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

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

Explanation

Solution

Hint : In order to determine the show the above given result , rewrite the cot⁡x\cot x as cos⁡xsin⁡x\dfrac{{\cos x}}{{\sin x}} and apply the quotient rule by considering f(x)=cos⁡xf\left( x \right) = \cos x and g(x)=sin⁡xg\left( x \right) = \sin x . Later use the identity of trigonometry sin⁡2x+cos⁡2x=1{\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 ddx(cot⁡x)=−(csc⁡x)2\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
=ddx(cot⁡x)= \dfrac{d}{{dx}}\left( {\cot x} \right)
Rewriting the above using the property of trigonometry cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}}
=ddx(cos⁡xsin⁡x)= \dfrac{d}{{dx}}\left( {\dfrac{{\cos x}}{{\sin x}}} \right)
Now applying the quotient rule ddx(f(x)g(x))=ddx(f(x))×g(x)−f(x)×ddx(g(x))(g(x))2\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(x)=cos⁡xf\left( x \right) = \cos x and g(x)=sin⁡xg\left( x \right) = \sin x
=ddx(cos⁡xsin⁡x)=ddx(cos⁡x)×sin⁡x−cos⁡x×ddx(sin⁡x)(sin⁡x)2= \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 ddx(sin⁡x)=cos⁡x\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x and ddx(cos⁡x)=−sin⁡x\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x . Rewriting the above we get
=(−sin⁡x)×sin⁡x−cos⁡x×(cos⁡x)sin⁡2x =−sin⁡2x−cos⁡2xsin⁡2x =−(sin⁡2x+cos⁡2x)sin⁡2x   = \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⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 to rewrite the expression above , we get
=−1sin⁡2x= \dfrac{{ - 1}}{{{{\sin }^2}x}}
=−csc⁡2x= - {\csc ^2}x
or
∴ddx(cot⁡x)=−(csc⁡x)2\therefore \dfrac{d}{{dx}}\left( {\cot x} \right) = - {\left( {\csc x} \right)^2}
Therefore, we have successfully verified the result ddx(cot⁡x)=−(csc⁡x)2\dfrac{d}{{dx}}\left( {\cot x} \right) = - {\left( {\csc x} \right)^2} .

Note : 1. Carefully identify the functions f(x),g(x)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 xx .