Question

Use the quotient rule and the derivative of the sine function to prove that $\frac{d}{dx}\left(csx\right)=-cscx.cotx$

Short answer

The function $\frac{d}{dx}\left(csx\right)=-cscx.cotx$has been proved.

Step-by-step solution

Step 1. Given Information.

The function:

$\frac{d}{dx}\left(csx\right)=-cscx.cotx$

Step 2. Quotient rule.

The quotient rule is:

$\left(\frac{f}{g}\right)\text{'}\left(x\right)=\frac{f\text{'}\left(x\right)g\left(x\right)-g\text{'}\left(x\right)f\left(x\right)}{(g{\left(x\right))}^2}$

Step 3. Obtain the derivative for the given function.

$$\begin{gathered} f\text{'}\left(cscx\right)=f\text{'}\left(\frac{1}{\sin x}\right) \\ =\frac{1\text{'}\left(\sin x\right)-1\left(\sin x\right)\text{'}}{{\left(\sin x\right)}^2} \\ =\frac{-\cos x}{{\sin }^{2}x} \\ =-\frac{1}{\sin x}.\frac{\cos x}{\sin x} \\ =-csc.cotx \end{gathered}$$