Are these derivatives correct?? Take take the function defined as $$f(x) =
\left\{
 \begin{array}{ll}
  exp(\dfrac{-1}{x^{2}})  & \mbox{if } x \neq 0 \\
  0 & \mbox{if } x = 0
 \end{array}
\right.
$$
Now I am asked to check that

I am pretty sure that there is a mistake in the derivatives given in the questions, as when I do it for example I get $$f'(x)=2x^{-3}exp(-\dfrac{1}{x^{2}})$$
and this leads to me getting different $f'' and f'''$ but I checked my answer multiple times. Is there a mistake in the picture above?
 A: we have
\begin{align}
f(x)=\exp\left(-\frac{1}{x^2}\right)=\exp\left(-x^{-2}\right)
\end{align}
Let $u=-x^{-2}$, then by chain rule, we get
\begin{align}
f'(x)&=\frac{d}{dx}\exp\left(-x^{-2}\right)\\
&=\frac{d}{du}\exp\left(u\right)\cdot\frac{du}{dx}\\
&=\exp\left(u\right)\cdot\frac{d}{dx}\left(-x^{-2}\right)\\
&=\exp\left(-x^{-2}\right)\left(-(-2)x^{-2-1}\right)\\
&=\frac{2}{x^3}\exp\left(-x^{-2}\right)
\end{align}
Hence, the first derivative in the question is wrong. Similarly with the higher order derivative, you can use chain rule to obtain the derivative.
A: The first derivative is
$$
f'(x) = 2x^{-3}\exp (-1/x^2)
$$
in general you have
$$
d(\exp(f(x)))/dx = f'(x) \exp(f(x))
$$
You are almost right. But the argument of the $\exp()$ function will not change. The others follow. It seems to me that there are mistakes also in the other derivatives (in the picture)...
If I am not mistaken you get
$$
f''(x) = 4x^{-6}\exp{(-1/x^2)}-6x^{-4}\exp{(-1/x^2)}
$$
and
$$
f'''(x)= 8x^{-9}\exp{(-1/x^2)}-36x^{-7}\exp{(-1/x^2)}+24x^{-5}\exp{(-1/x^2)}
$$
