You are given the function:

$g(x) = e^{2x} \left( \ln\left(x^2 + 4\right)^{\frac{1}{2}} \right)$

To work with this function, let’s simplify it and then, if needed, find its derivative.

Step 1: Simplify the logarithmic expression

The expression $\ln\left(x^2 + 4\right)^{\frac{1}{2}}$ can be simplified using the logarithmic identity:

$\ln(a^b) = b \ln(a)$

Thus:

$\ln\left(x^2 + 4\right)^{\frac{1}{2}} = \frac{1}{2} \ln\left(x^2 + 4\right)$

So the function simplifies to:

$g(x) = e^{2x} \left(\frac{1}{2} \ln\left(x^2 + 4\right)\right)$

$g(x) = \frac{1}{2} e^{2x} \ln\left(x^2 + 4\right)$

This is a more compact form of $g(x)$.

Step 2: Derivative of $g(x)$ (if needed)

To differentiate this function, we will use the product rule and the chain rule. The product rule states:

$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

Here, let:

• $f(x) = \frac{1}{2} \ln(x^2 + 4)$
• $h(x) = e^{2x}$

Now, let’s differentiate each part:

1. Derivative of $h(x) = e^{2x}$:

$\frac{d}{dx} e^{2x} = 2e^{2x}$

2. Derivative of $f(x) = \frac{1}{2} \ln(x^2 + 4)$: Using the chain rule for the logarithmic derivative:

$\frac{d}{dx} \ln(x^2 + 4) = \frac{2x}{x^2 + 4}$

Thus, the derivative of $f(x)$ is:

$\frac{d}{dx} \left( \frac{1}{2} \ln(x^2 + 4) \right) = \frac{1}{2} \cdot \frac{2x}{x^2 + 4} = \frac{x}{x^2 + 4}$

Step 3: Apply the product rule

Now, applying the product rule:

$g'(x) = \left(\frac{x}{x^2 + 4}\right) e^{2x} + \frac{1}{2} \ln(x^2 + 4) \cdot 2e^{2x}$

Simplifying:

$g'(x) = \frac{x e^{2x}}{x^2 + 4} + e^{2x} \ln(x^2 + 4)$

This is the derivative of the function $g(x)$.

Final result:

The derivative of the given function is:

$g'(x) = e^{2x} \left( \frac{x}{x^2 + 4} + \ln(x^2 + 4) \right)$

Would you like more details on any of these steps?

Relative Questions:

1. How do you apply the product rule in differentiation?
2. What is the chain rule and how is it applied here?
3. Why does the logarithmic identity simplify the original function?
4. What are the common derivatives involving exponential functions?
5. How would you integrate the simplified version of $g(x)$?

Tip:

When differentiating functions with products of terms, always identify the parts of the product to apply the product rule correctly.