In this question, the derivatives $\frac{dy}{dx}$ are calculated for each given function:

Part (a)

Given: $y = (2x^4 - 5)^3 - 3x^2$

Solution: Using the chain rule to differentiate $(2x^4 - 5)^3$ and the power rule for $-3x^2$: $\frac{dy}{dx} = 3(2x^4 - 5)^2 \cdot (8x^3) - 6x = 3(2x^4 - 5)^2 \cdot 8x^3 - 6x$

Part (b)

Given: $y = \ln(3x^4 + 2x - 3)$

Solution: Using the derivative of $\ln u = \frac{1}{u} \cdot \frac{du}{dx}$: $\frac{dy}{dx} = \frac{1}{3x^4 + 2x - 3} \cdot (12x^3 + 2) = \frac{12x^3 + 2}{3x^4 + 2x - 3}$

Part (c)

Given: $y = \cos(x^2 + 1) \sin^3 x$

Solution: This derivative requires the product rule: $(uv)' = u'v + uv'$. Define $u = \cos(x^2 + 1)$ and $v = \sin^3 x$. Then:

1. $u' = -\sin(x^2 + 1) \cdot 2x$
2. $v' = 3 \sin^2 x \cos x$

Therefore, $\frac{dy}{dx} = -\sin(x^2 + 1) \cdot 2x \cdot \sin^3 x + \cos(x^2 + 1) \cdot 3 \sin^2 x \cos x$

Part (d)

Given: $y = \frac{e^x}{1 - x^2}$

Solution: Using the quotient rule: $\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}$, where $f = e^x$ and $g = 1 - x^2$, $f' = e^x \quad \text{and} \quad g' = -2x$ Thus, $\frac{dy}{dx} = \frac{e^x (1 - x^2) - e^x \cdot (-2x)}{(1 - x^2)^2} = \frac{e^x (1 - x^2 + 2x)}{(1 - x^2)^2}$

The calculations in each part appear correct based on the provided work. Let me know if you'd like more details on any of these steps or if you have any questions.

Further Questions:

1. How is the chain rule applied in each differentiation?
2. Could you explore the product rule with different types of functions?
3. What are some common mistakes when applying the quotient rule?
4. How would the derivatives change if the exponents were altered?
5. How do these derivatives relate to real-world applications?

Tip:

Remember to carefully apply the chain rule when working with composite functions; differentiating the outer and inner functions in the right order is key.