b) $$ y=\csc ^{5}\left(2 x+x^{3}\right)^{4} ; \frac{d}{d x} $$

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Let 
$$
y=\left(\csc\left((2x+x^3)^4\right)\right)^5.
$$
We differentiate using the chain rule in several steps.

**Step 1.** Write
$$
y=\left(\csc u\right)^5,\quad \text{with } u=(2x+x^3)^4.
$$

**Step 2.** Differentiate $$y$$ with respect to $$u$$:
$$
\frac{dy}{du}=5\left(\csc u\right)^4\cdot\frac{d}{du}(\csc u).
$$
We know that
$$
\frac{d}{du}(\csc u)=-\csc u\,\cot u,
$$
so
$$
\frac{dy}{du}=-5\left(\csc u\right)^5 \cot u.
$$

**Step 3.** Differentiate $$u=(2x+x^3)^4$$ with respect to $$x$$. Let
$$
v=2x+x^3,\quad \text{so that } u=v^4.
$$
Then
$$
\frac{du}{dv}=4v^3,
$$
and
$$
\frac{dv}{dx}=2+3x^2.
$$
Thus,
$$
\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=4(2x+x^3)^3(2+3x^2).
$$

**Step 4.** Combine the results using the chain rule:
$$
\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=-5\left(\csc u\right)^5 \cot u\cdot 4(2x+x^3)^3(2+3x^2).
$$
Recall that $$u=(2x+x^3)^4$$. Hence,
$$
\frac{dy}{dx}=-20\,\csc^5\left((2x+x^3)^4\right)\cot\left((2x+x^3)^4\right)(2x+x^3)^3(2+3x^2).
$$
The derivative of $$ y = \csc^5\left((2x + x^3)^4\right) $$ with respect to $$ x $$ is: $$ \frac{dy}{dx} = -20\,\csc^5\left((2x + x^3)^4\right)\cot\left((2x + x^3)^4\right)(2x + x^3)^3(2 + 3x^2). $$

b) \( y=\csc ^{5}\left(2 x+x^{3}\right)^{4} ; \frac{d}{d x} \)

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