Chapter 8.8 Question 19.
Prove
\[ \frac{d}{dx}\sin(x)=\cos(x) \]
The Tailor Series for \(\sin(x)\) is
\[ \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!} \]
The derivative of sums is the sum of derivatives, so
\[ \frac{d}{dx}\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty\frac{d}{dx}(-1)^n\frac{x^{2n+1}}{(2n+1)!} \]
Looking inside the summation, we have
\[ \frac{d}{dx}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=\\ (-1)^n(2n+1)\frac{x^{2n}}{(2n+1)!}+0\cdot\frac{x^{2n+1}}{(2n+1)!}=\\ (-1)^n\frac{x^{2n}}{(2n)!} \]
This now gives us
\[ \frac{d}{dx}\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} \]
This is the Tailor Series for \(\cos(x)\), so
\[ \frac{d}{dx}\sin(x)=\cos(x) \]