To find $g'(1)$, we need to differentiate the function $g(x) = 2f(x) + (f^{-1}(x))^2$. Given that $f(x)$ is differentiable and $f^{-1}(x)$ is its inverse, let's go through the differentiation step-by-step.

Step 1: Differentiate $g(x)$

The function $g(x)$ is defined as: $g(x) = 2f(x) + \left(f^{-1}(x)\right)^2.$ We need to find $g'(x)$. Using the sum rule, we differentiate each part separately: $g'(x) = 2f'(x) + 2f^{-1}(x) \cdot \frac{d}{dx}[f^{-1}(x)].$

Step 2: Differentiate $f^{-1}(x)$

The derivative of the inverse function $f^{-1}(x)$ is given by: $\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'\left(f^{-1}(x)\right)}.$ So, the expression for $g'(x)$ becomes: $g'(x) = 2f'(x) + 2f^{-1}(x) \cdot \frac{1}{f'\left(f^{-1}(x)\right)}.$

Step 3: Evaluate $g'(1)$

We need to find $g'(1)$. First, we need to determine $f^{-1}(1)$. Since $f(1) = 1$, by the property of inverse functions, $f^{-1}(1) = 1$.

Now substitute $x = 1$ into the derivative: $g'(1) = 2f'(1) + 2f^{-1}(1) \cdot \frac{1}{f'\left(f^{-1}(1)\right)}.$ Substitute the values $f'(1) = \frac{1}{2}$ and $f^{-1}(1) = 1$: $g'(1) = 2 \cdot \frac{1}{2} + 2 \cdot 1 \cdot \frac{1}{f'(1)}.$ $g'(1) = 1 + 2 \cdot \frac{1}{\frac{1}{2}}.$ $g'(1) = 1 + 2 \cdot 2.$ $g'(1) = 1 + 4 = 5.$

Thus, the value of $g'(1)$ is $5$.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the geometric interpretation of the derivative of an inverse function?
2. How can the inverse function theorem be applied in different contexts?
3. What happens to $g'(x)$ if $f(x)$ is not differentiable at some point?
4. How does the chain rule apply when differentiating composite functions involving inverse functions?
5. How do you find the inverse of more complex functions analytically?

Tip: When working with inverse functions, remember that $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$. This is a crucial identity for differentiating expressions involving inverses.