Question

What is the equation of the line tangent to `f(x)=x^2/e^x-x/e^(x^2)` at `x=0`?

Answer 1

`f'(x)=-1`

Explanation:

Let's rewrite the function as `f(x)=x^2e^{-x}-xe^{-x^2}`

Reminding that the Mac Laurin of `e^x` is `1+x+x^2/2+o(x^2)`,
the one of `e^{-x}` is `1-x+x^2/2+o(x^2)`,
and the one of `e^{-x^2}` is `1-x^2+x^4/2+o(x^4)`
By replacing them, we get
`f(x)=x^2(1-x+x^2/2+o(x^2))-x(1-x^2+x^4/2+o(x^4))`.
By carrying out the multiplication we get
`-x+x^2+x^4/2-x^5/2+o(x^4)`
In the end `f'(0)=-1`, i.e. the coefficient of the linear term in the Mac Laurin polynomial for `f(x)`
Enjoy!