Are these derivatives correct??

Question

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?

Answer 1

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 $$ f'''(x)= 8x^{-9}\exp{(-1/x^2)}-36x^{-7}\exp{(-1/x^2)}+24x^{-5}\exp{(-1/x^2)} $$

Answer 2

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.