f(x) = x ln (x+1)

First let us find f'(x)

Let f(x) = u*v

u = x ==> u' = 1

v = ln (x+1) ==> v' = 1/(x+1)

Then:

f'(x) = u'v + uv'

= x/(x+1) + ln (x+ 1

f'(1) = 1/2 + ln 2

f'(0) = 0 + ln 1 = ln 1

==> f'(1) - f'(0) = 1/2 + ln 2 - ln 1

= 1/2 + ln 2/1

= 1/2 + ln 2

We notice that the expression to be calculated contains the first derivative of f(x), so, in order to evaluate the value of the derivative f'(0), we'll have to find first, the derivative of the function.

We'll use the product law, so that:

f'(x) = x'*ln(x+1) + x*ln(x+1)'

f'(x) = ln(x+1) + x/(x+1)

Now, we'll calculate f'(0), substituting x by 0, in the expression of f'(x).

f'(0) = ln(0+1) + 0/(0+1)

But ln1=0, so:

f'(0) = 0

Now, we'll calculate f(1), by substituting x by 1.

f(1) = 1*ln(1+1)

f(1) = ln 2

**So, the difference is:**

**f(1) - f'(0) = ln 2- 0 = ln 2.**

f(x) = xln(x+1)

To find f(1) -f'(0).

f(1) = 1ln (1+1) = ln2.

f'(x) = (xlnx)' = (x)' lnx +x(lnx)'

= 1ln(x+1)+x/(x+1) = ln(x+1) + x/(x+1)

f'(0) = ln(0+1)) +0/(0+1) = ln1 +0/1

= 0, as ln1 = 0.

Therefore f(1) - f'(0) = ln2 - 0.

f(1) = ln2