Derivative of cot(x) - Proof and Explanation

Proof

We start by defining $\cot (x)$ as $\frac{\cos (x)}{\sin (x)}$. 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 $u=\cos (x)$ and $v=\sin (x)$. The derivative of $\cos (x)$ is $-\sin (x)$, and the derivative of $\sin (x)$ is $\cos (x)$.

Applying the quotient rule:

Using the Pythagorean identity ${\sin }^2(x)+{\cos }^2(x)=1$:

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

Explanation

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

When finding the derivative of $\cot (x)$, 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 $u=\cos (x)$ and $v=\sin (x)$. The derivative of $\cos (x)$ with respect to $x$ is $-\sin (x)$, and the derivative of $\sin (x)$ is $\cos (x)$.

Applying the quotient rule:

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

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

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

Q.E.D.