## Proof

We start by defining $\mathrm{csc}(x)$ as $\frac{1}{\mathrm{sin}(x)}$. To find the derivative, we use the quotient rule, which states that the derivative of a quotient $\frac{u}{v}$ is $\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$.

Here, let $u=1$ and $v=\mathrm{sin}(x)$. The derivative of $u$ with respect to $x$ is $0$ since it’s a constant, and the derivative of $v=\mathrm{sin}(x)$ is $\mathrm{cos}(x)$.

Applying the quotient rule, we have:

$$\frac{\text{d}}{\text{d}x}\mathrm{csc}(x)=\frac{0\xb7\mathrm{sin}(x)-1\xb7\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}=-\frac{\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}$$Next, we simplify $-\frac{\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}$. This can be rewritten as $-\frac{1}{\mathrm{sin}(x)}\xb7\frac{\mathrm{cos}(x)}{\mathrm{sin}(x)}$, which simplifies to $-\mathrm{csc}(x)\mathrm{cot}(x)$.

Thus, the derivative of $\mathrm{csc}(x)$ is:

$$\frac{\text{d}}{\text{d}x}\mathrm{csc}(x)=\overline{)-\mathrm{csc}(x)\xb7\mathrm{cot}(x)}$$## Explanation

To understand this derivative, we first recognize that $\mathrm{csc}(x)$ is the reciprocal of the sine function, defined as $\mathrm{csc}(x)=\frac{1}{\mathrm{sin}(x)}$. This means that for any angle $x$, $\mathrm{csc}(x)$ represents the ratio of the hypotenuse to the opposite side in a right triangle.

When finding the derivative of $\mathrm{csc}(x)$, we use the quotient rule because it involves the division of two functions. According to the quotient rule, the derivative of a function expressed as $\frac{u}{v}$ is $\frac{{u}^{\prime}v-u{v}^{\prime}}{{v}^{2}}$, where $u$ and $v$ are functions of $x$.

In our case, we choose $u=1$ (a constant function) and $v=\mathrm{sin}(x)$. The derivative of a constant $(1)$ is $0$, and the derivative of $\mathrm{sin}(x)$ is $\mathrm{cos}(x)$.

Applying these derivatives in the quotient rule, we find:

$$\frac{\text{d}}{\text{d}x}\mathrm{csc}(x)=\frac{0\xb7\mathrm{sin}(x)-1\xb7\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}$$This simplifies to $-\frac{\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}$ since the term $0\xb7\mathrm{sin}(x)$ is zero and $-1\xb7\mathrm{cos}(x)$ is $-\mathrm{cos}(x)$.

Next, we simplify $-\frac{\mathrm{cos}(x)}{{\mathrm{sin}}^{2}(x)}$. It can be expressed as $-\frac{1}{\mathrm{sin}(x)}\xb7\frac{\mathrm{cos}(x)}{\mathrm{sin}(x)}$. Here, $\frac{1}{\mathrm{sin}(x)}$ is the definition of $\mathrm{csc}(x)$, and $\frac{\mathrm{cos}(x)}{\mathrm{sin}(x)}$ is $\mathrm{cot}(x)$. Therefore, the expression simplifies to $-\mathrm{csc}(x)\mathrm{cot}(x)$.

Therefore, the derivative of $\mathrm{csc}(x)$ with respect to $x$ is $\overline{)-\mathrm{csc}(x)\xb7\mathrm{cot}(x)}$.