),x>0 Differentiating w.r.t. x, we get f′(x)=‌
d
dx
(x)−‌
d
dx
‌log(‌
1+x
x
) =1−‌
1
1+x
x
⋅‌
d
dx
(‌
1+x
x
)‌‌ [from chain rule]
=1−‌
x
1+x
⋅{‌
x
d
dx
(1+x)−(1+x)
d
dx
(x)
x2
} =1−‌
x
1+x
{‌
x−(1+x)
x2
}=1−‌
x
1+x
{−‌
1
x2
}
f′(x)=1+‌
1
x(x+1)
Since x>0, therefore there is no possibility of f′(x) to be zero at any value of x. So, for the given function f′(x)≠0 Also, for ∀x∈R‌‌f′(x)>0 i.e., function f(x) is strictly increasing its defined domain. So, reason correctly explains given assertion.