https://file.aiquickdraw.com/mathbot/file/1730439023269evy9jkko.jpg

Let's solve each part step-by-step.

### Part (a)
Given:
\[
g(x) = 3x^2 + 4
\]
We are asked to find \( g'(1) \).

1. **Differentiate \( g(x) \) with respect to \( x \):**
   \[
   g'(x) = \frac{d}{dx}(3x^2 + 4) = 3 \cdot 2x = 6x
   \]

2. **Evaluate \( g'(1) \):**
   \[
   g'(1) = 6 \cdot 1 = 6
   \]

**Answer:** \( g'(1) = 6 \).

---

### Part (b)
Given:
\[
h(r) = r^2 + 2^r
\]
We are asked to find \( h'(r) \).

1. **Differentiate \( h(r) \) with respect to \( r \):**
   - For \( r^2 \), the derivative is \( 2r \).
   - For \( 2^r \), we use the rule for exponential functions: \( \frac{d}{dr}(a^r) = a^r \ln(a) \).
   
   Thus:
   \[
   h'(r) = 2r + 2^r \ln(2)
   \]

**Answer:** \( h'(r) = 2r + 2^r \ln(2) \).

---

### Part (c)
Given:
\[
z = 9^{4 \log_9(x)}
\]
We are asked to find \( z'' \), the second derivative of \( z \) with respect to \( x \).

1. **Simplify \( z \):**
   Using properties of exponents and logarithms, \( 9^{4 \log_9(x)} \) simplifies as follows:
   \[
   z = x^4
   \]

2. **Find the first derivative, \( z' \):**
   \[
   z' = \frac{d}{dx}(x^4) = 4x^3
   \]

3. **Find the second derivative, \( z'' \):**
   \[
   z'' = \frac{d}{dx}(4x^3) = 12x^2
   \]

**Answer:** \( z'' = 12x^2 \).

---

### Part (d)
Given:
\[
y(x) = \sqrt{e^{2x} + \ln(x^2)}
\]
We are asked to find \( y'(x) \).

1. **Rewrite \( y(x) \) for easier differentiation:**
   \[
   y(x) = \left(e^{2x} + \ln(x^2)\right)^{1/2}
   \]

2. **Differentiate \( y(x) \) using the chain rule:**
   Let \( u = e^{2x} + \ln(x^2) \), so \( y = u^{1/2} \). Then,
   \[
   y' = \frac{1}{2} u^{-1/2} \cdot u'
   \]
   
   Now, find \( u' \):
   - For \( e^{2x} \), the derivative is \( 2e^{2x} \).
   - For \( \ln(x^2) \), use the chain rule: \( \frac{d}{dx} \ln(x^2) = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \).

   So,
   \[
   u' = 2e^{2x} + \frac{2}{x}
   \]

3. **Combine the results:**
   \[
   y' = \frac{1}{2} \cdot \frac{1}{\sqrt{e^{2x} + \ln(x^2)}} \cdot \left(2e^{2x} + \frac{2}{x}\right)
   \]
   Simplifying further:
   \[
   y' = \frac{e^{2x} + \frac{1}{x}}{\sqrt{e^{2x} + \ln(x^2)}}
   \]

**Answer:** \( y' = \frac{e^{2x} + \frac{1}{x}}{\sqrt{e^{2x} + \ln(x^2)}} \).

---

Would you like further details on any of these solutions, or do you have specific questions? Here are some related questions to expand on these concepts:

1. How do you find higher-order derivatives of polynomial functions?
2. What is the chain rule, and how is it applied in derivative calculations?
3. How can we simplify expressions involving logarithmic and exponential functions before differentiating?
4. Why do exponential functions of the form \( a^x \) have derivatives involving \( \ln(a) \)?
5. How does implicit differentiation work, and when would you use it?

**Tip:** When differentiating complex functions, try simplifying the expression first—it often makes the differentiation process easier.

Find the derivatives as indicated in the following:

a) g(x) = 3x^2 + 4 ; g'(1)

b) h(r) = r^2 + 2^r ; h'(r)

c) z = 9^{4 log_9(x)} ; z''

d) y(x) = sqrt(e^{2x} + ln(x^2)) ; y'(x)

This page provides solutions to derivative problems involving polynomial, exponential, and logarithmic functions. Topics include applying the chain rule, differentiating functions with exponents and roots, and using logarithmic properties for simplification. Ideal for college-level calculus students.

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