
How do you find the derivative of \[\cot x\]?
Answer
548.7k+ 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.
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