Question

Differentiate with respect to x, y= (x+1)to power x

Answer 1

To differentiate the function \( y = (x+1)^x \) with respect to \( x \), we will use logarithmic differentiation. Here are the steps:

1. **Take the natural logarithm of both sides**:

\[
\ln y = \ln((x+1)^x)
\]

2. **Use the properties of logarithms**:

\[
\ln y = x \ln(x+1)
\]

3. **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:

\[
\frac{d}{dx}(x \ln(x+1)) = \ln(x+1) + x \cdot \frac{1}{x+1} \cdot 1 = \ln(x+1) + \frac{x}{x+1}
\]

4. **Set the derivatives equal to each other**:

\[
\frac{1}{y} \frac{dy}{dx} = \ln(x+1) + \frac{x}{x+1}
\]

5. **Multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \)**:

\[
\frac{dy}{dx} = y \left(\ln(x+1) + \frac{x}{x+1}\right)
\]

6. **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:

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