Let \(f(x) = \dfrac{e^{x} - e^{-x}}{2}\), \(x \in \mathbb{R}\). Let \(f^{(k)}(a)\) denote the \(k^{th}\) derivative of \(f\) evaluated at \(a\). What is the value of \(f^{(10)}(0)\)? (Note: ! denotes factorial)
Answer
(A)
Explanation
Differentiating \(f\) once gives \(f'(x) = \frac{e^{x} + e^{-x}}{2}\), and differentiating again brings back \(f''(x) = \frac{e^{x} - e^{-x}}{2} = f(x)\). So every even-order derivative equals \(f\) itself: \(f^{(10)}(x) = f(x)\), and \(f^{(10)}(0) = \frac{1 - 1}{2} = 0\).
Another view: \(f\) (the hyperbolic sine) is an odd function, so its Maclaurin series \(x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \cdots\) has only odd powers. The coefficient of \(x^{10}\), which equals \(\frac{f^{(10)}(0)}{10!}\), is therefore 0. Options C and D look like series coefficients rather than derivatives.
Answer: A