Question

23: Show that \(\frac{d}{{dx}}\left( {\csc x} \right) = - \csc x\cot x\).

Short answer

It is proved that \(\frac{d}{{dx}}\left( {\csc x} \right) = - \csc x\cot x\).

Step-by-step solution

 Derivative of Trigonometric Functions

The derivative of the trigonometric functions are shown below:

\(\begin{aligned}\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\csc x} \right) & = - \csc x\cot x\\\frac{d}{{dx}}\left( {\cos x} \right) & = - \sin x\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\sec x} \right) & = \sec x\tan x\\\frac{d}{{dx}}\left( {\tan x} \right) & = {\sec ^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\cot x} \right) & = - {\csc ^2}x\end{aligned}\)

Show that \(\frac{d}{{dx}}\left( {\csc x} \right) =  - \csc x\cot x\)

Differentiate the given function with respect to \(x\) as shown below:

\(\begin{aligned}\frac{d}{{dx}}\left( {\csc x} \right) &= \frac{d}{{dx}}\left( {\frac{1}{{\sin x}}} \right)\\ & = \frac{{\left( {\sin x} \right)\left( 0 \right) - 1\left( {\cos x} \right)}}{{{{\sin }^2}x}}\\ & = \frac{{ - \cos x}}{{{{\sin }^2}x}}\\ & = - \frac{1}{{\sin x}} \cdot \frac{{\cos x}}{{\sin x}}\\ & = - \csc x\cot x\end{aligned}\)

Thus, it is proved that \(\frac{d}{{dx}}\left( {\csc x} \right) = - \csc x\cot x\).