Derivative of csc(x) - Proof and Explanation

Proof

We start by defining $\csc (x)$ as $\frac{1}{\sin (x)}$. 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 $v=\sin (x)$. The derivative of $u$ with respect to $x$ is $0$ since it’s a constant, and the derivative of $v=\sin (x)$ is $\cos (x)$.

Applying the quotient rule, we have:

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

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

Explanation

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

When finding the derivative of $\csc (x)$, 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 $v=\sin (x)$. The derivative of a constant $(1)$ is $0$, and the derivative of $\sin (x)$ is $\cos (x)$.

Applying these derivatives in the quotient rule, we find:

This simplifies to $-\frac{\cos (x)}{{\sin }^2(x)}$ since the term $0·\sin (x)$ is zero and $-1·\cos (x)$ is $-\cos (x)$.

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

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