The image contains the same derivative problems and solutions as before, with the focus on finding $\frac{dy}{dx}$ for each function in the sub-questions (a) through (d). The solutions for parts (a), (b), and (d) are correct and verified. Part (c) remains incomplete.

Let's solve part (c), which is the derivative of: $y = \cos(x^2 + 1) \sin^3(x)$

To find $\frac{dy}{dx}$ for part (c), we’ll use the product rule combined with the chain rule.

Solution for Part (c)

The product rule states that if $y = u \cdot v$, then $y' = u'v + uv'$.

Let:

• $u = \cos(x^2 + 1)$
• $v = \sin^3(x)$

Step 1: Differentiate $u = \cos(x^2 + 1)$

Using the chain rule: $u' = -\sin(x^2 + 1) \cdot \frac{d}{dx}(x^2 + 1) = -\sin(x^2 + 1) \cdot 2x = -2x \sin(x^2 + 1)$

Step 2: Differentiate $v = \sin^3(x)$

Using the chain rule again: $v' = 3 \sin^2(x) \cdot \cos(x)$

Step 3: Apply the Product Rule

Now, substitute $u$, $u'$, $v$, and $v'$ into the product rule formula: $y' = u'v + uv'$ Substitute the values: $y' = \left(-2x \sin(x^2 + 1)\right) \sin^3(x) + \cos(x^2 + 1) \cdot \left(3 \sin^2(x) \cos(x)\right)$

Simplify each term: $y' = -2x \sin(x^2 + 1) \sin^3(x) + 3 \cos(x^2 + 1) \sin^2(x) \cos(x)$

So the derivative for part (c) is: $\frac{dy}{dx} = -2x \sin(x^2 + 1) \sin^3(x) + 3 \cos(x^2 + 1) \sin^2(x) \cos(x)$

Would you like further clarification on this solution or on any of the other parts?

Additional Questions

1. How can we apply the product rule in functions with trigonometric and exponential terms?
2. Why is the chain rule important when differentiating compositions of functions?
3. What steps can simplify differentiating trigonometric functions raised to a power?
4. How can we combine chain rule and product rule in multi-term derivatives?
5. What would the derivative look like if we replaced $\sin(x)$ with another function, like $\tan(x)$?

Tip: Remember to identify each component separately before applying rules like the product or chain rule to avoid mistakes in complex derivatives.