Like it says on the tin. I thought that the rule for deriving $sin{x}$ and $\cos x$ was simply the chain rule; $\displaystyle\frac d{dx}f(g(x))=f'(g(x))g'(x)$; applying to $\cos(x)$, this appears to becomes $-\sin{(-x)}(-1)=\sin(x)$.
However, I know that $\cos(-x)=\cos(x)$, and that $\displaystyle\frac d{dx}\cos(-x)=-\sin{(-x)}$.
What am I missing? Thank you!
On
$$\frac{d \{\cos(-x)\}}{dx}=\frac{d \{\cos(-x)\}}{d(-x)}\frac{d(-x)}{dx}$$
$$=-\sin(-x)\cdot(-1)=\sin(-x)=-\sin x$$
On
I don't understand how you apply the Chain Rule for the computation of $\frac{d}{dx}\cos(x)$. I you want to use the Chain Rule for $\cos(x)$, you can write
$\frac{d}{dx}(\cos(x))=\frac{d}{dx}(\sin(\frac{\pi}{2}-x))=\cos(\frac{\pi}{2}-x)\frac{d}{dx}(\frac{\pi}{2}-x)=-\cos(\frac{\pi}{2}-x)=-\sin(x)$.
Moreover,
$\frac{d}{dx}(\cos(-x))=-\sin(-x)\frac{d}{dx}(-x)=\sin(-x)=-\sin(x)$ since $\sin$ is an odd function. Of course, we obtain the same result that $\frac{d}{dx}(\cos(x))$ since $\cos$ is an even function.
Your computation with the chain rule is correct, but the simplification is not. Since $\sin(-x) = -\sin x$, you have
$$- \sin(-x)(-1) = (-1)^3 \sin x = - \sin x$$
agreeing with the other calculation.