https://file.aiquickdraw.com/mathbot/file/1729986637115ls0nx1rb.jpg

The question requires calculating the derivatives \( \frac{dy}{dx} \) for four different functions \( y \) as shown in the image. Let’s go over each part and verify if the solutions are correct.

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### Part (a)
\[ y = (2x^4 - 5)^3 - 3x^2 \]
To find \( \frac{dy}{dx} \), we apply the chain rule and differentiate each term separately:

1. **Derivative of** \( (2x^4 - 5)^3 \):  
   Using the chain rule:
   \[
   \frac{d}{dx} \left( (2x^4 - 5)^3 \right) = 3(2x^4 - 5)^2 \cdot \frac{d}{dx}(2x^4 - 5)
   \]
   where
   \[
   \frac{d}{dx}(2x^4 - 5) = 8x^3
   \]
   So,
   \[
   \frac{d}{dx} \left( (2x^4 - 5)^3 \right) = 3(2x^4 - 5)^2 \cdot 8x^3 = 24x^3 (2x^4 - 5)^2
   \]

2. **Derivative of** \( -3x^2 \):
   \[
   \frac{d}{dx}(-3x^2) = -6x
   \]

Combining these:
\[
\frac{dy}{dx} = 24x^3 (2x^4 - 5)^2 - 6x
\]
The solution provided matches this answer.

---

### Part (b)
\[ y = \ln(3x^4 + 2x - 3) \]
To differentiate, we use the chain rule with the natural logarithm:

1. **Derivative of** \( \ln(3x^4 + 2x - 3) \):
   \[
   \frac{dy}{dx} = \frac{1}{3x^4 + 2x - 3} \cdot \frac{d}{dx}(3x^4 + 2x - 3)
   \]
   where
   \[
   \frac{d}{dx}(3x^4 + 2x - 3) = 12x^3 + 2
   \]
   So,
   \[
   \frac{dy}{dx} = \frac{12x^3 + 2}{3x^4 + 2x - 3}
   \]

The provided solution also matches this answer.

---

### Part (c)
\[ y = \cos(x^2 + 1) \sin^3(x) \]
To differentiate, we use the product rule, letting:
   - \( u = \cos(x^2 + 1) \)
   - \( v = \sin^3(x) \)

1. **Derivative of** \( u = \cos(x^2 + 1) \):
   \[
   \frac{du}{dx} = -\sin(x^2 + 1) \cdot 2x = -2x \sin(x^2 + 1)
   \]

2. **Derivative of** \( v = \sin^3(x) \):
   \[
   \frac{dv}{dx} = 3 \sin^2(x) \cdot \cos(x) = 3 \sin^2(x) \cos(x)
   \]

Using the product rule:
\[
\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}
\]
\[
= \cos(x^2 + 1) \cdot 3 \sin^2(x) \cos(x) - \sin^3(x) \cdot 2x \sin(x^2 + 1)
\]
\[
= 3 \cos(x^2 + 1) \sin^2(x) \cos(x) - 2x \sin^3(x) \sin(x^2 + 1)
\]

---

### Part (d)
\[ y = \frac{e^x}{1 - x^2} \]
To differentiate, we use the quotient rule, letting:
   - \( u = e^x \)
   - \( v = 1 - x^2 \)

1. **Derivative of** \( u = e^x \):
   \[
   \frac{du}{dx} = e^x
   \]

2. **Derivative of** \( v = 1 - x^2 \):
   \[
   \frac{dv}{dx} = -2x
   \]

Using the quotient rule:
\[
\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]
\[
= \frac{(1 - x^2)e^x - e^x(-2x)}{(1 - x^2)^2}
\]
\[
= \frac{e^x(1 - x^2 + 2x)}{(1 - x^2)^2}
\]
This matches the answer given.

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Would you like further details on any specific part? 

### Related Questions:
1. How is the chain rule applied to more complex functions?
2. What are some advanced uses of the product rule in calculus?
3. How does the quotient rule differ from other differentiation rules?
4. Can logarithmic differentiation simplify problems like part (b)?
5. How does trigonometric differentiation apply in physics and engineering?

#### Tip:
When working with derivatives, identifying the type of function (product, quotient, composite) helps in selecting the correct differentiation rule.

Q5: Find dy/dx for the given functions: (a) y = (2x^4 - 5)^3 - 3x^2, (b) y = log_e(3x^4 + 2x - 3), (c) y = cos(x^2 + 1) * sin^3(x), (d) y = e^x / (1 - x^2)

Step-by-step solution for finding dy/dx of four complex functions using differentiation rules: chain rule, product rule, and quotient rule. Covers logarithmic differentiation and trigonometric functions, ideal for students in Grades 11-12.

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