Fix a metric $g$ on a smooth, closed manifold $\mathcal{M}$. Take a finite subcover of the manifold from its atlas. Is it true that any smooth partition of unity subordinate to this cover has uniformly bounded derivatives in $L^p (\mathcal{M})$ for each $p \geq 1$ ?
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A counterexample: Let $M$ be the unit circle. Let the two charts be the arcs $A = (0,2\pi)$ and $B = (\pi/2, 5\pi/2)$. For each $n$, consider the partition of unity subordinate to $\{A,B\}$ given by $$ \psi_{A,n} = \sin^2 ( (2n+1) \theta ) $$ and $$ \psi_{B,n} = \cos^2 ( (2n+1) \theta ) $$ Check that $\psi_{A,n} (0) = 0 = \psi_{B,n} (\pi/2)$, and clearly $\psi_{A,n} + \psi_{B,n} = 1$. By the scaling property it is easy to check that $$ \ \partial^k\psi_{A,n} \ = (2n+1)^k \ \partial^k\psi_{A,1} \ $$ And using that $$ \sin^2(\theta) = \frac12 (1\cos 2\theta) $$ you see immediately that your desired uniform bound is impossible. 

