We are tasked with using logarithmic differentiation to find $\frac{dy}{dx}$ for the function:

$y = \frac{x^x}{x + 1}$ at $x = 1$, and then determining the equation of the tangent line at that point.

Step 1: Logarithmic Differentiation of the Function

Given $y = \frac{x^x}{x + 1}$, we will first take the natural logarithm of both sides:

$\ln y = \ln \left( \frac{x^x}{x + 1} \right) = \ln(x^x) - \ln(x + 1)$

We know that $\ln(x^x) = x \ln x$, so:

$\ln y = x \ln x - \ln(x + 1)$

Now differentiate both sides with respect to $x$, using implicit differentiation on the left and applying the product and chain rules on the right:

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} \left( x \ln x \right) - \frac{d}{dx} \left( \ln(x + 1) \right)$

The derivative of $x \ln x$ is:

$\frac{d}{dx} (x \ln x) = \ln x + 1$

The derivative of $\ln(x + 1)$ is:

$\frac{d}{dx} \ln(x + 1) = \frac{1}{x + 1}$

So the equation becomes:

$\frac{1}{y} \cdot \frac{dy}{dx} = (\ln x + 1) - \frac{1}{x + 1}$

Now solve for $\frac{dy}{dx}$:

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

Since $y = \frac{x^x}{x + 1}$, substitute this back in:

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

Step 2: Evaluate at $x = 1$

Now substitute $x = 1$ into the equation:

1. For $x^x = 1^1 = 1$,
2. $\ln 1 = 0$,
3. So the expression becomes:

$\frac{dy}{dx} = \frac{1}{1 + 1} \left( 0 + 1 - \frac{1}{1 + 1} \right) = \frac{1}{2} \left( 1 - \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

Thus, $\frac{dy}{dx}$ at $x = 1$ is $\frac{1}{4}$.

Step 3: Find the Equation of the Tangent Line

The equation of the tangent line is given by:

$y - y_1 = m(x - x_1)$

where $m = \frac{dy}{dx}$, and $(x_1, y_1)$ is the point on the curve at $x = 1$.

To find $y_1$, evaluate $y$ at $x = 1$:

$y_1 = \frac{1^1}{1 + 1} = \frac{1}{2}$

So the point is $(1, \frac{1}{2})$, and the slope is $m = \frac{1}{4}$.

Thus, the equation of the tangent line is:

$y - \frac{1}{2} = \frac{1}{4}(x - 1)$

Simplifying:

$y = \frac{1}{4}(x - 1) + \frac{1}{2}$

$y = \frac{1}{4}x + \frac{1}{4}$

Final Answer

The derivative is $\frac{dy}{dx} = \frac{1}{4}$ at $x = 1$, and the equation of the tangent line is:

$y = \frac{1}{4}x + \frac{1}{4}$

Would you like further details or explanations on any part?

Here are 5 related questions:

1. How does logarithmic differentiation simplify finding derivatives?
2. What is the significance of the natural logarithm in calculus?
3. Can you explain how the product and chain rules apply here?
4. How do we derive the equation of a tangent line in general?
5. What are other methods to differentiate functions like $x^x$?

Tip: Logarithmic differentiation is especially useful for functions where variables appear both as a base and exponent.