The function presented is:

$y = \left(\ln(x + 5)\right)^x$

We are asked to find the derivative of this function using logarithmic differentiation. Let's proceed step by step.

Steps for Logarithmic Differentiation:

1. Take the natural logarithm of both sides:

   Apply $\ln$ to both sides to simplify the product and exponent: $\ln y = \ln\left[\left(\ln(x + 5)\right)^x\right]$

2. Use logarithm properties:

   We can bring the exponent $x$ down in front, applying the logarithmic property $\ln(a^b) = b\ln(a)$: $\ln y = x \cdot \ln(\ln(x + 5))$

3. Differentiate both sides implicitly:

   • The left side, $\ln y$, differentiates to $\frac{1}{y} \cdot \frac{dy}{dx}$.
   • On the right side, we will need to apply the product rule. Let’s differentiate $x \cdot \ln(\ln(x + 5))$.

   Using the product rule for $(x) \cdot (\ln(\ln(x + 5)))$: [ \frac{d}{dx} \left[ x \cdot \ln(\ln(x + 5)) \right] = 1 \cdot \ln(\ln(x + 5)) + x \cdot \frac{d}{dx}\left[\ln(\ln(x + 5))\right] ]

   Now, for the derivative of $\ln(\ln(x + 5))$, use the chain rule: $\frac{d}{dx} \left[\ln(\ln(x + 5))\right] = \frac{1}{\ln(x + 5)} \cdot \frac{1}{x + 5}$

   So, the full derivative on the right-hand side becomes: $\ln(\ln(x + 5)) + x \cdot \frac{1}{\ln(x + 5)} \cdot \frac{1}{x + 5}$

4. Solve for $\frac{dy}{dx}$:

   Now, multiply both sides by $y$ to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} = y \cdot \left[ \ln(\ln(x + 5)) + \frac{x}{\ln(x + 5)(x + 5)} \right]$

5. Substitute $y = (\ln(x + 5))^x$:

   Since $y = (\ln(x + 5))^x$, we can now substitute it back into the equation: $\frac{dy}{dx} = (\ln(x + 5))^x \cdot \left[ \ln(\ln(x + 5)) + \frac{x}{\ln(x + 5)(x + 5)} \right]$

This gives the derivative of the function $y = \left(\ln(x + 5)\right)^x$ using logarithmic differentiation.

Let me know if you'd like further details or clarification!

Here are 5 related questions for further exploration:

1. How does logarithmic differentiation simplify taking derivatives of functions with variable exponents?
2. What are other cases where logarithmic differentiation is useful?
3. Can logarithmic differentiation be applied to trigonometric functions?
4. How would you compute the second derivative of this function?
5. What happens if we differentiate a logarithmic function with multiple variables?

Tip: Logarithmic differentiation is particularly useful when the function involves both products and powers, especially when the exponent itself is a function of $x$.