Derivative of Cosine Function/Proof 1

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Theorem

$D_x \left({\cos x}\right) = -\sin x$


Proof

From the definition of the cosine function, we have:

$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!}$

Then:

\(\displaystyle D_x \left({\cos x}\right)\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^n 2n \frac {x^{2n - 1} }{\left({2n}\right)!}\) Power Series is Differentiable on Interval of Convergence
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \frac {x^{2n - 1} }{\left({2n - 1}\right)!}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^{n+1} \frac {x^{2n + 1} }{\left({2n + 1}\right)!}\) changing summation index
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n + 1} }{\left({2n + 1}\right)!}\)


The result follows from the definition of the sine function.

$\blacksquare$


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