If $$g$$ is the inverse of a function $$f$$ and $$f'(x) = \frac{1}{1+x^5}$$, then $$g'(x)$$ is equal to:

We wish to find the derivative of the inverse function. Let us denote the given function by $$y=f(x)$$ and its inverse by $$x=g(y)$$, so that $$g=f^{-1}$$.

Because $$g$$ and $$f$$ are inverses, we have the relation $$f\!\bigl(g(y)\bigr)=y$$ for every value of $$y$$ that lies in the domain of $$g$$.

Now we differentiate the identity $$f\!\bigl(g(y)\bigr)=y$$ with respect to $$y$$. Applying the chain rule, we obtain

$$f'\!\bigl(g(y)\bigr)\,\cdot\,g'(y)=1.$$

We want an explicit expression for $$g'(y)$$, so we solve for it algebraically. Dividing both sides by $$f'\!\bigl(g(y)\bigr)$$ we get

$$g'(y)=\frac{1}{f'\!\bigl(g(y)\bigr)}.$$

We now substitute the given derivative of $$f$$. According to the statement of the problem,

$$f'(x)=\frac{1}{1+x^{5}}.$$

Replacing the variable $$x$$ by $$g(y)$$ in this formula gives

$$f'\!\bigl(g(y)\bigr)=\frac{1}{1+\{g(y)\}^{5}}.$$

Substituting this result into our expression for $$g'(y)$$, we obtain

$$g'(y)=\frac{1}{\dfrac{1}{1+\{g(y)\}^{5}}}.$$

Simplifying the complex fraction (since dividing by a fraction is the same as multiplying by its reciprocal), we find

$$g'(y)=1+\{g(y)\}^{5}.$$

Because the independent variable in the final answer is conventionally written as $$x$$, we can replace $$y$$ by $$x$$ without changing the meaning:

$$g'(x)=1+\{g(x)\}^{5}.$$

This matches Option B in the list provided.

Hence, the correct answer is Option B.