2. If \( f(x)=\left\{\begin{array}{ll}x^{3} \cos \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right. \), then (a) \( f^{\prime}(0)=-1 \) (b) \( f(x) \) is not differentiable at \( x=0 \) (c) \( f^{\prime}(0)=\frac{1}{2} x \) (d) \( f^{\prime}(0)=0 \) (e) \( f^{\prime}(0)=1 \)

To determine the derivative of the function \( f(x) \) at \( x = 0 \), we need to compute the limit that defines the derivative: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] Given that \( f(0) = 0 \), we can simplify this to: \[ f'(0) = \lim_{h \to 0} \frac{f(h)}{h} \] For \( h \neq 0 \), we have: \[ f(h) = h^3 \cos \frac{1}{h} \] Thus, we can write: \[ f'(0) = \lim_{h \to 0} \frac{h^3 \cos \frac{1}{h}}{h} = \lim_{h \to 0} h^2 \cos \frac{1}{h} \] Now, we analyze the limit: 1. The term \( h^2 \) approaches \( 0 \) as \( h \to 0 \). 2. The term \( \cos \frac{1}{h} \) oscillates between \(-1\) and \(1\) as \( h \) approaches \( 0\). Since \( h^2 \) approaches \( 0 \) and \( \cos \frac{1}{h} \) is bounded, we can conclude that: \[ \lim_{h \to 0} h^2 \cos \frac{1}{h} = 0 \] Therefore, we find that: \[ f'(0) = 0 \] Now, we can evaluate the options given: (a) \( f^{\prime}(0)=-1 \) - Incorrect (b) \( f(x) \) is not differentiable at \( x=0 \) - Incorrect (c) \( f^{\prime}(0)=\frac{1}{2} x \) - Incorrect (d) \( f^{\prime}(0)=0 \) - Correct (e) \( f^{\prime}(0)=1 \) - Incorrect Thus, the correct answer is: \[ \boxed{d} \]