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

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

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:

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

Factor.

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

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 $-1$.

References:

[1] https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-7-derivatives-of-sine-and-cosine/MIT18_01SCF10_Ses7b.pdf