Derivative of Sine Function

Theorem
If $$y = \sin(x)\,$$, then $$\frac{dy}{dx}=\cos(x)$$

Proof
From the definition of the sine function, we have $$\sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!}$$.

From Power Series Differentiable on Interval of Convergence, we have:

$$D_x \left({\sin x}\right) = \sum_{n=1}^\infty \left({-1}\right)^n \left({2n+1}\right)\frac {x^{2n}}{\left({2n+1}\right)!} = \sum_{n=1}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!}$$.

The result follows from the definition of the cosine function.

Alternative Proof
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