Let $f(x)={\int }_0^x\frac{e^t}{1+t}dt$. Find $f^{\prime \prime }(0)$.

Solution

1. By the Fundamental Theorem of Calculus, $f^{\prime }(x)=\frac{e^x}{1+x}$.
2. Differentiating again using the quotient rule, $f^{\prime \prime }(x)=\frac{(1+x)e^x-e^x(1)}{(1+x{)}^2}$.
3. Simplifying, $f^{\prime \prime }(x)=\frac{x{e}^x}{(1+x{)}^2}$.
4. Evaluating at $x=0$, $$\begin{array}{rl}{f}^{\prime \prime }(0)& =\frac{0\cdot 1}{1^2}\\ & =0\end{array}$$ is wrong; $$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{e^x(1+x)-e^x}{(1+x{)}^2}\\ & =\frac{x{e}^x}{(1+x{)}^2}\end{array}$$.
5. Hence the answer is (A).