Question

How do you find the derivative of $\cot \left(x\right)$ ?

Answer 1

Step 1: Apply quotient rule of differentiation

We know that $\cot \left(x\right)=\frac{\cos \left(x\right)}{\sin \left(x\right)}$

applying the quotient rule of differentiation

$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}$

putting $u=\cos \left(x\right)$ and $v=\sin \left(x\right)$

$\frac{d}{dx}\left(\frac{\cos \left(x\right)}{\sin \left(x\right)}\right)=\frac{\sin \left(x\right){\displaystyle \frac{d\cos \left(x\right)}{dx}}-\cos \left(x\right){\displaystyle \frac{d\sin \left(x\right)}{dx}}}{{\sin }^2\left(x\right)}$

Step 2: (Solve for differentiation)

$$\begin{gathered} \frac{d}{dx}\left(\frac{\cos \left(x\right)}{\sin \left(x\right)}\right)=\frac{\sin \left(x\right)\times -{\displaystyle \sin \left(x\right)}-\cos \left(x\right){\displaystyle \times \cos \left(x\right)}}{{\sin }^2\left(x\right)}\left[\therefore \frac{d}{dx}\cos \left(x\right)=-\sin xand\frac{d}{dx}\sin \left(x\right)=\cos \left(x\right)\right] \\ \frac{d}{dx}\left(\frac{\cos \left(x\right)}{\sin \left(x\right)}\right)=\frac{-{\sin }^2\left(x\right)-{\cos }^2\left(x\right)}{{\sin }^2\left(x\right)} \\ \frac{d}{dx}\left(\frac{\cos \left(x\right)}{\sin \left(x\right)}\right)=\frac{-\left[{\sin }^2\left(x\right)+{\cos }^2\left(x\right)\right]}{{\sin }^2\left(x\right)} \\ \frac{d}{dx}\left(cot\left(x\right)\right)=\frac{-1}{{\sin }^2\left(x\right)}\left[\therefore {\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1\right] \\ \frac{d}{dx}\left(cot\left(x\right)\right)=-cose{c}^2\left(x\right)\left[\therefore \frac{1}{\sin \left(x\right)}=\csc \left(x\right)\right] \end{gathered}$$

Hence, the derivative of $\cot \left(x\right)$ is $-cose{c}^2\left(x\right)$.