t fn(x) = sin^n(x) for all x [0,pi]. Prove that fn' is not uniformly convergent on [0,pi].    Hint: Suppose false and apply a contradiction using the theorem:    Suppose fn is defined on a finite interval I and fn is continuous on (I). Suppose fn' converges uniformly on I. Suppose moreover that there exists at least one point a (an element) of I such that fn(a) is a convergent sequence of real numbers. Then there exists a differentiable function f such that fn->f uniformly on I and    f'(x)= lim(n>infinity) fn'(x)

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