Find (d^ny)/(dx^n) for x = e^-at, y = e^at
Find $\dfrac{d^{n}y}{dx^{n}}$ for $x = e^{-at},\ y = e^{at}$.
Explanation of Method
Every step multiplies by −n and adds one to the exponent.
Step 1: The first three derivatives
From Problem 319: \[y'_{x} = -e^{2at}, \qquad y''_{xx} = 2e^{3at}, \qquad y'''_{xxx} = -6e^{4at}.\]
The coefficients are $-1,\,2,\,-6$, i.e. $\bigl(-1\bigr)^{n}n!$, and the exponent is $\bigl(n+1\bigr)at$.
Step 2: The conjecture
\[ \frac{d^{n}y}{dx^{n}} = \bigl(-1\bigr)^{n}\,n!\,e^{\,(n+1)at}. \tag{1} \]
Step 3: Proof by induction
Base case. $n = 1$ gives $\bigl(-1\bigr)\cdot1!\cdot e^{2at} = -e^{2at}$.
Inductive step. Assume (1). Differentiating with respect to $t$ brings down $\bigl(n+1\bigr)a$:
\[ \left(\frac{d^{n}y}{dx^{n}}\right)'_{t} = \bigl(-1\bigr)^{n}n!\,\bigl(n+1\bigr)a\,e^{\,(n+1)at}, \]
and dividing by $\dot x = -ae^{-at}$ multiplies by $-\dfrac{1}{a}e^{at}$: $\frac{d^{\,n+1}y}{dx^{\,n+1}} = \bigl(-1\bigr)^{n+1}\bigl(n+1\bigr)!\,e^{\,(n+2)at}$.
which is (1) at order $n+1$.
Step 4: Verification
\[xy = 1 \;\Longrightarrow\; y = \frac1x, \qquad \frac{d^{n}}{dx^{n}}\left(\frac1x\right) = \frac{\bigl(-1\bigr)^{n}n!}{x^{\,n+1}}\]
\[x^{\,n+1} = e^{-(n+1)at} \;\Longrightarrow\; \frac{d^{n}y}{dx^{n}} = \bigl(-1\bigr)^{n}n!\,e^{\,(n+1)at} \;\;\checkmark\]
\[n = 1,2,3:\ -e^{2at},\ 2e^{3at},\ -6e^{4at} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{d^{n}y}{dx^{n}} = \bigl(-1\bigr)^{n}\,n!\,e^{\,(n+1)at}\,}\]
Remark
Remark. This is Problem 290 in parametric dress: the $n$-th derivative of $1/x$ is $\bigl(-1\bigr)^{n}n!\,x^{-(n+1)}$, and the parametrisation merely rewrites $x^{-(n+1)}$ as $e^{(n+1)at}$. It is a useful check on the parametric machinery that it reproduces a known answer exactly, including the alternating sign and the factorial.