## Proof

We start by defining $\mathrm{cot}(x)$ as $\frac{\mathrm{cos}(x)}{\mathrm{sin}(x)}$. To find the derivative, we use the quotient rule, which states that the derivative of a quotient $\frac{u}{v}$ is $\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$.

Let $u=\mathrm{cos}(x)$ and $v=\mathrm{sin}(x)$. The derivative of $\mathrm{cos}(x)$ is $-\mathrm{sin}(x)$, and the derivative of $\mathrm{sin}(x)$ is $\mathrm{cos}(x)$.

Applying the quotient rule:

$$\frac{\text{d}}{\text{d}x}\mathrm{cot}(x)=\frac{(-\mathrm{sin}(x))\xb7\mathrm{sin}(x)-\mathrm{cos}(x)\xb7\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}$$ $$=\frac{-{\mathrm{sin}}^{2}(x)-{\mathrm{cos}}^{2}(x)}{{\mathrm{sin}}^{2}(x)}$$Using the Pythagorean identity ${\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x)=1$:

$$=\frac{-1}{{\mathrm{sin}}^{2}(x)}=-{\mathrm{csc}}^{2}(x)$$Thus, the derivative of $\mathrm{cot}(x)$ is:

$$\frac{\text{d}}{\text{d}x}\mathrm{cot}(x)=\overline{)-{\mathrm{csc}}^{2}(x)}$$## Explanation

To understand this derivative, let’s start by recognizing that $\mathrm{cot}(x)$ is defined as $\frac{\mathrm{cos}(x)}{\mathrm{sin}(x)}$, which is the ratio of the cosine function to the sine function. This means that for any angle $x$, $\mathrm{cot}(x)$ gives the ratio of the adjacent side to the opposite side in a right triangle.

When finding the derivative of $\mathrm{cot}(x)$, we use the quotient rule. This rule is used when differentiating a quotient of two functions. It states that if you have a function expressed as $\frac{u}{v}$, the derivative is $\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$, where $u$ and $v$ are functions of $x$.

Here, we choose $u=\mathrm{cos}(x)$ and $v=\mathrm{sin}(x)$. The derivative of $\mathrm{cos}(x)$ with respect to $x$ is $-\mathrm{sin}(x)$, and the derivative of $\mathrm{sin}(x)$ is $\mathrm{cos}(x)$.

Applying the quotient rule:

$$\frac{\text{d}}{\text{d}x}\mathrm{cot}(x)=\frac{(-\mathrm{sin}(x))\xb7\mathrm{sin}(x)-\mathrm{cos}(x)\xb7\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}$$This simplifies to $\frac{-{\mathrm{sin}}^{2}(x)-{\mathrm{cos}}^{2}(x)}{{\mathrm{sin}}^{2}(x)}$. Using the Pythagorean identity ${\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x)=1$, we simplify the numerator to $-1$, resulting in:

$$\frac{-1}{{\mathrm{sin}}^{2}(x)}$$This can be rewritten as $-{\mathrm{csc}}^{2}(x)$, since $\mathrm{csc}(x)=\frac{1}{\mathrm{sin}(x)}$.

Thus, the derivative of $\mathrm{cot}(x)$ with respect to $x$ is $\overline{)-{\mathrm{csc}}^{2}(x)}$.

*Q.E.D.*