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In a problem of a simple pendulum, we solve the differential equation $x'' + \sin x = 0$ with the initial condition $x(0) = A$ and $ x'(0) = 0$. The solution is $$dt = \frac{\pm dx}{\sqrt{2(\cos x-\cos A)}}, A \in (0, \pi) $$ Hence the period is $$T(A) = \int_{0}^{T(A)} dt = 4\int_{0}^{A} \frac{dx}{\sqrt{2(\cos x-\cos A)}} $$ Here my problem is: How to prove T(A) is continuously differentiable on $(0, \pi)$ and $T'(A) > 0$ for every $x \in (0, \pi)$.

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Your $x'$ seems to imply $$ x'' = \frac{d}{dx} \left( \sqrt{2(\cos x - \cos A)} \right) = \frac{\sin x}{\sqrt{2(\cos x - \cos A)}} $$ which is a different ODE.

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  • $\begingroup$ The solution is from my textbook, so I'm sure there's no problem. We are only asked to solve the "continuously differentiable" problem as homework. $\endgroup$ – Rubisco Lee Oct 19 '16 at 10:03

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