## Proof

$$\frac{d}{dx}[\phantom{\rule{0.167em}{0ex}}\mathrm{sin}(x)\phantom{\rule{0.167em}{0ex}}]\&={lim}_{h\to 0}\frac{\mathrm{sin}(x+h)-\mathrm{sin}(x)}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&={lim}_{h\to 0}\frac{\mathrm{sin}(x)\mathrm{cos}(h)+\mathrm{sin}(h)\mathrm{cos}(x)-\mathrm{sin}(x)}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&={lim}_{h\to 0}\frac{\mathrm{sin}(x)(\mathrm{cos}(h)-1)+\mathrm{sin}(h)\mathrm{cos}(x)}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&={lim}_{h\to 0}\frac{\mathrm{sin}(x)(\mathrm{cos}(h)-1)}{h}+{lim}_{h\to 0}\frac{\mathrm{sin}(h)\mathrm{cos}(x)}{h}\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&=\mathrm{sin}(x)\xb70+1\xb7\mathrm{cos}(x)\phantom{\rule{0ex}{0ex}}[\mathrm{2ex}]\&=\overline{)\mathrm{cos}(x)}$$## 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 sine with respect to $x$, which is the limit as $h$ approaches $0$ of $\frac{\mathrm{sin}(x+h)-\mathrm{sin}(x)}{h}$.

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

The numerator is then rearranged by collecting the terms containing $\mathrm{sin}(x)$. Specifically, $\mathrm{sin}(x)$ is factored out from the terms involving it and $\mathrm{sin}(x)\mathrm{cos}(h)-\mathrm{sin}(x)$ is factored to $\mathrm{sin}(x)(\mathrm{cos}(h)-1)$, and $\mathrm{sin}(h)\mathrm{cos}(x)$ is left as is. 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 $\frac{\mathrm{sin}(x)(\mathrm{cos}(h)-1)}{h}$ and another for $\frac{\mathrm{sin}(h)\mathrm{cos}(x)}{h}$.

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 $\mathrm{sin}(x)$ and $\mathrm{cos}(x)$ respectively, we get $\mathrm{sin}(x)\xb70$ and $1\xb7\mathrm{cos}(x)$.

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

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