx}(5x) = 5.
$$
Thus,
$$
\frac{du}{dx} = \frac{d}{dx}[\cot(5x)] = -\csc^2(5x) \cdot 5 = -5\csc^2(5x).
$$

**Step 3. Apply the Chain Rule.**

Using the chain rule,
$$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx},
$$
substitute the expressions we found:
$$
\frac{dy}{dx} = 2u \cdot \left(-5\csc^2(5x)\right).
$$
Substitute back $$u = \cot(5x)$$:
$$
\frac{dy}{dx} = 2\cot(5x) \cdot \left(-5\csc^2(5x)\right).
$$

**Step 4. Simplify the expression.**

Multiplying the constants,
$$
\frac{dy}{dx} = -10\cot(5x)\csc^2(5x).
$$

Thus, the derivative of the function is
$$
\boxed{\frac{dy}{dx} = -10\cot(5x)\csc^2(5x)}.
$$
The derivative of $$ y = \cot^2(5x) $$ is $$ \frac{dy}{dx} = -10\cot(5x)\csc^2(5x) $$.

\( 2.1 \quad y=\cot ^{2} 5 x \)

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