Question

Find the derivative: \[\cot x\].

Answer 1

Hint:Here, we will convert the given trigonometric function in terms of the other trigonometric functions. Then, we will use the quotient rule of derivatives to further solve the differentiation and find the required derivative of \[\cot x\].

Formula Used:
We will use the following formulas:
According to the quotient rule,
\[\dfrac{{dy}}{{dx}} = \dfrac{{\left( {II \times \dfrac{d}{{dx}}I} \right) - \left( {I \times \dfrac{d}{{dx}}II} \right)}}{{I{I^2}}}\]
\[{\sin ^2}x + {\cos ^2}x = 1\]

Complete step-by-step answer:
In order to find the derivative of \[\cot x\], first of all, we know that \[\cot x = \dfrac{1}{{\tan x}}\]
Also, we know that \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
Hence, we will write \[\cot x\] as:
\[\cot x = \dfrac{1}{{\tan x}}\]
\[\cot x = \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}}}\]
This can be written as: \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
Now, since we are required to find the derivative.
Therefore, we will use the quotient rule of derivatives.
According to the quotient rule, if \[y = \dfrac{I}{{II}}\], then
\[\dfrac{{dy}}{{dx}} = \dfrac{{\left( {II \times \dfrac{d}{{dx}}I} \right) - \left( {I \times \dfrac{d}{{dx}}II} \right)}}{{I{I^2}}}\]
Hence, using quotient rule in \[\cot x = \dfrac{{\cos x}}{{\sin x}}\], and differentiating both sides with respect to \[x\], we get,
\[\dfrac{{dy}}{{dx}}\cot x = \dfrac{{\left( {\sin x \times \dfrac{d}{{dx}}\cos x} \right) - \left( {\cos x \times \dfrac{d}{{dx}}\sin x} \right)}}{{{{\sin }^2}x}}\]
Now using the differentiation formula \[\dfrac{{dy}}{{dx}}\sin x = \cos x\] and \[\dfrac{{dy}}{{dx}}\cos x = - \sin x\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}\cot x = \dfrac{{\left( {\sin x \times \left( { - \sin x} \right)} \right) - \left( {\cos x \times \cos x} \right)}}{{{{\sin }^2}x}}\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}\cot x = \dfrac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}} = \dfrac{{ - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x}}\]
Using the trigonometric identity, \[{\sin ^2}x + {\cos ^2}x = 1\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}\cot x = \dfrac{{ - 1}}{{{{\sin }^2}x}}\]
Using the reciprocal trigonometric function \[\cos ecx = \dfrac{1}{{\sin x}}\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}\cot x = - \cos e{c^2}x\]

Therefore, we get that the derivative of \[\cot x\] is \[ - \cos e{c^2}x\]

Hence, this is the required answer.

Note: In mathematics, the rate of change of a function with respect to a variable is known as its derivative. Integration is the opposite of differentiation and hence it is called antiderivative. According to the quotient rule, it means that when two parts of a function are being divided, then, the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.