Discussion
If operator \(\hat{D}^{n}\) (n times differentiate with respect to x) operates on eigenfunction eax, the eigenvalue would be :
Answer: C
The eigenvalue is an, printed in the options as αn.
Each differentiation of eax brings down one factor of a:
\[ \begin{aligned} \hat{D}\,e^{ax} &= a\,e^{ax} \\ \hat{D}^{n}e^{ax} &= a^{n}e^{ax} \end{aligned} \]
So eax is an eigenfunction of the n-th derivative operator with eigenvalue an. The options write the constant as α. A product na would come from adding the factor n times, na swaps base and exponent, and n! would arise only for differentiating a power xn, not an exponential.
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