#### To determine

h′(e)

#### Answer

h′(e)=1e+1

#### Explanation

**Given:**

f(x)=ex+lnx, h(x)=f−1(x)

Given, f(x)=ex+lnx

Differentiate the above function with respect to *x* and get:

f′(x)=ex+1x

It is given that h(x)=f−1(x)

Also, f(x)=ex+lnx

This implies:

f(1)=e1+ln1=e+0=e

This implies:

f−1(e)=1h(e)=1

So,

h′(e)=1f′(1)=1e+1

**Conclusion:**

Hence, the value of h′(e)=1e+1.