To determine when the function \(f(x) = x^p \cos(1/x)\) is differentiable at \(x = 0\), let's examine the definition of differentiability. A function \(f(x)\) is differentiable at \(x = 0\) if the limit:
\(\lim_{{h \to 0}} \frac{f(h) - f(0)}{h}\)
exists. Given that \(f(0) = 0\), this simplifies to:
\(\lim_{{h \to 0}} \frac{h^p \cos(1/h)}{h}\)
or equivalently,
\(\lim_{{h \to 0}} h^{p-1} \cos(1/h)\)
The cosine function varies between -1 and 1, so:
\(-h^{p-1} \leq h^{p-1} \cos(1/h) \leq h^{p-1}\)
Both \(-h^{p-1}\) and \(h^{p-1}\) have limits of 0 as \(h \to 0\) if \(p-1 > 0\) or \(p > 1\). In this case, by Squeeze Theorem, the limit:
\(\lim_{{h \to 0}} h^{p-1} \cos(1/h) = 0\)
exists and equals 0. Thus, the function \(f(x)\) is differentiable at \(x=0\) when \(p>1\).
For other values of \(p\), specifically when \(p \leq 1\), \(h^{p-1}\) does not tend to zero, and thus, the limit does not exist or is not zero. Therefore, the function is not differentiable at \(x = 0\) in those cases.
Hence, the function \(f(x)\) is differentiable at \(x=0\) when \(p>1\).
Therefore, the correct answer is: \(p > 1\).