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

To differentiate the function \( y = \csc^{5}(2x + x^{3})^{4} \), we first apply the chain rule. Starting with \( y = u^{5} \) where \( u = \csc((2x + x^{3})^{4}) \). The derivative \( \frac{dy}{dx} \) will involve \( \frac{du}{dx} \) as well. 1. Differentiate using the chain rule: \[ \frac{dy}{dx} = 5u^{4} \cdot \frac{du}{dx} \] 2. Now differentiate \( u = \csc(v) \) where \( v = (2x + x^{3})^{4} \): \[ \frac{du}{dx} = -\csc(v)\cot(v) \cdot \frac{dv}{dx} \] 3. For \( v = (2x + x^{3})^{4} \), use the chain rule again: \[ \frac{dv}{dx} = 4(2x + x^{3})^{3} \cdot (2 + 3x^{2}) \] Plugging everything back in gives us the full derivative. Make sure to substitute \( u \) and \( v \) correctly in the final answer. This approach showcases the beauty of calculus and how interconnected various concepts are. Happy differentiating!