To differentiate the function \( y = (x+1)^x \) with respect to \( x \), we will use logarithmic differentiation. Here are the steps:
- Take the natural logarithm of both sides:
\[ \ln y = \ln((x+1)^x) \]
- Use the properties of logarithms:
\[ \ln y = x \ln(x+1) \]
- Differentiate both sides with respect to \( x \) using the product rule on the right side:
- The left side requires implicit differentiation:
\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \]
- The right side requires the product rule and chain rule:
\[ \frac{d}{dx}(x \ln(x+1)) = \ln(x+1) + x \cdot \frac{1}{x+1} \cdot 1 = \ln(x+1) + \frac{x}{x+1} \]
- Set the derivatives equal to each other:
\[ \frac{1}{y} \frac{dy}{dx} = \ln(x+1) + \frac{x}{x+1} \]
- Multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left(\ln(x+1) + \frac{x}{x+1}\right) \]
- Substitute back for \( y \):
Since \( y = (x+1)^x \), we replace \( y \):
\[ \frac{dy}{dx} = (x+1)^x \left(\ln(x+1) + \frac{x}{x+1}\right) \]
So, the final result for the derivative is:
\[ \frac{dy}{dx} = (x+1)^x \left(\ln(x+1) + \frac{x}{x+1}\right) \]