Let $$f$$ and $$g$$ be differentiable functions on $$R$$ such that $$fog$$ is the identity function. If for some $$a, b \in R$$, $$g'(a) = 5$$ and $$g(a) = b$$, then $$f'(b)$$ is equal to:

We are told that $$f \circ g$$ is the identity function on $$\mathbb R$$. In symbols this means

$$f\bigl(g(x)\bigr)=x \quad\text{for every real }x.$$

To connect the derivatives of $$f$$ and $$g$$ we differentiate both sides of this equality with respect to $$x$$. First we recall the Chain Rule:

The Chain Rule states that if $$h(x)=u(v(x)),$$ then $$h'(x)=u'\bigl(v(x)\bigr)\,v'(x).$$

Applying the Chain Rule to $$h(x)=f\bigl(g(x)\bigr)$$ we obtain

$$\bigl(f \circ g\bigr)'(x)=f'\bigl(g(x)\bigr)\,g'(x).$$

Because $$f \circ g$$ is the identity function, its derivative is simply $$1$$ for every $$x$$, so we have

$$f'\bigl(g(x)\bigr)\,g'(x)=1.$$

Now we substitute the specific value $$x=a$$ that is mentioned in the question. This gives

$$f'\bigl(g(a)\bigr)\,g'(a)=1.$$

The problem states that $$g'(a)=5$$ and that $$g(a)=b$$, so we replace these quantities:

$$f'(b)\,\bigl(5\bigr)=1.$$

To isolate $$f'(b)$$ we divide both sides by $$5$$:

$$f'(b)=\frac{1}{5}.$$

Hence, the correct answer is Option A.