Understanding a Cosine Derivative Let $c$ be a constant. Why is it that

$$
D_x \left(- \frac{\cos(cx)}{c} \right) = \sin(cx)?
$$

I understand that $D_x \cos(x) = - \sin(x)$.  So what trigonometric identity is allowing us to infer the above?
 A: By De Moivre,
$$\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^n.$$
Then by the derivative of a power,
$$(\cos(nx)+i\sin(nx))'\\
=\left((\cos(x)+i\sin(x))^n\right)'\\
=n(\cos(x)+i\sin(x))^{n-1}\left(\cos(x)+i\sin(x)\right)'\\
=n\left((\cos((n-1)x)+i\sin((n-1)x))^{n-1}\right)\left(-\sin(x)+i\cos(x)\right)\\
=n\left((-\cos((n-1)x)\sin(x)-\sin((n-1)x)\cos(x)\\
-i\sin((n-1)x)\sin(x)+)^{n-1}+i\cos((n-1)x)\cos(x)+)^{n-1}\right)\\
=n(-\sin(nx)+i\cos(nx)),$$
which establishes the claim for all integer $n$.
But all of this is a stupid sledgehammer compared to the simple use of the chain rule.
A: Let,$f\left( x \right) =\cos { x } ,g\left( x \right) =cx\quad $ then composition of $f$ and $g$ will be $f\circ g=f\left( g\left( x \right)  \right) \quad =\cos { \left( cx \right)  } $
we know from the chain rule that $$\\ f^{ \prime  }\left( g\left( x \right)  \right) =f^{ \prime  }\left( g\left( x \right)  \right) \quad g^{ \prime  }\left( x \right) $$ and consider the $D_{ x }\left( f\left( x \right)  \right) =-\sin { \left( cx \right)  } \\ { D }_{ x }\left( g\left( x \right)  \right) =c$
we get $$D_{ x }\left( -\frac { \cos  (cx) }{ c }  \right) =-\frac { 1 }{ c } { D }_{ x }\left( \cos { \left( cx \right)  }  \right) =-\frac { 1 }{ c } \cdot \left( -\sin { \left( cx \right)  }  \right) \cdot c=\sin { \left( cx \right)  } \\ $$
A: \begin{align}
y & = \cos(cx) \\[10pt]
y & = \cos u & u & = cx \\[10pt]
\frac{dy}{du} & = -\sin u & \frac{du}{dx} & = c
\end{align}
Now apply the chain rule:
$$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \Big(-\sin u\Big) \cdot c = \Big(-\sin(cx)\Big) \cdot c.
$$
A: You are just using the chain rule. Hence you get by using $D_x\cos(x) = -\sin(x)$ and $D_x(cx) = c$ that
$$
D_x \left(- \frac{\cos(cx)}{c} \right) = - c \frac{-\sin(cx)}{c} = \sin(cx)
$$
I hope that helps you :)
A: $$f(x) = {-\cos (cx) \over c}$$ 
$$\begin{align}  f^\prime(x) & = \lim_{h \to 0} {{{-\cos (c(x + h)) \over c} - {-\cos (cx) \over c}}\over h}\\
& = \lim_{h \to 0} {\cos (cx)-\cos (c(x + h)) \over hc}\\
& = \lim_{h \to 0} {-2 \sin ({cx + cx + ch \over 2})\times\sin ({cx - cx - ch \over 2}) \over hc}\\
& = \lim_{h \to 0} {2 \sin (cx + {ch \over 2})\times \sin ({ch \over 2}) \over hc}\\
& = \lim_{h \to 0} {\sin (cx + {ch \over 2})\times\sin ({ch \over 2}) \over {hc \over 2}}\\
& = \sin (cx)\times\lim_{h \to 0} {\sin ({ch \over 2}) \over {hc \over 2}}\\
& = \sin (cx)
 \end{align}$$ 
If you don't like chain rule :)
A: The easiest and most elegant way is to apply the chain rule literally:
$\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

$\lfrac{d(\cos(cx))}{dx} \overset{\text{chain rule}}= \lfrac{d(\cos(cx))}{d(cx)} \times \lfrac{d(cx)}{dx} = -\sin(cx) \times c$ for any variable $x$ and constant $c$.

Of course this then gives $\lfrac{d(-\frac1c\cos(cx))}{dx} = -\lfrac1c ( -\sin(cx) \times c ) = \sin(cx)$.
