What and how do derivate?

Question

How do I derive this function?

$f(x) = x(e^{-x^2})$

I need the first and second derivative.

Answer 1

Hint: Use product rule then chain rule.

Answer 2

First, remember the product and chain rules:

Product/Leibniz Rule: $(fg)'=f'g+fg'$

Chain Rule: $(f\circ g)'=(f'\circ g)g'$

First, we have to see what are the "outernmost" simple functions in the definition of $f(x)=x\cdot e^{-x^2}$: If we let $g(x)=x$ and $h(x)=e^{-x^2}$, then $f=gh$, so using the product rule: $$f'(x)=g'(x)h(x)+g(x)h'(x)$$ In the equation above, you know $g(x)$, $g'(x)$, and $h(x)$, sothe problem is to calculate $h'(x)$. Notice that $h'(x)=e^{-x^2}$. Letting $i(x)=e^x$ and $j(x)=-x^2$, then $h=i\circ j$, so, by the chain rule, $$h'(x)=i'(j(x))\cdot j'(x)$$ and you probably know the values of $i'(y)$ and $j'(x)$. Use the two equations above and find $f'(x)$.