Answer
$\dfrac{e^2+1}{4}$
Work Step by Step
The first $4$ non-zero terms of Taylor series can be written as: $e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+.......$
Now, we have:
$\dfrac{e^x-1}{x}=\dfrac{(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+.......)-1}{x}=\Sigma_{k=0}^{\infty} \dfrac{x^k}{(k+1)!}$
Next, $f'(x)=\dfrac{e^x(x-1)+1}{x^2}=\Sigma_{k=0}^{\infty} \dfrac{kx^{k-1}}{(k+1)!}$
Plug $x=1$ in the above equation to obtain:
$f'(2)=\Sigma_{k=0}^{\infty} \dfrac{k(2^{k-1})}{(k+1)!}=\dfrac{e^2+1}{4}$
