To solve this problem using logarithmic differentiation, we have the function:

$y = x^{(x+1)}$

Step-by-step solution:

1. Take the natural logarithm of both sides: $\ln(y) = \ln\left(x^{x+1}\right)$

2. Use the logarithmic identity $\ln(a^b) = b \cdot \ln(a)$: $\ln(y) = (x+1) \cdot \ln(x)$

3. Differentiate both sides with respect to $x$. We need to apply implicit differentiation on the left and product/chain rule on the right: $\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}\left((x+1) \ln(x)\right)$

4. Differentiate the right side:

   • Use product rule on $(x+1)\ln(x)$: $\frac{d}{dx}\left((x+1)\ln(x)\right) = \frac{d}{dx}(x+1) \cdot \ln(x) + (x+1) \cdot \frac{d}{dx}(\ln(x))$
   • $\frac{d}{dx}(x+1) = 1$
   • $\frac{d}{dx}(\ln(x)) = \frac{1}{x}$

   So: $\frac{dy}{dx} = 1 \cdot \ln(x) + (x+1) \cdot \frac{1}{x}$ Simplifying: $\frac{dy}{dx} = \ln(x) + 1 + \frac{1}{x}$

5. Multiply both sides by $y$ (which is $x^{(x+1)}$) to solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = x^{(x+1)} \left( \ln(x) + 1 + \frac{1}{x} \right)$

Final result:

$\frac{dy}{dx} = x^{(x+1)} \left( \ln(x) + 1 + \frac{1}{x} \right)$

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

Here are five related questions to expand your understanding:

1. How do you apply logarithmic differentiation to more complex functions, like $y = e^{x^2}$?
2. What is the importance of using the chain rule in implicit differentiation?
3. How does logarithmic differentiation help in simplifying functions with variables as exponents?
4. Can you differentiate the function $y = x^x$ using logarithmic differentiation?
5. What would change in the solution if the base of the power was a constant, like $y = 2^{(x+1)}$?

Tip:

Logarithmic differentiation is especially useful when the variable appears both in the base and the exponent, as it simplifies the differentiation process by turning multiplicative relationships into additive ones.