Question

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

Answer 1

Find the derivative of $cot\left(x\right)$:

First, use quotient identity to write the expression for $cot\left(x\right)$:

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

Write Quotient Rule of Derivatives:

$\frac{d}{dx}\left(\frac{u\left(x\right)}{v\left(x\right)}\right)=\frac{v\left(x\right)\frac{d}{dx}\left[u\left(x\right)\right]-u\left(x\right)\frac{d}{dx}\left[v\left(x\right)\right]}{{\left[v\left(x\right)\right]}^2}$

Consider $u\left(x\right)=\cos \left(x\right),v\left(x\right)=\sin \left(x\right)$:

Now, use the Quotient Rule of Derivatives:

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

Hence, the derivative of $cot\left(x\right)$ is $\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.