Derivative of Sine Function

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Theorem

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


Corollary

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


Proof 1

From the definition of the sine function, we have:

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

From Power Series over Factorial, this series converges for all $x$.


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

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


The result follows from the definition of the cosine function.

$\blacksquare$


Proof 2

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle D_x \left({\sin x}\right)\) \(=\) \(\displaystyle \) \(\displaystyle \lim_{h \to 0} \frac { \sin \left({x + h}\right) - \sin \left({x}\right) } h\) \(\displaystyle \) \(\displaystyle \)          Definition of derivative at a point          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \lim_{h \to 0} \frac { \sin \left({x}\right) \cos \left({h}\right) + \sin \left({h}\right) \cos \left({x}\right) - \sin \left({x}\right) } h\) \(\displaystyle \) \(\displaystyle \)          Sine of Sum          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \lim_{h \to 0} \frac { \sin \left({x}\right) \left({\cos \left({h}\right) - 1}\right) + \sin \left({h}\right) \cos \left({x}\right) } h\) \(\displaystyle \) \(\displaystyle \)          Collecting terms containing $\sin \left({x}\right)$ and factoring          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \lim_{h \to 0} \frac { \sin \left({x}\right) \left({\cos \left({h}\right) - 1}\right) } h + \lim_{h \to 0} \frac { \sin \left({h}\right) \cos \left({x}\right) } h\) \(\displaystyle \) \(\displaystyle \)          Sum Rule          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \sin \left({x}\right) \times 0 + 1 \times \cos \left({x}\right)\) \(\displaystyle \) \(\displaystyle \)          From Limit of Sine of X over X and Limit of (Cosine (X) - 1) over X          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \cos \left({x}\right)\) \(\displaystyle \) \(\displaystyle \)                    

$\blacksquare$


Proof 3

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle D_x \sin x\) \(=\) \(\displaystyle \) \(\displaystyle D_x \cos \left({\frac \pi 2 - x}\right)\) \(\displaystyle \) \(\displaystyle \)          Cosine of Complement equals Sine          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \sin \left({\frac \pi 2 - x}\right)\) \(\displaystyle \) \(\displaystyle \)          Derivative of Cosine Function and Chain Rule          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \cos x\) \(\displaystyle \) \(\displaystyle \)          Sine of Complement equals Cosine          

$\blacksquare$


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