Question

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

Answer 1

Hint: In order to determine the show the above given result , rewrite the $ \cot x $ as $ \dfrac{{\cos x}}{{\sin x}} $ and apply the quotient rule by considering $ f\left( x \right) = \cos x $ and $ g\left( x \right) = \sin x $ . Later use the identity of trigonometry $ {\sin ^2}x + {\cos ^2}x = 1 $ to simplify the result and obtain the required answer.

Complete step-by-step answer:
We are given a result as $ \dfrac{d}{{dx}}\left( {\cot x} \right) = - {\left( {\csc x} \right)^2} $ and we have to show that this result is true.
To Show the result is true we will first take the LHS part of the equation and solve it to make it equal to the RHS par of the equation.
Taking Left-Hand Side of the equation
 $ = \dfrac{d}{{dx}}\left( {\cot x} \right) $
Rewriting the above using the property of trigonometry $ \cot x = \dfrac{{\cos x}}{{\sin x}} $
 $ = \dfrac{d}{{dx}}\left( {\dfrac{{\cos x}}{{\sin x}}} \right) $
Now applying the quotient rule $ \dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) \times g\left( x \right) - f\left( x \right) \times \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} $ by considering $ f\left( x \right) = \cos x $ and $ g\left( x \right) = \sin x $
 $ = \dfrac{d}{{dx}}\left( {\dfrac{{\cos x}}{{\sin x}}} \right) = \dfrac{{\dfrac{d}{{dx}}\left( {\cos x} \right) \times \sin x - \cos x \times \dfrac{d}{{dx}}\left( {\sin x} \right)}}{{{{\left( {\sin x} \right)}^2}}} $
Using the rule of derivative $ \dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x $ and $ \dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x $ . Rewriting the above we get
 $
   = \dfrac{{\left( { - \sin x} \right) \times \sin x - \cos x \times \left( {\cos x} \right)}}{{{{\sin }^2}x}} \\
   = \dfrac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}} \\
   = \dfrac{{ - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x}} \;
  $
Now using the identity of trigonometry $ {\sin ^2}x + {\cos ^2}x = 1 $ to rewrite the expression above , we get
 $ = \dfrac{{ - 1}}{{{{\sin }^2}x}} $
 $ = - {\csc ^2}x $
or
 $ \therefore \dfrac{d}{{dx}}\left( {\cot x} \right) = - {\left( {\csc x} \right)^2} $
Therefore, we have successfully verified the result $ \dfrac{d}{{dx}}\left( {\cot x} \right) = - {\left( {\csc x} \right)^2} $ .

Note: 1. Carefully identify the functions $ f\left( x \right),g\left( x \right) $ with proper sign .
2. Avoid any jump of step in solving such types of questions as this will increase the chance of error in the solution.
3. Quotient rule is only applicable when both numerator and denominator both are some functions in variable $ x $ .