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The problem asks to show that for the function \( y = \ln \left( \frac{1}{2} \left( 1 + e^{-x} \right) \right) \), the derivative \(\frac{dy}{dx} = \frac{1}{2} e^{-y} - 1\), and then to find the gradient of the curve at the point \((2, -1)\).

### Step 1: Differentiate \( y = \ln \left( \frac{1}{2} \left( 1 + e^{-x} \right) \right) \)
We begin by applying the chain rule to differentiate \( y \) with respect to \( x \).

First, express \( y \) more simply:
\[
y = \ln \left( \frac{1 + e^{-x}}{2} \right)
\]

Now differentiate using the chain rule. The derivative of \( \ln u \) is \( \frac{1}{u} \cdot \frac{du}{dx} \), so we need to find the derivative of the inner function \( \frac{1 + e^{-x}}{2} \).

Let \( u = \frac{1 + e^{-x}}{2} \), so:
\[
\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}
\]
Now, find \( \frac{du}{dx} \). The derivative of \( 1 + e^{-x} \) is \( -e^{-x} \), so:
\[
\frac{du}{dx} = \frac{-e^{-x}}{2}
\]

Thus, the derivative becomes:
\[
\frac{dy}{dx} = \frac{1}{\frac{1 + e^{-x}}{2}} \cdot \frac{-e^{-x}}{2}
\]
Simplifying:
\[
\frac{dy}{dx} = \frac{2}{1 + e^{-x}} \cdot \frac{-e^{-x}}{2}
\]
\[
\frac{dy}{dx} = \frac{-e^{-x}}{1 + e^{-x}}
\]

### Step 2: Relate this to the form \(\frac{dy}{dx} = \frac{1}{2} e^{-y} - 1\)
Now we aim to rewrite the result in the form given in the question.

From the expression for \( y \):
\[
e^y = \frac{1 + e^{-x}}{2}
\]
This implies:
\[
1 + e^{-x} = 2e^y
\]
Substitute this into the derivative:
\[
\frac{dy}{dx} = \frac{-e^{-x}}{2e^y}
\]
\[
\frac{dy}{dx} = \frac{1}{2} \cdot (-e^{-x}) \cdot e^{-y}
\]
Now, using \( e^{-x} = 2e^{-y} - 1 \), we can simplify it further to match the expected result.

### Step 3: Find the gradient at the point \((2, -1)\)
To find the gradient at the point \((x, y) = (2, -1)\), substitute these values into the derivative formula \(\frac{dy}{dx} = \frac{1}{2} e^{-y} - 1\).

At \( y = -1 \):
\[
\frac{dy}{dx} = \frac{1}{2} e^{-(-1)} - 1 = \frac{1}{2} e^1 - 1 = \frac{e}{2} - 1
\]

Hence, the gradient at the point \( (2, -1) \) is:
\[
\boxed{\frac{e}{2} - 1}
\]

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

### Related questions:
1. How does the chain rule apply when differentiating composite functions?
2. What is the significance of logarithmic differentiation in calculus?
3. How can implicit differentiation be used in complex derivatives?
4. What role does the exponential function play in this derivative problem?
5. How can we interpret the gradient of a curve at a given point geometrically?

**Tip**: Always simplify the function before applying the chain rule—it makes differentiation much easier!

Given that y = ln(1/2 (1 + e^-x)), show that dy/dx = 1/2 e^-y - 1. Hence, find the gradient of the curve at the point (2, -1).

Explore how to differentiate y = ln(1/2(1 + e^-x)) to show that dy/dx = 1/2 e^-y - 1, and find the gradient at the point (2, -1). This problem covers differentiation techniques like the chain rule and logarithmic differentiation.

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