How do you find the derivative of \[\cot x\]?
Answer
584.4k+ views
Hint:
The given equation is based on derivatives and trigonometry. For solving the problem, we will apply the identities of trigonometry and the identities of derivatives as well. So, by applying all the identities which are suitable for the given problem and simplifying the equation we can easily conclude our result and find the solution of the above problem.
Complete Step by step Solution:
The above problem can be solved by using trigonometric identities and derivative identities.
We are given \[\cot x\], where \[\cot x\] is the function of trigonometry. We have to find the derivative of \[\cot x\]. By using the identity of \[\cot x = \dfrac{{\cos x}}{{\sin x}}\], we can solve the given problem.
We can write \[\dfrac{d}{{dx}}\left( {\cot x} \right)\]…….(A)
where \[\dfrac{d}{{dx}}\] represents the derivative function.
Substituting the value of above identity in equation (A), it becomes:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\cos x}}{{\sin x}}} \right)\]
We have to apply the \[\dfrac{u}{v}\] rule in the above equation as identity of \[\dfrac{u}{v}\] rule is given below:
\[ \Rightarrow \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]
Replacing \[u\] by \[\cos x\] and \[v\] by \[\sin x\] in above identity it becomes:
\[ \Rightarrow \dfrac{{\sin x\dfrac{d}{{dx}}\cos x - \cos x\dfrac{d}{{dx}}\sin x}}{{{{\sin }^2}x}}\]
As the identity of \[\dfrac{d}{{dx}}\left( {\cos x} \right)\] is \[ - \sin x\] and the identity of \[\dfrac{d}{{dx}}\left( {\sin x} \right)\] is \[\cos x\], so, applying the identities in the above equation, it becomes:
\[ \Rightarrow \dfrac{{\sin x\left( { - \sin x} \right) - \cos x\left( {\cos x} \right)}}{{{{\sin }^2}x}}\]
Multiplying \[\sin x\left( { - \sin x} \right)\], it becomes \[ - {\sin ^2}x\] and \[\cos x\left( {\cos x} \right)\] becomes \[{\cos ^2}x\]. Putting the value above, we get:
\[ \Rightarrow \dfrac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}\]
Taking minus \[\left( - \right)\] sign common from numerator, it becomes:
\[ \Rightarrow \dfrac{{ - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x}}\]
By using the identity \[{\sin ^2}x + {\cos ^2}x = 1\] in above equation, it becomes:
\[ - \dfrac{1}{{{{\sin }^2}x}}\]
Again, the identity of \[\dfrac{1}{{{{\sin }^2}x}}\] is \[\cos e{c^2}x\]. So, substituting the value above:
\[ = - \cos e{c^2}x\]
We calculated the solution of \[\dfrac{d}{{dx}}\left( {\cot x} \right)\], which is \[ - \cos e{c^2}x\]
Note:
The above problem is based on derivatives and trigonometric functions. The applications of derivatives are the rate of change of quantity. Derivatives determine concavity, curve sketching and optimization. The Astronomers use trigonometry to calculate how far stars and planets are from earth. Trigonometry can be used to roof a house to make the roof inclined and height of the roof in building.
The given equation is based on derivatives and trigonometry. For solving the problem, we will apply the identities of trigonometry and the identities of derivatives as well. So, by applying all the identities which are suitable for the given problem and simplifying the equation we can easily conclude our result and find the solution of the above problem.
Complete Step by step Solution:
The above problem can be solved by using trigonometric identities and derivative identities.
We are given \[\cot x\], where \[\cot x\] is the function of trigonometry. We have to find the derivative of \[\cot x\]. By using the identity of \[\cot x = \dfrac{{\cos x}}{{\sin x}}\], we can solve the given problem.
We can write \[\dfrac{d}{{dx}}\left( {\cot x} \right)\]…….(A)
where \[\dfrac{d}{{dx}}\] represents the derivative function.
Substituting the value of above identity in equation (A), it becomes:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\cos x}}{{\sin x}}} \right)\]
We have to apply the \[\dfrac{u}{v}\] rule in the above equation as identity of \[\dfrac{u}{v}\] rule is given below:
\[ \Rightarrow \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]
Replacing \[u\] by \[\cos x\] and \[v\] by \[\sin x\] in above identity it becomes:
\[ \Rightarrow \dfrac{{\sin x\dfrac{d}{{dx}}\cos x - \cos x\dfrac{d}{{dx}}\sin x}}{{{{\sin }^2}x}}\]
As the identity of \[\dfrac{d}{{dx}}\left( {\cos x} \right)\] is \[ - \sin x\] and the identity of \[\dfrac{d}{{dx}}\left( {\sin x} \right)\] is \[\cos x\], so, applying the identities in the above equation, it becomes:
\[ \Rightarrow \dfrac{{\sin x\left( { - \sin x} \right) - \cos x\left( {\cos x} \right)}}{{{{\sin }^2}x}}\]
Multiplying \[\sin x\left( { - \sin x} \right)\], it becomes \[ - {\sin ^2}x\] and \[\cos x\left( {\cos x} \right)\] becomes \[{\cos ^2}x\]. Putting the value above, we get:
\[ \Rightarrow \dfrac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}\]
Taking minus \[\left( - \right)\] sign common from numerator, it becomes:
\[ \Rightarrow \dfrac{{ - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x}}\]
By using the identity \[{\sin ^2}x + {\cos ^2}x = 1\] in above equation, it becomes:
\[ - \dfrac{1}{{{{\sin }^2}x}}\]
Again, the identity of \[\dfrac{1}{{{{\sin }^2}x}}\] is \[\cos e{c^2}x\]. So, substituting the value above:
\[ = - \cos e{c^2}x\]
We calculated the solution of \[\dfrac{d}{{dx}}\left( {\cot x} \right)\], which is \[ - \cos e{c^2}x\]
Note:
The above problem is based on derivatives and trigonometric functions. The applications of derivatives are the rate of change of quantity. Derivatives determine concavity, curve sketching and optimization. The Astronomers use trigonometry to calculate how far stars and planets are from earth. Trigonometry can be used to roof a house to make the roof inclined and height of the roof in building.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

