# 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?

• The chain rule. $\qquad$ – Michael Hardy Jul 27 '16 at 14:57
• Often I post something just because I think a question can be answered more simply than it is in the other answers. See my answer below. $\qquad$ – Michael Hardy Jul 27 '16 at 19:33

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 :)

$$f(x) = {-\cos (cx) \over c}$$ \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 :)

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.

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 $$D_{ x }\left( -\frac { \cos (cx) }{ c } \right) =-\frac { 1 }{ c } { D }_{ x }\left( \cos { \left( cx \right) } \right) =-\frac { 1 }{ c } \cdot \left( -\sin { \left( cx \right) } \right) \cdot c=\sin { \left( cx \right) } \\$$

\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: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \Big(-\sin u\Big) \cdot c = \Big(-\sin(cx)\Big) \cdot c.$$

• Possibly the lack of words? – Daniel R. Collins Jul 27 '16 at 16:04
• @Daniel, Sometimes "words get in the way", as Miss Estefan has once said. ;) – J. M. is a poor mathematician Jul 27 '16 at 17:15
• And yet here it is being downvoted. – Daniel R. Collins Jul 27 '16 at 18:42
• @DanielR.Collins : In some places in m.s.e. there is indeed insufficient use of words, and in some places there are far too many words, and often both problems occur in the same paragraph. However, I have now added some words. $\qquad$ – Michael Hardy Jul 27 '16 at 19:31
• Although i upvoted the answer, i don't get how this isn't a duplicate of the accepted answer. There are already 2 duplicates of the accepted answer. Anyway thanks for $\quad$. – A---B Jul 28 '16 at 16:49

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 $\lfrac{d(-\frac1c\cos(cx))}{dx} = -\lfrac1c ( -\sin(cx) \times c ) = \sin(cx)$.