Find the n -th derivative of y = e^x
Find the $n$-th derivative of $y = e^{x}$.
Explanation of Method
The function that is its own derivative.
Step 1: The first few derivatives
\[ y' = e^{x}, \qquad y'' = e^{x}, \qquad y''' = e^{x}. \]
Nothing changes, which makes the conjecture immediate.
Step 2: Proof by induction
Base case. $y^{(1)} = \bigl(e^{x}\bigr)' = e^{x}$.
Inductive step. If $y^{(k)} = e^{x}$ for some $k\ge1$, then
\[ y^{(k+1)} = \bigl(y^{(k)}\bigr)' = \bigl(e^{x}\bigr)' = e^{x}. \]
Hence the statement holds for every $n\ge1$.
Step 3: Verification
\[e^{x} = \sum_{k=0}^{\infty}\frac{x^{k}}{k!}, \qquad \text{differentiating term by term reproduces the same series} \;\;\checkmark\]
\[y^{(n)}(0) = 1 \text{ for every } n, \text{ giving the coefficients } \frac{1}{n!} \;\;\checkmark\]
\[\text{For } y = e^{ax} \text{ the same argument gives } y^{(n)} = a^{n}e^{ax} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y^{(n)} = e^{x}\,}\]
Remark
Remark. The property $y' = y$ with $y(0) = 1$ defines the exponential: no other function has it. Every one of the $n$-th-derivative formulas in the problems that follow — for $a^{x}$, for $e^{-x/2}$, for $e^{x}\sin x$ — is ultimately a consequence of this single fact combined with the chain rule.