In this post, the derivative of the cosine function is found. To do this, the steps in reference 1 are followed.

Start with a definition of a derivative, from this post:

$\frac{df(x)}{dx}\bigg|_{a^+} \equiv \lim_{\Delta x \rightarrow 0^+} \frac{ f(a + \Delta x) – f(a) }{\Delta x} $.

Since $f(x)$ and $\cos(x)$ are both functions of $x$, replace $f$ with $\cos$:

$\frac{d\cos(x)}{dx}\bigg|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{ \cos(a + \Delta x) – \cos(a) }{\Delta x} $.

Recall the angle addition identities. Specifically,

$ \cos(A+B) = \cos A \cos B – \sin A \sin B$.

Using this identity, the numerator of the previous limit can be rewritten:

$\frac{d\cos(x)}{dx}\bigg|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{ \cos a \cos \Delta x – \sin a \sin \Delta x – \cos(a) }{\Delta x} $.


$\frac{d\cos(x)}{dx}\bigg|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{ \cos a (\cos \Delta x – 1 )  – \sin a \sin \Delta x }{\Delta x} $.

From this limit property,

$\frac{d\cos(x)}{dx}\bigg|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{ \cos a (\cos \Delta x – 1 ) }{\Delta x} – \lim_{\Delta x \rightarrow 0^+}\frac{\sin a \sin \Delta x }{\Delta x} $.

Using this limit property,

$\frac{d\cos(x)}{dx}\bigg|_{a^+} = \cos a \lim_{\Delta x \rightarrow 0^+} \frac{  \cos \Delta x – 1 }{\Delta x} – \sin a \lim_{\Delta x \rightarrow 0^+}\frac{ \sin \Delta x }{\Delta x} $.

The limit with the cosine function is evaluated here, and the limit with the sine function is determined here. Therefore,

$\frac{d\cos(x)}{dx}\bigg|_{a^+} =  -\sin a $.

Replacing $a$ by $x$, the previous equation becomes

$ \boxed{ \frac{d\cos(x)}{dx}\bigg|_{x^+} =  -\sin x }$.

This derivative of cosine is sine multiplied by $-1$.



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