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Given that we know that $A(t) = \frac{1}{2}$ $\int_{t_0}^t r^2 \frac{d\theta}{dt}$.

To derive $\frac{dA}{dt}$, we should use the Fundamental Theorem of Calculus, taking the derivative of both sides. I'm confused on how the right side of the equation works out, since there's two derivatives. Any help would be great, I'm sure I'm just overthinking it.

The final answer should be: $\frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt}$.

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Note that $A(t)$ is given by

$$A(t)=\frac12\int_{t_0}^{t}r^2(s)\frac{d\theta(s)}{ds}\,ds$$

If the integrand $r^2(s)\frac{d\theta(s)}{ds}$ is continuous on $[t_0,t_1]$, then the fundamental theorem of calculus guarantees that $A(t)$ is differentiable on $(t_0,t_1)$ and $\frac{dA(t)}{dt}$ is given by

$$\frac{dA(t)}{dt}=\frac12 r^2(t)\frac{d\theta(t)}{dt}$$

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