How to find approximate value of $1.01e^{1.01({0.99) }^2} $?

Question

I want to find the approximate value of $1.01e^{1.01({0.99) }^2}$ by using derivative. I tried choosing x=1 and $\delta x=0.01$ it didnt work. How can I start?

Answer 1

Use $f(x)=(1+x)e^{(1+x)(1-x)^2}$, $f(0.01)\approx f(0)+0.01f'(0)$

Answer 2

Following frank000: Take a Taylor series of $f(x) = (1+x) e^{(1+x)(1-x)^2}$ around $x = 0$:

$e-\frac{3 e \delta ^2}{2}+\frac{4 e \delta ^3}{3}+\frac{7 e \delta ^4}{8}-\frac{9 e \delta ^5}{5}+O\left(\delta ^6\right)$

and substitute $\delta \to 0.01$ to find:

$2.71788$, which is also the value given by Mathematica for the explicit computation in the title.