## Proof

$$\frac{d}{dx}\mathrm{cos}x\&={lim}_{h\to 0}\frac{\mathrm{cos}(x+h)-\mathrm{cos}x}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&={lim}_{h\to 0}\frac{\mathrm{cos}x\mathrm{cos}h-\mathrm{sin}x\mathrm{sin}h-\mathrm{cos}x}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&={lim}_{h\to 0}\frac{\mathrm{cos}x(\mathrm{cos}h-1)-\mathrm{sin}x\mathrm{sin}h}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&=\mathrm{cos}x{lim}_{h\to 0}\frac{\mathrm{cos}h-1}{h}-\mathrm{sin}x{lim}_{h\to 0}\frac{\mathrm{sin}h}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&=\mathrm{cos}x\xb70-\mathrm{sin}x\xb71\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&=-\mathrm{sin}x$$## Explanation

The proof begins by stating the definition of the derivative of a real function at a point. In this case, it’s the derivative of $\mathrm{cos}(x)$ with respect to $x$, which is the limit as $h$ approaches $0$ of $\frac{\mathrm{cos}(x+h)-\mathrm{cos}(x)}{h}$.

The next step uses the trigonometric identity for the cosine of a sum: $\mathrm{cos}(A+B)=\mathrm{cos}(A)\mathrm{cos}(B)-\mathrm{sin}(A)\mathrm{sin}(B)$. Here, $A$ is $x$ and $B$ is $h$. Applying this identity to $\mathrm{cos}(x+h)$, we get: $\mathrm{cos}(x)\mathrm{cos}(h)-\mathrm{sin}(x)\mathrm{sin}(h)$.

The numerator is then rearranged by separating the terms involving $\mathrm{cos}(x)$ and $\mathrm{sin}(x)$. Specifically, $\mathrm{cos}(x)$ is factored out from the terms involving it, and we write the expression as $\mathrm{cos}(x)(\mathrm{cos}(h)-1)-\mathrm{sin}(x)\mathrm{sin}(h)$. The denominator $h$ remains unchanged.

The limit is split into two parts using the sum rule for limits. This rule states that the limit of a sum is equal to the sum of the limits, provided both limits exist. So we now have two limits: one for $\frac{\mathrm{cos}(x)(\mathrm{cos}(h)-1)}{h}$ and another for $\frac{-\mathrm{sin}(x)\mathrm{sin}(h)}{h}$.

We can evaluate each of these limits separately. The limit of $\frac{\mathrm{sin}(h)}{h}$ as $h$ approaches $0$ is equal to $1$ (this is a standard limit). The limit of $\frac{\mathrm{cos}(h)-1}{h}$ as $h$ approaches $0$ is equal to $0$ (this is another standard limit). When we multiply these limits by $\mathrm{cos}(x)$ and $-\mathrm{sin}(x)$ respectively, we get $\mathrm{cos}(x)\xb70$ and $-\mathrm{sin}(x)\xb71$.

Adding these together as per the sum rule for limits, we get $0-\mathrm{sin}(x)$, which simplifies to $-\mathrm{sin}(x)$.

* QED:* Therefore, the derivative of $\mathrm{cos}(x)$ with respect to $x$ is $\overline{)-\mathrm{sin}(x)}$.