Derivation of nth derivative(term) using mathmatical induction of$~y_{k}^{(k)}=(-1)^{k}x^{-(k+1)}\exp(1/x)~$ $$y=x^{n-1}\exp\left(\frac{1}{x}\right)$$
$$y_{n}=x^{n-1}\exp\left(\frac{1}{x}\right)$$
$$y_{1}^{\left(1\right)}=\left(-1\right)x^{-2}\exp\left(1/x\right)$$
$$y_{2}^{\left(2\right)}=x^{-3}\exp\left(1/x\right)$$
$$y_{3}^{\left(3\right)}=\left(-1\right)x^{-4}\exp\left(1/x\right)$$
$$y_{n}^{\left(n\right)}=\left(-1\right)^{n}x^{-\left(n+1\right)}\exp\left(1/x\right)~~\leftarrow~~\text{Assumption of nth derivative}$$
I will use mathematical induction.
$$\text{Suppose for any}~k\in\mathbb{N}~~,\text{the following is held}$$
$$y_{k}^{\left(k\right)}=\left(-1\right)^{k}x^{-\left(k+1\right)}\exp\left(1/x\right)$$
My attempt is as following
$$y_{k+1}^{\left(k+1\right)}=\left(x^{\left(k+1\right)-1}\exp\left(1/x\right)\right)^{\left(k+1\right)}$$
$$=\left(x^{k}\exp\left(1/x\right)\right)^{\left(k+1\right)}$$
$$=\left(\frac{d}{dx}\left(x^{k}\exp\left(1/x\right)\right)\right)^{\left(k\right)}$$
$$=\left(k x^{k-1}\exp\left(1/x\right)+\frac{d}{dx}\left(x^{-1}\right)\exp\left(1/x\right)x^{k}\right)^{\left(k\right)}$$
$$=\left(k x^{k-1}\exp\left(1/x\right)+\left(-1\right)x^{-2}\exp\left(1/x\right)x^{k}\right)^{\left(k\right)}$$
$$=\left(\exp\left(1/x\right)\left(k x^{k-1}-x^{k-2}\right)\right)^{\left(k\right)}$$
My hand has stopped from here. The following kth derivative should appear.
Should I proceed the calcultion from here?I am waiting for the kth derivative arises inside the k-differentiatior.
$$y_{k}^{\left(k\right)}=\left(-1\right)^{k}x^{-\left(k+1\right)}\exp\left(1/x\right)$$
Or should I change the fundamental approach?
 A: Assume $ \frac{d^k y_k}{dx^n}=(-1)^kx^{-(k+1)}\exp\left(\frac{1}{x}\right)$ holds for some $k \geq 1$.
The first step I took was split the derivative operator:
$$ \frac{d^{k+1} y_{k+1}}{dx^{k+1}} = \frac{d^k}{dx^k} \left( \frac{d}{dx} y_{k+1} \right)  = \frac{d^k}{dx^k} \left( \frac{d}{dx} x^k \exp\left(\frac{1}{x}\right) \right) $$
Computing the inner derivative, we have
$$ \begin{aligned}
\frac{d}{dx} x^k \exp\left(\frac{1}{x}\right) &= kx^{k-1} \exp\left(\frac{1}{x}\right) + x^k \exp\left(\frac{1}{x}\right) \left(-x^{-2}\right) \\
&= x^{k-1}\left(k-\frac{1}{x}\right) \exp\left(\frac{1}{x}\right)
\end{aligned} $$
Thus,
$$ \frac{d^{k+1} y_{k+1}}{dx^{k+1}} = \frac{d^k}{dx^k} \left( x^{k-1}\left(k-\frac{1}{x}\right) \exp\left(\frac{1}{x}\right)\right) $$
We can now use the linearity of the derivative to split this up.
$$ \frac{d^{k+1} y_{k+1}}{dx^{k+1}} = \frac{d^k}{dx^k} \left( kx^{k-1}\exp\left(\frac{1}{x}\right) \right)- \frac{d^k}{dx^k} \left( x^{k-2}\exp\left(\frac{1}{x}\right) \right) $$
I believe, at this point, you may have the necessary hints to keep going. If not, please leave a comment, and I will complete my answer with the full solution. Note that you will need to use the principle of strong induction, i.e., you will need to utilize both $y_k$ and $y_{k-1}$.
