Question

Use the quotient rule and the derivatives of the sine and cosine functions to prove that$\frac{d}{dx}\left(cotx\right)=-cs{c}^{2}x$.

Short answer

We proved$\frac{d}{dx}\left(cotx\right)=-cs{c}^{2}x$.

Step-by-step solution

Step 1. Given Information

We are given $cotx$and using the quotient rule and the derivatives of the sine and cosine functions we need to prove that$\frac{d}{dx}\left(cotx\right)=-cs{c}^{2}x$.

Step 2. Proving the expression

$$\begin{gathered} \frac{d}{dx}\left(cotx\right) \\ =\frac{d}{dx}\left(\frac{\cos x}{\sin x}\right) \end{gathered}$$

Using the quotient rule we get,

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