Question

25: Show that \(\frac{d}{{dx}}\left( {\cot x} \right) = - {\csc ^2}x\).

Short answer

It is proved that \(\frac{d}{{dx}}\left( {\cot x} \right) = - {\csc ^2}x\).

Step-by-step solution

 Derivative of Trigonometric Functions

The derivative of trigonometric functions is shown below:

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

Show that \(\frac{d}{{dx}}\left( {\cot x} \right) =  - {\csc ^2}x\)

Differentiate the function as shown below:

\(\begin{array}{c}\frac{d}{{dx}}\left( {\cot x} \right) = \frac{d}{{dx}}\left( {\frac{{\cos x}}{{\sin x}}} \right)\\ = \frac{{\left( {\sin x} \right)\left( { - \sin x} \right) - \left( {\cos x} \right)\left( {\cos x} \right)}}{{{{\sin }^2}x}}\\ = - \frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\cos }^2}x}}\\ = - \frac{1}{{{{\sin }^2}x}}\\ = - {\csc ^2}x\end{array}\)

Thus, it is proved that \(\frac{d}{{dx}}\left( {\cot x} \right) = - {\csc ^2}x\).