https://file.aiquickdraw.com/mathbot/file/1727303544447dwx6wn8o.jpg

Let's break down the two expressions in the image and find solutions:

### a) \( y = 7x^4 \cdot \sqrt{x^3 + 5} \)

Here, we are given the function \( y = 7x^4 \cdot \sqrt{x^3 + 5} \), and I assume the question is asking for the derivative \( \frac{dy}{dx} \).

We will use the **product rule** combined with the **chain rule** for differentiation.

#### Steps:

1. The first term is \( 7x^4 \), and its derivative is \( 28x^3 \).
2. The second term is \( \sqrt{x^3 + 5} \), which is equal to \( (x^3 + 5)^{1/2} \).
   - Applying the chain rule to differentiate \( (x^3 + 5)^{1/2} \), we get:
     \[
     \frac{d}{dx} (x^3 + 5)^{1/2} = \frac{1}{2}(x^3 + 5)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 5}}.
     \]

Using the **product rule**: 
\[
\frac{dy}{dx} = \frac{d}{dx} \left( 7x^4 \right) \cdot \left( \sqrt{x^3 + 5} \right) + \left( 7x^4 \right) \cdot \frac{d}{dx} \left( \sqrt{x^3 + 5} \right)
\]

So,
\[
\frac{dy}{dx} = 28x^3 \cdot \sqrt{x^3 + 5} + 7x^4 \cdot \frac{3x^2}{2\sqrt{x^3 + 5}}.
\]
This is the derivative for part (a).

### b) \( y = \frac{(2x + 7)^5}{x^2} \)

Here, we are given the function \( y = \frac{(2x + 7)^5}{x^2} \), and again, I assume the question is asking for the derivative \( \frac{dy}{dx} \).

We will use the **quotient rule** and the **chain rule** to differentiate this expression.

#### Steps:

Define:
\[
u = (2x + 7)^5, \quad v = x^2.
\]

The quotient rule states:
\[
\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}.
\]

1. \( u = (2x + 7)^5 \), and using the chain rule:
   \[
   \frac{du}{dx} = 5(2x + 7)^4 \cdot 2 = 10(2x + 7)^4.
   \]
2. \( v = x^2 \), and the derivative is:
   \[
   \frac{dv}{dx} = 2x.
   \]

Now, applying the quotient rule:
\[
\frac{dy}{dx} = \frac{x^2 \cdot 10(2x + 7)^4 - (2x + 7)^5 \cdot 2x}{x^4}.
\]

Simplifying:
\[
\frac{dy}{dx} = \frac{10x^2(2x + 7)^4 - 2x(2x + 7)^5}{x^4}.
\]
We can factor out \( 2x(2x + 7)^4 \):
\[
\frac{dy}{dx} = \frac{2x(2x + 7)^4 \left[ 5x - (2x + 7) \right]}{x^4}.
\]
Finally, simplify the expression in the brackets:
\[
5x - (2x + 7) = 3x - 7,
\]
which gives the final expression:
\[
\frac{dy}{dx} = \frac{2x(2x + 7)^4 (3x - 7)}{x^4}.
\]
Canceling out \( x \) gives:
\[
\frac{dy}{dx} = \frac{2(2x + 7)^4 (3x - 7)}{x^3}.
\]

---

Would you like further clarification on any of these steps? Feel free to ask questions. 

Here are 5 related questions to explore:
1. What is the general product rule for differentiation?
2. How can the chain rule be applied to more complex nested functions?
3. How do you simplify expressions after applying the quotient rule?
4. Can the second derivative \( \frac{d^2y}{dx^2} \) be computed for these functions?
5. How do you approach integration for functions like the ones given in parts (a) and (b)?

**Tip**: Always simplify your final derivative expressions to ensure they are in their cleanest form.

Given two functions: a) y = 7x^4 * sqrt(x^3 + 5) and b) y = (2x + 7)^5 / x^2, find the derivatives.

Explore how to differentiate complex functions like y = 7x^4 * sqrt(x^3 + 5) and y = (2x + 7)^5 / x^2 using the product, quotient, and chain rules. Suitable for students studying Calculus I.

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