#### To determine

**To find:**

The value of (f−1)′(π4) for the given function.

**Solution:** 23.

#### Explanation

**Given:**

f(x)=lnx+tan−1x

**Formulae used:** The (f−1)′=1f′(f−1(x))

Consider f(x)=lnx+tan−1x

The domain of f−1(x) is the range of f(x) and the range of f−1(x) is the domain of f(x)

The domain of f−1(x) is π4. It means that the range of the f(x) is π4

Therefore,

f(x)=lnx+tan−1xlnx+tan−1x=π4lnx+tan−1x=tan−11lnx=tan−11−tan−1x

Only x=1 is satisfied the above equation.

Thus, the domain of f(x) is 1. It means that the range of f−1(x) is also 1

Hence, use the formulae of (f−1)′=1f′(f−1(x))

Now according to the given expression, the value of the expression (f−1)′(π4)

1f′(f−1(π4))=1f′(1)f′(x)=1x+11+x2f′(1)=321f′(f−1(π4))=23

**Conclusion:** The value of expression (f−1)′(π4) is 23.