Question

Let \( f(x) = \frac{e^x - e^{-x}}{2}, \, x \in \mathbb{R} \). Let \( f^{(k)}(a) \) denote the \( k^{\text{th}} \) derivative of \( f \) evaluated at \( a \). What is the value of \( f^{(10)}(0)? \) (Note: \( ! \) denotes factorial)

• A. 0
• B. 1
• C. \( \frac{1}{10!} \)
• D. \( \frac{2}{10!} \)

Hint:

When differentiating hyperbolic functions, remember the periodicity: the derivatives of \( \sinh(x) \) and \( \cosh(x) \) alternate, with even derivatives corresponding to \( \sinh(x) \) and odd derivatives to \( \cosh(x) \).

Answer 1

The Correct Option is A

We are given the function \( f(x) = \frac{e^x - e^{-x}}{2} \), which is the definition of the hyperbolic sine function, \( \sinh(x) \), i.e., \[ f(x) = \sinh(x). \] The \( k^{\text{th}} \) derivative of \( f(x) = \sinh(x) \) is: \[ f^{(k)}(x) = \frac{d^k}{dx^k} \sinh(x). \] We know that the derivatives of \( \sinh(x) \) follow a periodic pattern:
\( f'(x) = \cosh(x) \)
\( f''(x) = \sinh(x) \)
\( f^{(3)}(x) = \cosh(x) \)
\( f^{(4)}(x) = \sinh(x) \), and so on. This pattern alternates between \( \sinh(x) \) and \( \cosh(x) \) for successive derivatives. 
Specifically:
For even derivatives, \( f^{(2n)}(x) = \sinh(x) \) 
For odd derivatives, \( f^{(2n+1)}(x) = \cosh(x) \) 
Since \( f^{(10)}(x) \) is an even derivative, it will be equal to \( \sinh(x) \). Evaluating this at \( x = 0 \): \[ f^{(10)}(0) = \sinh(0) = 0. \] Thus, the value of \( f^{(10)}(0) \) is 0.